SCHOLAR Study Guide CfE Advanced Higher Mathematics Course ... · TOPIC 1. PARTIAL FRACTIONS 3 1.1...

SCHOLAR Study Guide

CfE Advanced Higher MathematicsCourse materialsTopic 1: Partial fractions

Authored by:Fiona Withey (Stirling High School)

Karen Withey (Stirling High School)

Reviewed by:Margaret Ferguson

Previously authored by:Jane S Paterson

Dorothy A Watson

Heriot-Watt University

Edinburgh EH14 4AS, United Kingdom.

First published 2016 by Heriot-Watt University.

This edition published in 2017 by Heriot-Watt University SCHOLAR.

Members of the SCHOLAR Forum may reproduce this publication in whole or in part for educationalpurposes within their establishment providing that no profit accrues at any stage, Any other use of thematerials is governed by the general copyright statement that follows.

Heriot-Watt University accepts no responsibility or liability whatsoever with regard to the informationcontained in this study guide.

Distributed by the SCHOLAR Forum.

SCHOLAR Study Guide Course Materials: CfE Advanced Higher Mathematics

1. CfE Advanced Higher Mathematics Course Code: C747 77

AcknowledgementsThanks are due to the members of Heriot-Watt University's SCHOLAR team who planned and created thesematerials, and to the many colleagues who reviewed the content.

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The content of this Study Guide is aligned to the Scottish Qualifications Authority (SQA) curriculum.

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Topic 1

Partial fractions

Contents1.1 Looking back . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.1.1 Division by (x - a) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.1.2 Factor theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.1.3 Factorising polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.2 Introduction to partial fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.3 Linear factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.4 Repeated factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.5 Irreducible quadratic factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

1.6 Algebraic long division . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

1.7 Reduce improper rational functions by division . . . . . . . . . . . . . . . . . . . . . . . . . . 29

1.8 Learning points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

1.9 Extended information . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

1.10 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

2 TOPIC 1. PARTIAL FRACTIONS

Learning objective

By the end of this topic, you should be able to:

• use the method of partial fractions to express proper rational functions as a sum ofpartial fractions;

• use algebraic long division and factorise polynomials of up to degree 3.

TOPIC 1. PARTIAL FRACTIONS 3

1.1 Looking back

SummaryPolynomials

• A polynomial is an expression containing the sum or difference of algebraic terms with powersor the equivalent in factorised form.e.g. 2x3 − 4x2 − 3x + 10 and (x + 5)(x2 − 5x − 3)

• The degree of a polynomial is the value of the highest power.e.g. 2x3 − 4x + 10 has degree 3.

• Synthetic division is a method for factorising a polynomial.

◦ (x − a) is the divisor e.g. (3x3 + 5x + 4) ÷ (x − 1)

• If a polynomial divided by (x − a) has remainder 0 then (x − a) is a factor.

• If (x − a) is a factor then the remainder under division by (x − a) is 0.

• When trying a divisor of the form (x − a) it is usually a good idea to start with (x − 1).

◦ If that does not work then think of the possible factors of the constant on the end of yourpolynomial.

◦ Be systematic and don't rub out any attempts that do not work.

• Solving a polynomial is best done in factorised form and allows you to identify the roots (i.e.the places where the graph of the polynomial crosses the x-axis).

• To determine the equation of a polynomial from its graph:

◦ use the roots to determine the factors e.g. roots a, b, cgive factors (x − a)(x − b)(x − c);

◦ remember the polynomial may have a scalar ke.g. y = k(x − a)(x − b)(x − c);

◦ substitute the coordinates of the y-intercept to determine k;

1.1.1 Division by (x - a)

We can divide polynomials using the method of synthetic division.

Key point

The Remainder Theorem

If a polynomial f(x) is divided by (x− a) the remainder is f(a).

Example

Problem:

f(x) = 3x3 + 5x + 4Use synthetic division to find (3x3 + 5x + 4) ÷ (x − 1).

Solution:

(x − a) is the divisor and in this example a = 1. Using synthetic division we will be able tofind the quotient and the remainder.

Also, notice that we do not have a squared term so we must interpret this in a polynomial as0x2. The polynomial should be interpreted as 3x3 + 0x2 + 5x + 4.

So under division by (x − 1) the quotient is 3x2 + 3x + 8 and the remainder is 12.

We can express the function as f(x) = (x − 1)( 3x2 + 3x + 8 ) + 12.

Key point

When f(x) is divided by (x − a) we can say

f(x) = (x − a)Q(x) + R

where Q(x) is the quotient and R is the remainder.

Examples

1. Problem:

If f(x) = 2x4 + x2 − x + 1, divide f(x) by (x + 1).

Solution:

We must interpret the function as 2x4 + 0x3 + x2 − x + 1.

We can only divide by (x − a) so we interpret the divisor as (x − (−1)).

This gives,

So the quotient is 2x3 − 2x2 + 3x − 4 and the remainder is 5.

Hence f(x) = (x + 1)( 2x3 − 2x2 + 3x − 4) + 5.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2. Problem:

If f(x) = 2x3 + 5x2 − x − 1, divide f(x) by (2x − 1).

Solution:

We can only divide by (x − a) so we interpret the divisor as 2(x − 1/2).

This gives,

So the quotient is 2x2 + 6x + 2 and the remainder is 0

So f(x) = (x − 1/2)( 2x2 + 6x + 2) but we have to take the common factor of 2 out of thequotient and put it back into the divisor.(x − 1

)(2x2 + 6x + 2

(x − 1

(x2 + 3x + 1

(x − 1

)(x2 + 3x + 1

giving f(x) = (2x − 1)(x2 + 3x + 1).

Go onlineDivision by (x - a) practice

Q1: What is the remainder when (4x2 − 10x + 2) ÷ (x − 3)?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q2: What is the remainder when (5x3 − 7x2 + 14) ÷ (x − 3)?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q3: What is the remainder when (3x2 − 8x + 4) ÷ (x + 3)?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q4: Express (6x3 + 7x2 − 1) ÷ (3x − 1) in the form (3x− 1)Q(x) + R, where Q(x) isthe quotient and R is the remainder.

1.1.2 Factor theorem

The Factor Theorem states that if f(a) = 0 then (x − a) is a factor of f(x) and if (x − a) is a factorof f(x) then f(a) = 0.

Key point

If a polynomial divided by (x − a) has remainder 0 then (x − a) is a factor and if (x − a) isa factor then the remainder under division by (x − a) is equal to 0.

Example

Problem:

Is (x − 4) a factor of f(x) where f(x) = x3 + 7x2 − 26x − 72?

Solution:

Since the remainder is 0, (x − 4) is a factor of f(x) and f(x) = (x − 4)(x2 + 11x + 18).

Notice that we can factorise the quotient and f(x) = (x − 4)(x + 2)(x + 9) which givesthe function in its fully factorised form.

Go onlineFactor theorem practice

Q5: Is (x − 1) a factor of f(x) where f(x) = x4 + 2x3 − 7x2 − 8x + 12?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q6: Is (x − 2) a factor of f(x) where f(x) = x3 + 3x2 − 4x − 12?

1.1.3 Factorising polynomials

Key point

Synthetic division can be used to help fully factorise a polynomial.

Example

Problem:

Factorise fully x4 + 2x3 − 7x2 − 8x + 12.

Solution:

We are not given a divisor for this example so we have to use trial and error to find the firstfactor. The simplest divisor to try is (x − 1) giving,

Since the remainder is 0, (x − 1) is a factor ofx4 + 2x3 − 7x2 − 8x + 12 = (x − 1)(x3 + 3x2 − 4x − 12).Now we need to factorise the quotient x3 + 3x2 − 4x − 12. We could try (x − 1) again butlet's try (x − 2) giving,

Since the remainder is 0, (x − 2) is a factor ofx3 + 3x2 − 4x − 12 = (x − 2)(x2 + 5x − 6).

So it follows that x4 + 2x3 − 7x2 − 8x + 12 = (x − 1)(x − 2)( x2 + 5x − 6).

Now all we have to do is try to factorise x2 + 5x − 6. . . and so,x2 + 5x − 6 = (x + 2) (x + 3)

x4 + 2x3 − 7x2 − 8x + 12 = (x − 1) (x − 2) (x + 2) (x + 3)

= (x − 1) (x − 2) (x + 2) (x + 3)

Key point

When trying a divisor of the form (x − a) it is usually a good idea to start with (x − 1). If thatdoes not work then think of the possible factors of the constant on the end of your polynomial.

For the constant -12 the factors and divisors are:

1 and -12 (x − 1) and (x + 12)

2 and -6 (x − 2) and (x + 6)

3 and -4 (x − 3) and (x + 4)

4 and -3 (x − 4) and (x + 3)

6 and -2 (x − 6) and (x + 2)

12 and -1 (x − 12) and (x + 1)

Be systematic. Don't rub out any attempts that do not work just put a line through them. Thiswill allow you to see what you have tried.

Example

Problem:

Fully factorise x3 + 2x2 − 5x − 6.

Solution:

For the constant -6 the factors and divisors are:

1 and -6 (x − 1) and (x + 6)

2 and -3 (x − 2) and (x + 3)

3 and -2 (x − 3) and (x + 2)

6 and -1 (x − 6) and (x + 1)

Try (x − 1).

Evaluate f(1):

Since the remainder �= 0, (x − 1) is not a factor.

Try (x + 1).

Evaluate f(−1):

Since the remainder = 0, (x + 1) is a factor andx3 + 2x2 − 5x − 6 = (x + 1)

(x2 + x − 6

)= (x + 1) (x + 3) (x − 2)

Go onlineFactorising polynomials practice

Q7: Fully factorise x3 − 3x + 2.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q8: Fully factorise 2x3 − 3x2 − 11x + 6.

1.2 Introduction to partial fractions

There are some very complex looking algebraic equations. To try to differentiate or integrate themas they stand would be very difficult. In this section methods for splitting them into manageableterms are investigated.

The following definitions will help to make this section clearer.

Key point

If P (x) = anxn + an − 1x

n - 1 + an − 2xn - 2 + . . . + a2x

2 + a1x1 + a0

Key point continued

where a0, . . . , an ∈ R then P is a polynomial of degree n.

Example

x + 3 has degree 1;

x2 − 2x + 3 has degree 2;

4x3 + 2x2 − 5 has degree 3;

a constant such as 7 has degree 0

Key point

If P (x) and Q (x) are polynomials then P (x)Q(x) is called a rational function.

Key point

Let P (x) be a polynomial of degree n and Q (x) be a polynomial of degree m.

If n < m then P (x)Q(x) is a proper rational function.

For example,x2 + 2x − 1x3 − 3x + 4

Key point

Let P (x) be a polynomial of degree n and Q (x) be a polynomial of degree m.

If n ≥ m then P (x)Q(x) is an improper rational function.

For example,x3 + 2x − 1

x2 − 4or x2 + 2x − 1

x2 − 4

An improper rational function can always be expressed as a polynomial plus a proper rationalfunction (by using algebraic long division).

Note: The process of algebraic long division is covered in a later section in this topic.

We already know how to simplify algebraic fractions by finding a common denominator e.g.1

1 + x + 1x + 2 = (x + 2) + (1 + x)

(1 + x)(x + 2) = 2x + 3(1 + x)(x + 2)

The opposite process, for example, expressing 2x + 3(1 + x)(x + 2) as 1

1 + x + 1x + 2 , is called putting proper

rational functions into partial fractions.

Key point

The process of taking a proper rational function and splitting it into separate terms each witha factor of the original denominator as its denominator is called expressing the function aspartial fractions.

The way in which the rational function splits up depends on whether the denominator is a quadraticequation or a cubic equation.

It also depends on whether this denominator has linear, repeated linear or quadratic factors (with noreal roots).

The different ways in which a proper rational function with a denominator of degree at most threecan be split into partial fractions are now explained.

Examples

1. Linear factors

Regardless of whether the denominator is a quadratic or a cubic, if it can be factorised intodistinct factors that are linear (have a degree of 1) then its partial fractions take the followingform:

This has a denominator of a quadratic with two distinct factors.···

(x + a)(x + b) ≡ A(x + a) + B

(x + b)

This has a denominator of a cubic with three distinct factors.···

(x + a)(x + b)(x + c) ≡ A(x + a) + B

(x + b) + C(x + c)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2. Repeated linear factors

This has a denominator of a quadratic with two repeated factors.···

(x + a)2≡ A

(x + a) + B(x + a)2

This has a denominator of a cubic with three repeated factors.···

(x + a)3≡ A

(x + a) + B(x + a)2

+ C(x + a)3

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3. Irreducible quadratic factors

This has an irreducible quadratic denominator (i.e. the denominator has no real roots).···

x2 + bx + c ≡ Ax + Bx2 + bx + c

This has a cubic denominator. One factor is linear and the other is an irreducible quadratic.···

(x + a)(x2 + bx + c) ≡ Ax + a + Bx + C

x2 + bx + c

1.3 Linear factors

Before we begin the process of expressing a function as a sum of partial fractions we must ensurethat we are working with a proper rational function, where the degree of the numerator is lessthan the degree of the denominator. If this is not the case, algebraic long division must be carriedout before the process of partial fraction is applied.

Note: The process of algebraic long division is covered in a later section in this topic.

Key point

A linear polynomial is one which has at most a degree of 1, graphically this would berepresented by a straight line.

The following examples will examine proper rational functions taking the general form shown belowwith linear factors and constants A, B . . . N to be determined. Each constant is written over onefactor from the original denominator and added to the next term. Resulting in a sum of terms on theRHS.

···(x + a)(x + b)···(x + n) ≡ A

(x + a) + B(x + b) + · · · + N

(x + n)

Note that you will be expected to factorise both quadratic and cubic polynomials.

Examples

1. Problem:

Express in partial fraction form 7x + 1x2 + x − 6

Solution:

Step 1: Factorise the denominator.

x2 + x − 6 = (x − 2)(x + 3)

Step 2: Identify the type of partial fraction. In this case it has linear factors.7x + 1

(x − 2)(x + 3) = A(x − 2) + B

(x + 3)

Step 3: Obtain the fractions with a common denominator.7x + 1

(x − 2)(x + 3) = A(x + 3)(x − 2)(x + 3) + B(x − 2)

(x − 2)(x + 3)

Step 4: Equate numerators since the denominators are equal.

7x + 1 ≡ A(x + 3) + B(x − 2)

Step 5: Choose two values of x to find the values of the two constants.

Whilst any two values for x will work, we can make the calculations easier by choosing x = 2and x = − 3 because either A or B will be multiplied by 0.

When x = 2:7 × 2 + 1 = A(2 + 3) + B(2 − 2)

15 = 5A

When x = − 3:7 × ( - 3) = A( - 3 + 3) + B( - 3 − 2)

- 20 = - 5B

So our answer is,7x + 1

x2 + x − 6≡ 3

(x − 2) + 4(x + 3)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

This example will demonstrate an alternative method to selecting values of x called equatingcoefficients. This method of working out the constants A, B, C, ... has advantages in certaincircumstances and this will be explored in further examples.

Problem:

Express x + 4(x2 − 7x + 10) in partial fractions.

Solution:

x + 4(x2 − 7x + 10) ≡ x + 4

(x − 2)(x − 5)

Step 2: Identify the type of partial fraction. In this case it has linear factors.

Let x + 4(x2 − 7x + 10)

≡ A(x − 2) + B

(x − 5)

Step 3: Obtain the fractions with a common denominator.

x + 4(x2 − 7x + 10)

≡ A(x − 5)(x − 2)(x − 5) + B(x − 2)

(x − 2)(x − 5)

x + 4 = A(x − 5) + B(x − 2) (*)

Step 5: Select values of x and substitute into (*)

Whilst we can choose any value of x we can make the calculations easier by choosing it suchthat (x − 5) = 0 and (x − 2) = 0 since then A and B will be multiplied by zero, respectively.

Hence the values taken are x = 5 and x = 2

Let x = 5:x+ 4 = A(x− 5) +B(x− 2)

5 + 4 = A(5− 5) +B(5− 2)

9 = 3B

Let x = 2:x+ 4 = A(x− 5) +B(x− 2)

2 + 4 = A(2− 5) +B(2− 2)

6 = −3A

A = −2

By either method the result follows x + 4(x2 − 7x + 10)

≡ −2(x − 2) + 3

(x − 5)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3. Problem:

Express x2 − 13x3 − 7x + 6

in partial fractions.

Solution:

Step 1: The denominator can be factorised using synthetic division.x2 − 13

x3 − 7x + 6 ≡ x2 − 13(x − 1)(x − 2)(x + 3)

Step 2: Identify the type of partial fraction. In this case again it has linear factors.

Let x2 − 13(x − 1)(x − 2)(x + 3) ≡ A

(x − 1) + B(x − 2) + C

(x + 3)

Step 3: Obtain the fractions with a common denominator.x2 − 13

(x − 1)(x − 2)(x + 3) ≡ A(x − 2)(x + 3)(x − 1)(x − 2)(x + 3) + B(x − 1)(x + 3)

(x − 1)(x − 2)(x + 3) + C(x − 1)(x − 2)(x − 1)(x − 2)(x + 3)

x2 − 13 = A(x − 2)(x + 3) + B(x − 1)(x + 3) + C(x − 1)(x − 2) (*)

Step 5: Select values of x and substitute into (*).

Whilst we can choose any value of x we can make the calculations easier by choosing it suchthat (x − 2) = 0, (x + 3) = 0 and (x − 1) = 0 since then A, B and C will be multipliedby zero respectively.

x2 − 13 = A(x − 2)(x + 3) + B(x − 1)(x + 3) + C(x − 1)(x − 2)

Let x = 2 then,−9 = 0 + 5B + 0

−9 = 5B

B = −9

5Let x = 1 then,−12 = −4A

Let x = − 3 then,−4 = 20C

C = −1

Therefore, x2 − 13x3 − 7x + 6

≡ 3(x − 1) − 9

5(x − 2) − 15(x + 3)

Go onlineLinear factors practice

Q9: Express x + 7x2 − x − 2

into partial fractions.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q10: Express 7x − 42x2 − 3x − 2

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q11: Express 8xx3 + 3x2 − x − 3

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q12: Express −11x − 34x3 + 5x2 + 2x − 8

Go onlineLinear factors exercise

Q13: Express 2x + 18x2 + 2x − 15 in partial fractions.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q14: Express −7x + 11−x2 + x + 6 in partial fractions.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q15: Express −x − 5x2 − 1

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q16: Express −4x − 10x2 + 2x − 8

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q17: Express 5x − 13x2 − 5x + 6

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q18: Express 5x2 + 6x + 7x3 − 2x2 − x + 2

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q19: Express 6x2 + 4x − 6x3 − 7x − 6

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q20: Express −7x + 9x3 − 2x2 − 9x + 18

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q21: Express 2x2 − 3x3 + 7x2 + 14x + 8

1.4 Repeated factors

When factorising quadratic and cubic functions factors may be repeated. This section will examineproper rational functions taking the form below with repeated factors on the denominator. ConstantsA, B, . . . , N are to be determined.

···(x + a)n ≡ A

(x + a) + B(x + a)2

+ · · · + N(x + a)n

Examples

1. Problem:

Express in partial fraction form 5x + 18x2 + 8x + 16

Solution:

x2 + 8x + 16 = (x + 4)(x + 4)

Step 2: Identify the type of partial fraction. In this case it has repeated linear factors.5x + 18(x + 4)2

= Ax + 4 + B

(x + 4)2

Step 3: Obtain the fractions with a common denominator.5x + 18(x + 4)2

= A(x + 4)

(x + 4)2+ B

(x + 4)2

5x + 18 = A (x + 4) + B (*)

Whilst any value for x will work, we can make the calculations easier by choosing x = − 4as A will be multiplied by zero.

Let x = − 4:

5x + 18 = A (x + 4) + B

5(−4) + 18 = A (−4 + 4) + B

−2 = B

B = −2

Now substitute B = − 2 back into (*) and select a value of x to simplify the calculation.

Let x = 0:5x + 18 = A (x + 4) + B

5(0) + 18 = A (0 + 4) − 2

18 + 2 = 4A

We have A = 5 and B = − 2.

Therefore: 5x + 18x2 + 8x + 16

= 5x + 4 − 2

(x + 4)2

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2. Problem:

Express in partial fraction form 2x − 10x3 + 6x2 + 12x + 8

Solution:

Step 1: Factorise the denominator. Synthetic division can be used to do this.

x3 + 6x2 + 12x + 8 = (x + 2)3

Step 2: Identify the type of partial fraction. In this case it has repeated linear factors.2x − 10(x + 2)3

= A(x + 2) + B

(x + 2)2+ C

(x + 2)3

2x − 10(x + 2)3

= A(x + 2)2

(x + 2)3+ B(x + 2)

(x + 2)3+ C

(x + 2)3

2x − 10 = A(x + 2)2 + B (x + 2) + C (*)

Step 5: Select values of x and equate coefficients.

First we will select a value for x and substitute into (*).

Whilst any value for x will work, we can make the calculations easier by choosing x = − 2as A and B will be multiplied by zero.

Let x = − 2:

2x − 10 = A(x + 2)2 + B (x + 2) + C

2(−2) − 10 = A(−2 + 2)2 + B (−2 + 2) + C

−14 = C

C = −14

In this case choosing another value of x would not eliminate A or B. Instead we use themethod of equating coefficients. This is done in the following way.

Substitute C = − 14 back into (*), expand the brackets and collect like terms.

2x − 10 = A(x + 2)2 + B (x + 2) + C

2x − 10 = A(x2 + 4x + 4

)+ Bx + 2B − 14

2x − 10 = Ax2 + 4Ax + 4A + Bx + 2B − 14

2x − 10 = Ax2 + (4A + B)x + 4A + 2B − 14

Set the coefficient in front of x2 on the LHS equal to the coefficient in front of x2 on the RHS.

In this case there is no x2 on the LHS. Its coefficient is therefore zero. The correspondingterm on the RHS is Ax2. The coefficient of x2 is A. We therefore equate zero and A.

Set the coefficients in front of x on the LHS equal to the coefficients in front of x on the RHS.

2 = 4A + B

2 = 4(0) + B

We have A = 0, B = 2 and C = − 14.

Therefore: 2x − 10x3 + 6x2 + 12x + 8

≡ 2(x + 2)2

− 14(x + 2)3

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3. Problem:

Express in partial fraction form 5x + 2x3 + x2

Solution:

x3 + x2 = x2(x + 1)

Step 2: Identify the type of partial fraction. In this case it has distinct linear and repeatedlinear factors.

5x + 2x2(x + 1)

= Ax + B

x2 + Cx + 1

x2(x + 1)= Ax(x + 1)

x2(x+1)+ B(x + 1)

x2(x+1)+ Cx2

x2(x+1)

5x + 2 = Ax (x + 1) + B (x + 1) + Cx2 (*)

Whilst any values for x will work, we can make the calculations easier by choosing x = − 1,x = 0 and x = 1.

Let x = − 1:

5x + 2 = Ax (x + 1) + B (x + 1) + Cx2

5(−1) + 2 = A(−1) (−1 + 1) + B (−1 + 1) + C(−1)2

−3 = C

C = −3

Let x = 0:5x + 2 = Ax (x + 1) + B (x + 1) + Cx2

5(0) + 2 = A(0) (0 + 1) + B (0 + 1) + C(0)2

Now substitute B = 2 and C = − 3 back into (*).

Let x = 1:5x + 2 = Ax (x + 1) + B (x + 1) + Cx2

5(1) + 2 = A(1) (1 + 1) + 2 (1 + 1) − 3(1)2

7 = 2A + 1

We have A = 3, B = 2 and C = − 3.

Therefore: 5x + 2x3 + x2 = 3

x + 2x2 − 3

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

4. Problem:

Express x + 3x2 − 4x + 4

Solution:

Step 1: Factorise the denominator.x + 3

x2 − 4x + 4≡ x + 3

(x − 2)2

Step 2: Identify the type of partial fraction. In this case it has repeated linear factors.x + 3

(x − 2)2≡ A

(x − 2) + B(x − 2)2

Step 3: Obtain the fractions with a common denominator.x + 3

(x − 2)2≡ A(x − 2)

(x − 2)2+ B

(x − 2)2

x + 3 = A(x − 2) + B (*)

Let x = 2 :

x + 3 = A (x − 2) + B

2 + 3 = A (2 − 2) + B

Since we know B and have only A left to work out we can choose any value of x and substituteit along with the value of B into (*). Choose x to make the calculation easy. Here we havechosen x = 0.

Let x = 0 :

x + 3 = A (x − 2) + B

0 + 3 = A (0 − 2) + 5

3 = −2A + 5

−2 = −2A

Therefore: x + 3(x − 2)2

≡ 1(x − 2) + 5

(x − 2)2

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

5. Problem:

Express 2x2 + 5x + 3x3 + 6x2 + 12x + 8 in partial fractions.

Solution:

Step 1: Use synthetic division to factorise the denominator.2x2 + 5x + 3

x3 + 6x2 + 12x + 8 = 2x2 + 5x + 3(x + 2)3

Step 2: Identify the type of partial fraction. In this case it has repeated linear factors.2x2 + 5x + 3

(x + 2)3= A

(x + 2) + B(x + 2)2

+ C(x + 2)3

2x2 + 5x + 3(x + 2)3

= A(x + 2)2

(x + 2)3+ B(x + 2)

(x + 2)3+ C

(x + 2)3

2x2 + 5x + 3 = A(x + 2)2 + B (x + 2) + C (*)

Let x = − 2:

2x2 + 5x + 3 = A(x + 2)2 + B (x + 2) + C

2(−2)2 + 5(−2) + 3 = A(−2 + 2)2 + B (−2 + 2) + C

8 − 10 + 3 = C

Now substitute C = 1 back into (*), expand the brackets and collect like terms.

2x2 + 5x + 3 = A(x + 2)2 + B (x + 2) + C

2x2 + 5x + 3 = Ax2 + 4Ax + 4A + Bx + 2B + 1

2x2 + 5x + 3 = Ax2 + (4A + B)x + (4A + 2B + 1)

x2 : 2 = A

Now, set the coefficient in front of x on the LHS equal to the coefficient in front of x on theRHS.

x : 5 = 4A + B

Now, substitute A = 2 and evaluate for B.

5 = 4(2) + B

5 = 8 + B

B = −3

We have the values of A, B and C so we do not need to equate the constants.

We have A = 2, B = − 3 and C = 1.

Therefore: 2x2 + 5x + 3x3 + 6x2 + 12x + 8

= 2(x + 2) − 3

(x + 2)2+ 1

(x + 2)3

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

6. Problem:

Express 4x2 + 9x3 + 4x2 − 3x − 18 in partial fractions.

Solution:

Step 1: Use synthetic division to factorise the denominator.4x2 + 9

x3 + 4x2 − 3x − 18 = 4x2 + 9(x − 2)(x + 3)2

Step 2: Identify the type of partial fraction. In this case it has a distinct linear factor andrepeated linear factors.

4x2 + 9(x − 2)(x + 3)2

= A(x − 2) + B

(x + 3) + C(x + 3)2

4x2 + 9(x − 2)(x + 3)2

= A(x + 3)2

(x − 2)(x + 3)2+ B(x − 2)(x + 3)

(x − 2)(x + 3)2+ C(x − 2)

(x − 2)(x + 3)2

4x2 + 9 = A(x + 3)2 + B (x − 2) (x + 3) + C (x − 2) (*)

Let x = − 3:

4x2 + 9 = A(x + 3)2 + B (x − 2) (x + 3) + C (x − 2)

4(−3)2 + 9 = A(−3 + 3)2 + B (−3 − 2) (−3 + 3) + C (−3 − 2)

36 + 9 = C (−5)

45 = −5C

C = −9

Let x = 2:4x2 + 9 = A(x + 3)2 + B (x − 2) (x + 3) + C (x − 2)

4(2)2 + 9 = A(2 + 3)2 + B (2 − 2) (2 + 3) + C (2 − 2)

16 + 9 = A(5)2

25 = 25A

Now substitute A = 1 and C = −9 back into (*), expand the brackets and collect like terms.

4x2 + 9 = A(x + 3)2 + B (x − 2) (x + 3) + C (x − 2)

4x2 + 9 = (x + 3)2 + B (x − 2) (x + 3) − 9 (x − 2)

4x2 + 9 = x2 + 6x + 9 + B(x2 + x − 6

) − 9x + 18

4x2 + 9 = x2 + 6x + 9 + Bx2 + Bx − 6B − 9x + 18

4x2 + 9 = x2 + Bx2 + Bx − 3x − 6B + 27

4x2 + 9 = (1 + B) x2 + (B − 3)x + (−6B + 27)

x2 : 4 = 1 +B

We have A = 1, B = 3 and C = − 9.

Therefore: 4x2 + 9(x − 2)(x + 3)2

= 1(x − 2) + 3

(x + 3) − 9(x + 3)2

Key point

When working out the values of A, B, C... always substitute values for x first. When this isexhausted equate coefficients.

Go onlineRepeated factors practice

Q22: Express 2x + 2(x + 3)(x + 3) in partial fractions.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q23: Express x2 − 7x + 2x3 − 3x2 + 3x − 1 in partial fractions.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q24: Express −3x2 + 6x + 20x3 − x2 − 8x + 12

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q25: Express x2 + 11x + 15x3 + 3x2 − 4

Go onlineRepeated factors exercise

Q26: Express 3x + 11(x + 1)2

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q27: Express 2x + 11x2 + 8x + 16

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q28: Express 3x + 5(x + 2)(x + 2) in partial fractions.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q29: Express x2 − 2x + 3(x + 1)3

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q30: Express 2x2 + 7x + 8x3 + 6x2 + 12x + 8

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q31: Express 6x2 + 5x + 2x3 − 3x − 2

1.5 Irreducible quadratic factors

Not all quadratic functions can be factorised into linear factors. When the resulting quadratic cannotbe factorised to give real roots it is called irreducible. This section will examine proper rationalfunctions taking the form below with irreducible quadratic factors on the denominator. Constants Aand B are to be determined.

···ax2 + bx + c ≡ Ax + B

ax2 + bx + c

Examples

1. Problem:

Express in partial fraction form x − 12x3 + 4x

Solution:

Step 1: Factorise the denominator.x − 12x3 + 4x

= x − 12x(x2 + 4)

Step 2: Identify the type of partial fraction. In this case it has a distinct linear factor and anirreducible quadratic factor. This can be checked by working out b2 − 4ac.

b2 − 4ac = 02 − 4 × 1× 4 = −16 which is less than zero therefore x2 + 4 is a irreduciblequadratic factor.

The partial fraction form is therefore x − 12x(x2 + 4)

= Ax + Bx + C

x2 + 4.

x − 12x(x2 + 4) =

A(x2 + 4)x + (Bx + C)x

x2 + 4

x − 12 = A(x2 + 4

)+ (Bx + C) x (*)

Let x = 0:0 − 12 = A (0 + 4) + (0B + C) 0

−12 = 4A

A = −3

Now substitute A = − 3 back into (*), expand the brackets and collect like terms.

x − 12 = −3(x2 + 4

)+ (Bx + C)x

x − 12 = −3x2 − 12 + Bx2 + Cx

x − 12 = (−3 + B) x2 + Cx − 12

Set the coefficients in front of x2 on the LHS equal to the coefficients in front of x2 on theRHS.0 = −3 + B

Set the coefficients in front of x on the LHS equal to the coefficients in front of x on the RHS.

We have A = − 3, B = 3 and C = 1.

Therefore: x − 12x3 + 4x = −3

x + 3x + 1x2 + 4

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2. Problem:

Express 4x + 1x3 − x2 + x − 6

Solution:

Step 1: Factorise the denominator.4x + 1

x3 − x2 + x − 6 = 4x + 1(x − 2)(x2 + x + 3)

Step 2: Identify the type of partial fraction. In this case it has a distinct linear factor and anirreducible quadratic factor.

4x + 1(x − 2)(x2 + x + 3)

= Ax − 2 + Bx + C

x2 + x + 3

4x + 1(x − 2)(x2 + x + 3)

=A(x2 + x + 3)

(x − 2)(x2 + x + 3)+ (Bx + C)(x − 2)

(x − 2)(x2 + x + 3)

4x + 1 = A(x2 + x + 3

)+ (Bx + C) (x − 2) (*)

Let x = 2:4x + 1 = A

(x2 + x + 3

)+ (Bx + C) (x − 2)

4(2) + 1 = A(22 + 2 + 3

)+ (2B + C) (2 − 2)

9 = 9A

Now substitute A = 1 back into (*) and select a value of x to simplify the calculation.

Let x = 0:4x + 1 = A

(x2 + x + 3

)+ (Bx + C) (x − 2)

4(0) + 1 = 1(02 + 0 + 3

)+ (0B + C) (0 − 2)

1 = 3 − 2C

2C = 2

Now substitute A = 1 and C = 1 back into (*) and select another value of x to simplify thecalculation.

Let x = 1:4x + 1 = A

(x2 + x + 3

)+ (Bx + C) (x − 2)

4(1) + 1 = 1(12 + 1 + 3

)+ (1B + 1) (1 − 2)

5 = 5 − B − 1

B = −1

We have A = 1, B = − 1 and C = 1.

Therefore: 4x + 1(x − 2)(x2 + x + 3) = 1

x − 2 + −x + 1x2 + x + 3

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3. Problem:

Express −7x + 5x3 − x2 + x + 3 in partial fractions.

Solution:

Step 1: Use synthetic division to factorise the denominator.−7x + 5

x3 − x2 + x + 3 = −7x + 5(x + 1)(x2 − 2x + 3)

−7x + 5(x + 1)(x2 − 2x + 3) = A

x + 1 + Bx + Cx2 − 2x + 3

−7x + 5(x + 1)(x2 − 2x + 3)

=A(x2 − 2x + 3)

(x + 1)(x2 − 2x + 3)+ (Bx + C)(x + 1)

(x + 1)(x2 − 2x + 3)

−7x + 5 = A(x2 − 2x + 3

)+ (Bx + C) (x + 1) (*)

Let x = − 1:

−7x + 5 = A(x2 − 2x + 3

)+ (Bx + C) (x + 1)

−7(−1) + 5 = A((−1)2 − 2(−1) + 3

)+ (B(−1) + C) (−1 + 1)

12 = 6A

Now substitute A = 2 back into (*) and select another value of x to simplify the calculation.

Let x = 0:−7x + 5 = A

(x2 − 2x + 3

)+ (Bx + C) (x + 1)

−7(0) + 5 = 2(02 − 2(0) + 3

)+ (B(0) + C) (0 + 1)

5 = 6 + C

C = −1

Now substitute A = 2 and C = − 1 back into (*) and select another value of x to simplifythe calculation.

Let x = 1:−7x + 5 = A

(x2 − 2x + 3

)+ (Bx + C) (x + 1)

−7(1) + 5 = 2(12 − 2(1) + 3

)+ (B(1) − 1) (1 + 1)

−2 = 4 + 2B − 2

2B = −4

B = −2

We have A = 2, B = − 2 and C = − 1.

Therefore: −7x + 5(x + 1)(x2 − 2x + 3)

= 2x + 1 + −2x − 1

x2 − 2x + 3

Go onlineIrreducible quadratic factors practice

Q32: Express x2 − 4x + 14x3 − x2 +2x − 8

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q33: Express 10x2 + 16x + 92x3 + 7x2 + 5x + 6

Go onlineIrreducible quadratic factors exercise

Q34: Express 5x2 − 4x + 1x3 − 2x2 − 2x + 1 in partial fractions.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q35: Express x + 4x3 − 4x2 + 7x − 6

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q36: Express 2x2 + 11x + 22x3 + x − 10

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q37: Express x2 + 9x2x3 − 3x2 − x − 2

Go onlinePartial fraction formation

Given 5x − 1x2 − x − 2

, this has a denominator of a quadratic with two distinct factors.

Factorise the denominator:5x − 1

x2 − x − 2→ 5x − 1

(x − 2)(x + 1)

Equate with the general form ofthis type:

5x − 1(x − 2)(x + 1) → A

(x − 2) + B(x + 1)

Obtain a common denominatorfor both fractions:

A(x − 2) + B

(x + 1) → A(x + 1)(x − 2)(x + 1) + B(x − 2)

(x − 2)(x + 1)

Equate numerator: 5x − 1 = A (x + 1) + B (x − 2)

Select values of x and substituteto solve for A and B:

Let x = −1 then −6 = A (0) +B (−3)

Let x = 2 then 9 = A (3) +B (0)

Hence A = 3 and B = 2 which gives5x − 1

x2 − x − 2= 3

x − 2 +2

Given 2x − 5x2 − 6x + 9 , this has a denominator of a quadratic with two repeated factors.

Factorise the denominator:2x − 5

x2 − 6x + 9 → 2x − 5(x − 3)2

2x − 5(x − 3)2

→ A(x − 3) + B

(x − 3)2

A(x − 3) + B

(x − 3)2→ A(x − 3)

(x − 3)2+ B

(x − 3)2

Equate numerator: 2x − 5 = A (x − 3) + B

Let x = 3 then −6 = A (0) +B (−3)

Let x = 0 and B = 1 then −5 = A (−3) + 1

Hence A = 2 and B = 1 which gives2x − 5

x2 − 6x + 9 = 2x−3 +

1(x−3)2

Given 5x2 + 6x + 4x3 + 2x2 − 2x − 1 , this has a cubic denominator. One factor is linear and the other is an

irreducible quadratic.

Factorise the denominator: 5x2 + 6x + 4x3 + 2x2 − 2x − 1

→ 5x2 + 6x + 4(x − 1)(x2 + 3x + 1)

5x2 + 6x + 4(x − 1)(x2 + 3x + 1)

→ A(x − 1) + Bx + C

(x2 + 3x + 1)

(x − 1)+

Bx + C

(x2 + 3x + 1)

↓A(x2 + 3x + 1

)(x − 1) (x2 + 3x + 1)

+Bx + C (x − 1)

(x − 1) (x2 + 3x + 1)

Equate numerator:5x2 + 6x + 4 =A(x2 + 3x + 1

)+ (Bx + C) (x − 1)

Let x = 1 then 12 = A (5) + (B (1) + C) (0)

Let x = 0 andA = 3

then 4 = 3 (1) + (B (0) + C) (−1)

Let x = −1, A = 3 and

C = −1

then 3 = 3 (−1) + (B (−1) + (−1)) (−2)

Hence A = 3, B = 2 and C = − 1 which gives5x2 + 6x + 4

x3 + 2x2 − 2x − 1= 3

x − 1 + 2x − 1x2 + 3x + 1

1.6 Algebraic long division

Key point

The dividend in a long division calculation is the expression which is being divided. As afraction it is the numerator.

Key point

The divisor is the expression which is doing the dividing.

It is the expression outside the division sign. As a fraction it is the denominator.

Key point

The quotient is the answer to the division not including the remainder.

Key point

The remainder is what is left over after dividing.

Before performing long division with polynomials a numerical example will be used to illustrate themethod.

Examples

1. 351 ÷ 8 = 43 r 7

The dividend is 351.The divisor is 8.The quotient is 43 and the remainder is 7.

In long division style this is written as:

)3 5 1

3 2 ↓3 1

This would be written in fraction terms as 43 78 .

The same technique can be used for dividing polynomials.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2. Problem:

Divide x3 − 2x + 5 by x2 + 2x − 3

Solution:

Step 1:Lay out the long division taking account of 'missing terms'.

Step 2:Divide the first term of the divisor (x2) into the first of the dividend (x3) and write the answerat the top (x).

Step 3:Multiply each of the terms in the divisor by the first term of the quotient and write underneaththe dividend.

Step 4:Subtract to give a new last line in the dividend.

Step 5:Divide the first term of the divisor (x2) into the first term of the last line (−2x2) and write theanswer at the top.

Step 6:Multiply each of the terms in the divisor by the 2nd term of the quotient (−2) and writeunderneath the divisor.

Step 7:Subtract to give a new last line in the dividend (5x − 1).The division stops here in this case as the degree of the divisor (2) is greater than the degreeof the last line (1).

Therefore x3 − 2x + 5x2 + 2x − 3 = x − 2 + 5x − 1

x2 + 2x − 3

(x − 2) is the quotient and (5x − 1) is the remainder.

Go onlineAlgebraic long division practice

Q38: Divide 3x3 − 2x2 + 6 by x2 + 4

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q39: Divide x3 + 8x2 + 13x − 10 by x + 5.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q40: Evaluate x4 + 7x2 + 13x − 4x2 + 4x

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q41: What is x5 − x3

x5 + 1?

Go onlineAlgebraic long division exercise

Q42: Divide 3x6 − 4x3 + 9 by x3 + 4.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q43: Divide x5 − 2x4 + 5x + 3 by x2 − 2.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q44: Divide x4 − 2x + 5 by x2 + 4.

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Q45: Find the remainder when x5 − 4x3 is divided by x − 4.

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Q46: Divide x3 + 3x2 + 7 by x2 − 2x.

1.7 Reduce improper rational functions by division

A variation on the previous problems occurs when the initial expression is an improper rationalfunction. In these circumstances it is necessary to divide through and obtain a polynomial anda proper rational function first. This rational function can then be expressed as a sum of partialfractions using the appropriate method from the types given in the previous sections.

Summary of partial fraction forms

Before we looking at reducing improper rational functions by division. Let's look back at thetypes of partial fractions from the previous sections. This activity provides a summary of thedifferent partial fraction forms and when they should used.

Example : Partial fraction forms

In the denominator each linear factor e.g. 2x + 1, requires one partial fraction of the form:A

2x + 1

each repeated linear factor e.g. (2x + 1)2 requires two partial fractions: A2x + 1 and B

(2x + 1)2

each irreducible quadratic factor e.g. x2 + 5, requires a partial fraction of the form: Ax + Bx2 + 5

For the following fractions where the denominator has already been factorised identify thecorrect form of partial fractions.

1) 1(x + 1)(x − 2) a) A

(x + 1) + B(x − 2) + C

(x − 2)2

2) 1(x2 + 3) b) A

(x + 1) + B(x + 1)2

+ C(x − 2)

3) 1(x + 1)2 c) Ax + B

(x2 − 6) + C(x + 1)

4) 1(x + 3)(x − 1) d) Ax + B

x2 + 3

5) 1(x2 − 6)(x + 1) e) A

(x + 1) + B(x − 2)

6) 1(x + 1)(x − 2)2 f) Ax + B

(x2 + 1) + C(x − 2) + D

(x − 2)2

7) 1(x + 1)2(x − 2) g) A

(x + 1) + B(x + 1)2

8) 1(x2 + 1)(x − 2)2 h) A

(x + 3) + B(x − 1)

Examples

1. Improper rational functions

Problem:

Express in partial fraction form x2 + 8x − 5x2 + x − 6

Solution:

It is clear that we are working with an improper rational function as the degree of thenumerator = degree of the denominator. We must therefore use algebraic long division beforewriting it as the sum of partial fractions.

Step 1: Divide using algebraic long division.

This gives x2 + 8x − 5x2 + x − 6

= 1 + 7x + 1x2 + x − 6

Step 2: Now express 7x + 1x2 + x − 6 in partial fractions by first factorising the denominator.

7x + 1x2 + x − 6 = 7x + 1

(x + 3)(x − 2)

Step 3: Now express 7x + 1(x + 3)(x − 2) in partial fractions. In this case it has distinct linear factors.

7x + 1(x + 3)(x − 2) = A

x + 3 + Bx − 2

(x + 3)(x − 2) = A(x − 2)(x + 3)(x − 2) + B(x + 3)

(x + 3)(x − 2)

7x + 1 = A (x − 2) + B (x + 3) (*)

Let x = − 3:

7x + 1 = A (x − 2) + B (x + 3)

7(−3) + 1 = A (−3 − 2) + B (−3 + 3)

−20 = −5A

Let x = 2:7x + 1 = A (x − 2) + B (x + 3)

7(2) + 1 = A (2 − 2) + B (2 + 3)

15 = 5B

Therefore:7x + 1

(x + 3)(x − 2) = 4x + 3 + 3

x − 2

Step 7: Substitute partial fractions back into the expression.

x2 + 8x − 5

x2 + x − 6= 1 +

7x + 1

x2 + x − 6

= 1 +4

x + 3+

x − 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2. Problem:

Express 2x3 − 7x2 − 11x + 10(x + 2)(x − 4) in partial fractions.

Solution:

If we expand the denominator it becomes x2 − 2x − 8. It is then clear that we are working withan improper rational function as the degree of the numerator > degree of the denominator.

This gives 2x3 − 7x2 − 11x + 10(x + 2)(x − 4) = 2x − 3 − x + 14

(x + 2)(x − 4) .

Step 2: Now express x + 14(x + 2)(x − 4) in partial fractions. In this case it has a distinct linear

factors.x + 14

(x + 2)(x − 4) =A

(x + 2) +B

(x − 4)

(x + 2)(x − 4) =A(x − 4)

(x + 2)(x − 4) +B(x + 2)

(x + 2)(x − 4)

x + 14 = A (x − 4) + B (x + 2) (*)

Let x = − 2:

x + 14 = A (x − 4) + B (x + 2)

−2 + 14 = A (−2 − 4) + B (−2 + 2)

12 = −6A

A = −2

Let x = 4:x + 14 = A (x − 4) + B (x + 2)

4 + 14 = A (4 − 4) + B (4 + 2)

18 = 6B

Therefore:x + 14

(x + 2)(x − 4) =−2

x + 2 + 3x − 4

2x3 − 7x2 − 11x + 10

(x + 2) (x − 4)= 2x − 3 − x + 14

(x + 2) (x − 4)

= 2x − 3 −[ −2

x + 2+

x − 4

= 2x − 3 +2

x + 2− 3

x − 4

Note that the sum of partial fraction is being subtracted and so the signs of the partial fractionschanges.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3. Problem:

Express x3 − 5x2 + 6x − 8x2 − 2x − 3

Solution:

It is clear that we are working with an improper rational function as the degree of thenumerator > degree of the denominator.

This gives x3 − 5x2 + 6x − 8x2 − 2x − 3

= x − 3 + 3x − 17x2 − 2x − 3

Step 2: Now express 3x − 17x2 − 2x − 3

in partial fractions by first factorising the denominator.3x − 17

x2 − 2x − 3= 3x − 17

(x + 1)(x − 3)

Step 3: Identify the type of partial fraction. In this case it has a distinct linear factors.3x − 17

(x + 1)(x − 3) =A

(x + 1) +B

(x − 3)

Step 4: Obtain the fractions with a common denominator.3x − 17

(x + 1)(x − 3) =A(x − 3)

(x + 1)(x − 3) +B(x + 1)

(x + 1)(x − 3)

3x − 17 = A (x − 3) + B (x + 1) (*)

Let x = − 1:

3x − 17 = A (x − 3) + B (x + 1)

3(−1) − 17 = A (−1 − 3) + B (−1 + 1)

−20 = −4A

Let x = 3:3x − 17 = A (x − 3) + B (x + 1)

3(3) − 17 = A (3 − 3) + B (3 + 1)

−8 = 4B

B = −2

Therefore:3x − 17

(x + 1)(x − 3) =5

(x + 1) − 2(x − 3)

x3 − 5x2 + 6x − 8

x2 − 2x − 3= x − 3 +

3x − 17

x2 − 2x − 3

= x − 3 +5

x + 1− 2

x − 3

Go onlineReduce improper rational functions by division practice

Q48: Express x3

(x + 1)(x + 2) in partial fractions.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q49: Express x3 + 3x2 + 4x + 3(x + 1)(x + 2) in partial fractions.

Go onlineReduce improper rational functions by division exercise

Q50: Express 5x2 − 6x − 1(x − 1)2

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q51: Express 2x3 − 5x2 − x + 3(x − 1)(x − 2) in partial fractions.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q52: Express −7x2 − 63x − 110−x2 − 8x − 12

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q53: Express 3x4+12x3+16x2 + 7x + 1x3 + 4x2 + 5x + 2

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q54: Express x5 − 2x4 + x3 + x2 − 10x + 5x3 − 2x2 − x + 2

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q55: Express x3 − 5x2 + 9x − 11(x − 2)3

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q56: Express 2x4 + 9x3 + 22x2 + 27x + 152x3 + 7x2 + 5x + 6

1.8 Learning points

Partial fractionsRational functions

• A proper rational function of two polynomials is when the degree of the numerator < thedegree of the denominator.x2 + 2x − 1x3 − 3x + 4

• An improper rational function of two polynomials is when the degree of the numerator ≥ thedegree of the denominator.x3 + 2x − 1

x2 + 4or x2 + 2x − 1

x2 + 4

Algebraic long division

• The method of partial fractions is applied to proper rational functions only. If the function isimproper, algebraic long division must be carried out first.

• When setting up algebraic long division each power of the variable must be accounted for inthe dividend.In the example above 0x2 has been written in the dividend to retain the position of the x2

terms.

Linear factors

• Any rational functions that have distinct linear factors on the denominator take the followingform:

···(x + a)(x + b)(x + c) ≡ A

(x + a) + B(x + b) + C

(x + c)

• Note that this rule can be extended to any number of distinct linear factors on the denominator.

Repeated linear factors

• Any rational functions that have repeated linear factors on the denominator take the followingform:

···(x + a)3

≡ A(x + a) + B

(x + a)2+ C

(x + a)3

• Note that this rule can be extended to any number of distinct linear factors on the denominator.

Irreducible quadratic factors

• An Irreducible Quadratic is a quadratic polynomial that cannot be factorised to give real rootsi.e. b2 − 4ac < 0

• Any rational functions that have irreducible quadratic factors on the denominator take thefollowing form:

···x2 + bx + c

≡ Ax + Bx2 + bx + c

• Note that this rule can be extended to any irreducible polynomial factor on the denominator.

1.9 Extended information

Learning objective

To encourage an interest in related topics

The following web links should serve as an insight to the wealth of information and encouragereaders to explore the subject further.The authors do not maintain these web links and no guarantee can be given as to theireffectiveness at a particular date.

http://www.quickmath.com/webMathematica3/quickmath/page.jsp?s1=algebra&s2=partialfractions&s3=basicAn extremely useful site for checking answers and solving problems.

1.10 End of topic test

Go onlineEnd of topic 1 test

Linear Factors

Q57: Express 5x − 8x2 − 3x + 2

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q58: Express −6x − 16x2 + 4x + 3

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q59: Express −4x − 10x2 + 2x − 8

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q60: Express 4x2 + x + 1x3 − 7x − 6

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q61: Express 5x2 + 6x + 7x3 − 2x2 − x + 2

Repeated Linear Factors

Q62: Express 2x − 7x2 − 4x + 4

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q63: Express 2x + 8x2 + 6x + 9 in partial fractions.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q64: Express x2 + 7x + 19x3 + 3x2 − 4 in partial fractions.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q65: Express 2x2 − 5x + 6x3 − 6x2 +12x − 8 in partial fractions.

Irreducible Factors

Q66: Express 4x2 + 15x + 23x3 + 5x2 + 11x + 7 in partial fractions.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q67: Express 8x − 15x3 − 3x2 + 5x − 3

Improper Fractions

Q68: Use algebraic division to express x3

x2 + 3x + 2as the sum of a polynomial and a proper

fraction.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q69: Now express x3

x2 + 3x + 2 in partial fractions.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q70: Use algebraic division to express 2x4 − 2x3 − 2x2 − 15x + 20x2 − 3x + 2

as the sum of a polynomialand a proper fraction.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q71: Now express 2x4 − 2x3 − 2x2 − 15x + 20x2 − 3x + 2

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q72: Use algebraic division to express 4x3 − 16x2 + 20x − 11x2 − 4x + 4

as the sum of a polynomial and aproper fraction.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q73: Now express 4x3 − 16x2 + 20x − 11x2 − 4x + 4

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q74: Use algebraic division to express −3x4 + 11x3 − 17x2 − 4x + 18x3 − 2x2 + 2x + 5

as the sum of a polynomialand a proper fraction.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q75: Now express −3x4 + 11x3 − 17x2 − 4x + 18x3 − 2x2 + 2x + 5 in partial fractions.

GLOSSARY 39

Glossary

dividend

the dividend in a long division calculation is the expression which is being divided; as afraction it is the numerator

divisor

the divisor is the expression which is doing the dividing; it is the expression outside the divisionsign; as a fraction it is the denominator

improper rational function

let P (x) be a polynomial of degree n and Q (x) be a polynomial of degree m

If n ≥ m then P (x)Q(x) is an improper rational function.

For examplex3 + 2x − 1

x2 − 4 or x2 + 2x − 1x2 − 4

linear polynomial

a linear polynomial is one which has at most a degree of 1, graphically this would berepresented by a straight line

partial fractions

the process of taking a proper rational function and splitting it into separate terms each with afactor of the original denominator as its denominator is called expressing the function inpartial fractions

polynomial of degree n

if P (x) = anxn + an − 1x

n - 1 + an − 2xn - 2 + . . . + a2x

2 + a1x1 + a0

where a0, . . . , an ∈ R

then P is a polynomial of degree n.

proper rational function

let P (x) be a polynomial of degree n and Q (x) be a polynomial of degree m

If n < m then P (x)Q(x) is a proper rational function.

For example,x2 + 2x − 1x3 − 3x + 4

quotient

the quotient is the answer to the division but not including the remainder

rational function

if P (x) and Q (x) are polynomials then P (x)Q(x) is called a rational function

remainder

the remainder is what is left over after dividing

40 ANSWERS: UNIT 1 TOPIC 1

Answers to questions and activities

Topic 1: Partial fractions

Division by (x - a) practice (page 5)

Q2: 86

Q3: 55

Remember we do not have an x term so 6x3 + 7x2 − 1 must be interpreted as 6x3 + 7x2 + 0x − 1and the divisor is factorised to give 3

(x − 1

So f(x) = (x − 1/3)( 6x2 + 9x + 3) but we have to take the common factor of 3 out of the

quotient and put it back into the divisor.(x − 1

)(6x2 + 9x + 3

(x − 1

(2x2 + 3x + 1

(x − 1

)(2x2 + 3x + 1

giving, f(x) = (3x − 1)(2x2 + 3x + 1).

So the quotient is 2x2 + 3x + 1 and the remainder = 0.

Factor theorem practice (page 6)

Q5: Yes

Q6: Yes

Factorising polynomials practice (page 8)

Evaluate f(1):

Since the remainder = 0, (x − 1) is a factor and,x3 − 3x + 2 = (x − 1)

(x2 + x − 2

)= (x − 1) (x + 2) (x − 1)

= (x + 2) (x − 1)2

ANSWERS: UNIT 1 TOPIC 1 41

Evaluate f(1):

Evaluate f(−1):

Since the remainder �= 0, (x + 1) is not a factor.

Evaluate f(2):

Evaluate f(−2):

Since the remainder = 0, (x + 2) is a factor and,2x3 − 3x2 − 11x + 6 = (x + 2)

(2x2 − 7x + 3

)= (x + 2) (2x − 1) (x − 3)

Linear factors practice (page 13)

Step 1: Factorise the denominator.x + 7

x2 − x − 2= x + 7

(x + 1)(x − 2)

Step 2: Identify the type of partial fraction. In this case it has linear factors.x + 7

(x + 1)(x − 2) = A(x + 1) + B

(x − 2)

(x + 1)(x − 2) = A(x − 2)(x + 1)(x − 2) + B(x + 1)

(x + 1)(x − 2)

x + 7 = A(x − 2) + B(x + 1) (*)

Let x = 2:x + 7 = A(x − 2) + B(x + 1)

2 + 7 = A(2 − 2) + B(2 + 1)

9 = 3B

Let x = − 1:

x + 7 = A(x − 2) + B(x + 1)

−1 + 7 = A(−1 − 2) + B(−1 + 1)

6 = −3A

A = −2

Therefore: x + 7x2 − x − 2

= 3x − 2 − 2

Step 1: Factorise the denominator.7x − 4

2x2 − 3x − 2 = 7x − 4(2x + 1)(x − 2)

Step 2: Identify the type of partial fraction. In this case it has linear factors.7x − 4

(2x + 1)(x − 2) = A(2x + 1) + B

(x − 2)

(2x + 1)(x − 2) = A(x − 2)(2x + 1)(x − 2) + B(2x + 1)

(2x + 1)(x − 2)

7x − 4 = A(x − 2) + B(2x + 1) (*)

Let x = 2:7x − 4 = A(x − 2) + B(2x + 1)

7(2) − 4 = A(2 − 2) + B(2(2) + 1)

10 = 5B

Let x = − 12 :

7x − 4 = A(x − 2) + B(2x + 1)

7(−1

2) − 4 = A(−1

2− 2) + B(2(−1

2) + 1)

2− 4 = −2

2= −5

5A = 3

Therefore: 7x − 42x2 − 3x − 2

= 32x + 1 + 2

x − 2

Step 1: Factorise the denominator using synthetic division.8x

x3 + 3x2 − x − 3= 8x

(x + 1)(x − 1)(x + 3)

Step 2: Identify the type of partial fraction. In this case it has linear factors.8x

(x + 1)(x − 1)(x + 3) = A(x + 1) +

B(x − 1) +

C(x + 3)

(x+ 1) (x− 1) (x+ 3)=

A (x− 1) (x+ 3)

(x+ 1) (x− 1) (x+ 3)+

B (x+ 1) (x+ 3)

(x+ 1) (x − 1) (x+ 3)+

C (x+ 1) (x− 1)

(x+ 1) (x− 1) (x+ 3)

8x = A(x − 1)(x + 3) + B(x + 1)(x + 3) + C(x + 1)(x − 1) (*)

Let x = 1:8x = A (x− 1) (x+ 3) +B (x+ 1) (x+ 3) + C (x+ 1) (x− 1)

8(1) = A (1− 1) (1 + 3) +B (1 + 1) (1 + 3) + C (1 + 1) (1− 1)

8 = 8B

Let x = − 1:

8x = A (x− 1) (x+ 3) +B (x+ 1) (x+ 3) + C (x+ 1) (x− 1)

8(−1) = A (−1− 1) (−1 + 3) +B (−1 + 1) (−1 + 3) + C (−1 + 1) (−1− 1)

−8 = −4A

Let x = − 3:

8x = A (x− 1) (x+ 3) +B (x+ 1) (x+ 3) + C (x+ 1) (x− 1)

8(−3) = A (−3− 1) (−3 + 3) +B (−3 + 1) (−3 + 3) + C (−3 + 1) (−3− 1)

−24 = 8C

C = −3

Therefore: 8x2x3 + 3x2 − x − 3 = 2

x + 1 + 1x − 1 − 3

Step 1: Factorise the denominator using synthetic division.−11x − 34

x3 + 5x2 + 2x − 8= −11x − 34

(x + 2)(x − 1)(x + 4)

Step 2: Identify the type of partial fraction. In this case it has linear factors.−11x − 34

(x + 2)(x − 1)(x + 4) = A(x + 2) +

B(x − 1) +

C(x + 4)

Step 3: Obtain the fractions with a common denominator.−11x−34

(x+2)(x−1)(x+4) =A(x−1)(x+4)

(x+2)(x−1)(x+4) +B(x+2)(x+4)

(x+2)(x−1)(x+4) +C(x+2)(x− 1)

(x+2)(x−1)(x+4)

−11x − 34 = A(x − 1)(x + 4) + B(x + 2)(x + 4) + C(x + 2)(x − 1) (*)

Let x = 1:−11x− 34 = A (x− 1) (x+ 4) +B (x+ 2) (x+ 4) + C (x+ 2) (x− 1)

−11(1) − 34 = A (1− 1) (1 + 4) +B (1 + 2) (1 + 4) + C (1 + 2) (1− 1)

−45 = 15B

B = −3

Let x = − 2:

−11x− 34 = A (x− 1) (x+ 4) +B (x+ 2) (x+ 4) + C (x+ 2) (x− 1)

−11(−2) − 34 = A (−2− 1) (−2 + 4) +B (−2 + 2) (−2 + 4) + C (−2 + 2) (−2− 1)

−12 = −6A

Let x = − 4:−11x− 34 = A (x− 1) (x+ 4) +B (x+ 2) (x+ 4) + C (x+ 2) (x− 1)

−11(−4) − 34 = A (−4− 1) (−4 + 4) +B (−4 + 2) (−4 + 4) + C (−4 + 2) (−4− 1)

10 = 10C

Therefore: −11x − 34x3 + 5x2 + 2x − 8

= 2x + 2 − 3

x − 1 + 1x + 4

Linear factors exercise (page 14)

x2 + 2x − 15 = 2x + 18(x + 5)(x − 3)

Step 2: Identify the type of partial fraction. In this case it has linear factors.2x + 18

(x + 5)(x − 3) =A

(x + 5) + B(x − 3)

(x + 5)(x − 3) =A(x − 3)

(x + 5)(x − 3) + B(x + 5)(x + 5)(x − 3)

2x + 18 = A (x − 3) + B (x + 5) (*)

Let x = − 5:2x + 18 = A (x − 3) + B (x + 5)

2(−5) + 18 = A (−5 − 3) + B (−5 + 5)

8 = −8A

A = −1

Let x = 3:2x + 18 = A (x − 3) + B (x + 5)

2(3) + 18 = A (3 − 3) + B (3 + 5)

24 = 8B

Therefore: 2x + 18x2 + 2x − 15

= −1x + 5 + 3

x − 3

Step 1: Factorise the denominator.−7x + 11

−x2 + x + 6 = −7x + 11(−x + 3)(x + 2)

Step 2: Identify the type of partial fraction. In this case it has linear factors.−7x + 11

(−x + 3)(x + 2) =A

(−x + 3) +B

(x + 2)

Step 3: Obtain the fractions with a common denominator.−7x + 11

(−x + 3)(x + 2) =A(x + 2)

(−x + 3)(x + 2) +B(−x + 3)

(−x + 3)(x + 2)

−7x + 11 = A (x + 2) +B (−x + 3) (*)

Let x = − 2:−7x + 11 = A (x + 2) +B (−x + 3)

−7(−2) + 11 = A (−2 + 2) +B (−(−2) + 3)

25 = 5B

Let x = 3:−7x + 11 = A (x + 2) +B (−x + 3)

−7(3) + 11 = A (3 + 2) +B (−3 + 3)

−10 = 5A

A = −2

Therefore: −7x + 11−x2 + x + 6

−2−x + 3 + 5

Step 1: Factorise the denominator.−x − 5x2 − 1

= −x − 5(x + 1)(x − 1)

Step 2: Identify the type of partial fraction. In this case it has linear factors.−x − 5

(x + 1)(x − 1) =A

(x + 1) +B

(x − 1)

Step 3: Obtain the fractions with a common denominator.−x − 5

(x + 1)(x − 1) =A(x − 1)

(x + 1)(x − 1) +B(x + 1)

(x + 1)(x − 1)

−x − 5 = A (x − 1) +B (x + 1) (*)

Let x = − 1:

−x − 5 = A (x − 1) +B (x + 1)

− (−1) − 5 = A (−1 − 1) +B (−1 + 1)

−4 = −2A

Let x = 1:−x − 5 = A (x − 1) +B (x + 1)

− (1) − 5 = A (1 − 1) +B (1 + 1)

−6 = 2B

B = −3

Therefore: −x − 5x2 − 1

= 2x + 1 − 3

x − 1

Step 1: Factorise the denominator.−4x − 10

x2 + 2x − 8= −4x − 10

(x + 4)(x − 2)

Step 2: Identify the type of partial fraction. In this case it has linear factors.−4x − 10

(x + 4)(x − 2) =A

(x + 4) +B

(x − 2)

Step 3: Obtain the fractions with a common denominator.−4x − 10

(x + 4)(x − 2) =A(x − 2)

(x + 4)(x − 2) +B(x + 4)

(x + 4)(x − 2)

−4x − 10 = A (x − 2) +B (x + 4) (*)

Let x = − 4:

−4x − 10 = A (x − 2) +B (x + 4)

−4 (−4) − 10 = A (−4 − 2) +B (−4 + 4)

6 = −6A

A = −1

Let x = 2:−4x − 10 = A (x − 2) +B (x + 4)

−4 (2) − 10 = A (2 − 2) +B (2 + 4)

−18 = 6B

B = −3

Therefore: −4x − 10x2 + 2x − 8

= −1x + 4 − 3

x − 2

Step 1: Factorise the denominator.5x − 13

x2 − 5x + 6= 5x − 13

(x − 3)(x − 2)

Step 2: Identify the type of partial fraction. In this case it has linear factors.5x − 13

(x − 3)(x − 2) =A

(x − 3) +B

(x − 2)

(x − 3)(x − 2) =A(x − 2)

(x − 3)(x − 2) +B(x − 3)

(x − 3)(x − 2)

5x − 13 = A (x − 2) +B (x − 3) (*)

Let x = 2:5x − 13 = A (x − 2) +B (x − 3)

5(2) − 13 = A (2 − 2) +B (2 − 3)

−3 = −B

Let x = 3:5x − 13 = A (x − 2) +B (x − 3)

5(3) − 13 = A (3 − 2) +B (3 − 3)

Therefore: 5x − 13x2 − 5x + 6 = 2

x − 3 + 3x − 2

Step 1: Factorise the denominator.5x2 + 6x + 7

x3 − 2x2 − x + 2= 5x2 + 6x + 7

(x − 1)(x − 2)(x + 1)

Step 2: Identify the type of partial fraction. In this case it has linear factors.5x2 + 6x + 7

(x − 1)(x − 2)(x + 1) =A

(x − 1) +B

(x − 2) +C

(x + 1)

Step 3: Obtain the fractions with a common denominator.5x2 + 6x + 7

(x − 1)(x − 2)(x + 1) =A(x − 2)(x + 1)

(x − 1)(x − 2)(x + 1) +B(x − 1)(x + 1)

(x − 1)(x − 2)(x + 1) +C(x − 1)(x − 2)

(x − 1)(x − 2)(x + 1)

5x2 + 6x + 7 = A (x − 2) (x + 1) +B (x − 1) (x + 1) + C (x − 1) (x − 2) (*)

Let x = − 1:

5x2 + 6x + 7 = A (x − 2) (x + 1) +B (x − 1) (x + 1) + C (x − 1) (x − 2)

5(−1)2 + 6 (−1) + 7 = A (−1 − 2) (−1 + 1) +B (−1 − 1) (−1 + 1) + C (−1 − 1) (−1 − 2)

5 − 6 + 7 = C (−2) (−3)

6 = 6C

Let x = 1:5x2 + 6x + 7 = A (x − 2) (x + 1) +B (x − 1) (x + 1) + C (x − 1) (x − 2)

5(1)2 + 6 (1) + 7 = A (1 − 2) (1 + 1) +B (1 − 1) (1 + 1) + C (1 − 1) (1 − 2)

18 = A (−1) (2)

18 = −2A

A = −9

Let x = 2:5x2 + 6x + 7 = A (x − 2) (x + 1) +B (x − 1) (x + 1) + C (x − 1) (x − 2)

5(2)2 + 6 (2) + 7 = A (2 − 2) (2 + 1) +B (2 − 1) (2 + 1) + C (2 − 1) (2 − 2)

39 = B (1) (3)

39 = 3B

B = 13

Therefore: 5x2 + 6x + 7x3 − 2x2 − x + 2

= −9x − 1 + 13

x − 2 + 1x + 1

Step 1: Factorise the denominator.6x2 + 4x − 6x3 − 7x − 6

= 6x2 + 4x − 6(x + 1)(x + 2)(x − 3)

Step 2: Identify the type of partial fraction. In this case it has linear factors.6x2 + 4x − 6

(x + 1)(x + 2)(x − 3) =A

(x + 1) +B

(x + 2) +C

(x − 3)

Step 3: Obtain the fractions with a common denominator.6x2 + 4x − 6

(x + 1)(x + 2)(x − 3) =A(x + 2)(x − 3)

(x + 1)(x + 2)(x − 3) +B(x + 1)(x − 3)

(x + 1)(x + 2)(x − 3) +C(x + 1)(x + 2)

(x + 1)(x + 2)(x − 3)

6x2 + 4x − 6 = A (x + 2) (x − 3) +B (x + 1) (x − 3) + C (x + 1) (x + 2) (*)

Let x = − 2:

6x2 + 4x − 6 = A (x + 2) (x − 3) +B (x + 1) (x − 3) + C (x + 1) (x + 2)

6(−2)2 + 4 (−2) − 6 = A (−2 + 2) (−2 − 3) +B (−2 + 1) (−2 − 3) + C (−2 + 1) (−2 + 2)

24 − 8 − 6 = B (−1) (−5)

10 = 5B

Let x = − 1:6x2 + 4x − 6 = A (x + 2) (x − 3) +B (x + 1) (x − 3) + C (x + 1) (x + 2)

6(−1)2 + 4 (−1) − 6 = A (−1 + 2) (−1 − 3) +B (−1 + 1) (−1 − 3) + C (−1 + 1) (−1 + 2)

6 − 4 − 6 = A (1) (−4)

−4 = −4A

Let x = 3:6x2 + 4x − 6 = A (x + 2) (x − 3) +B (x + 1) (x − 3) + C (x + 1) (x + 2)

6(3)2 + 4 (3) − 6 = A (3 + 2) (3 − 3) +B (3 + 1) (3 − 3) + C (3 + 1) (3 + 2)

54 + 12 − 6 = C (4) (5)

60 = 20C

Therefore: 6x2 + 4x − 6x3 − 7x − 6

= 1x + 1 + 2

x + 2 + 3x − 3

Step 1: Factorise the denominator.−7x + 9

x3 − 2x2 − 9x + 18 = −7x + 9(x + 3)(x − 2)(x − 3)

Step 2: Identify the type of partial fraction. In this case it has linear factors.−7x + 9

(x + 3)(x − 2)(x − 3) =A

(x + 3) +B

(x − 2) +C

(x − 3)

(x + 3)(x − 2)(x − 3) =A(x − 2)(x − 3)

(x + 3)(x − 2)(x − 3) +B(x + 3)(x − 3)

(x + 3)(x − 2)(x − 3) +C(x + 3)(x − 2)

(x + 3)(x − 2)(x − 3)

−7x + 9 = A (x − 2) (x − 3) +B (x + 3) (x − 3) + C (x + 3) (x − 2) (*)

Let x = − 3:−7x + 9 = A (x − 2) (x − 3) +B (x + 3) (x − 3) + C (x + 3) (x − 2)

−7 (−3) + 9 = A (−3 − 2) (−3 − 3) +B (−3 + 3) (−3 − 3) + C (−3 + 3) (−3 − 2)

21 + 9 = A (−5) (−6)

30 = 30A

Let x = 2:−7x + 9 = A (x − 2) (x − 3) +B (x + 3) (x − 3) + C (x + 3) (x − 2)

−7 (2) + 9 = A (2 − 2) (2 − 3) +B (2 + 3) (2 − 3) + C (2 + 3) (2 − 2)

−14 + 9 = B (5) (−1)

−5 = −5B

Let x = 3:−7x + 9 = A (x − 2) (x − 3) +B (x + 3) (x − 3) + C (x + 3) (x − 2)

−7 (3) + 9 = A (3 − 2) (3 − 3) +B (3 + 3) (3 − 3) + C (3 + 3) (3 − 2)

−21 + 9 = C (6) (1)

−12 = 6C

C = −2

Therefore: −7x + 9x3 − 2x2 − 9x + 18

= 1x + 3 + 1

x − 2 − 2x − 3

Step 1: Factorise the denominator.2x2 − 3

x3 + 7x2 + 14x + 8= 2x2 − 3

(x + 1)(x + 2)(x + 4)

Step 2: Identify the type of partial fraction. In this case it has linear factors.2x2 − 3

(x + 1)(x + 2)(x + 4) =A

(x + 1) +B

(x + 2) +C

(x + 4)

Step 3: Obtain the fractions with a common denominator.2x2 − 3

(x + 1)(x + 2)(x + 4) =A(x + 2)(x + 4)

(x + 1)(x + 2)(x + 4) +B(x + 1)(x + 4)

(x + 1)(x + 2)(x + 4) +C(x + 1)(x + 2)

(x + 1)(x + 2)(x + 4)

2x2 − 3 = A (x + 2) (x + 4) +B (x + 1) (x + 4) + C (x + 1) (x + 2) (*)

Let x = − 4:2x2 − 3 = A (x + 2) (x + 4) +B (x + 1) (x + 4) + C (x + 1) (x + 2)

2(−4)2 − 3 = A (−4 + 2) (−4 + 4) +B (−4 + 1) (−4 + 4) + C (−4 + 1) (−4 + 2)

32 − 3 = C (−3) (−2)

29 = 6C

6Let x = − 2:

2x2 − 3 = A (x + 2) (x + 4) +B (x + 1) (x + 4) + C (x + 1) (x + 2)

2(−2)2 − 3 = A (−2 + 2) (−2 + 4) +B (−2 + 1) (−2 + 4) + C (−2 + 1) (−2 + 2)

8 − 3 = B (−1) (2)

5 = −2B

B = −5

2Let x = − 1:

2x2 − 3 = A (x + 2) (x + 4) +B (x + 1) (x + 4) + C (x + 1) (x + 2)

2(−1)2 − 3 = A (−1 + 2) (−1 + 4) +B (−1 + 1) (−1 + 4) + C (−1 + 1) (−1 + 2)

2 − 3 = A (1) (3)

−1 = 3A

A = −1

Therefore: 2x2 − 3x3 + 7x2 + 14x + 8

= −13(x + 1) − 5

2(x + 2) + 296(x + 4)

Repeated factors practice (page 20)

Step 1: The denominator has already been factorised and can be recognised as a repeated linearfactor.

2x + 2(x + 3)(x + 3) =

2x + 2(x + 3)2

Step 2: Identify the type of partial fraction. In this case it has repeated linear factors.2x + 2(x + 3)2

= A(x + 3) +

B(x + 3)2

= A(x + 3)

(x + 3)2+ B

(x + 3)2

2x + 2 = A (x + 3) + B (*)

Let x = − 3:

2x + 2 = A (x + 3) + B

2(−3) + 2 = A (−3 + 3) + B

−6 + 2 = B

B = −4

Now substitute B = − 4 back into (*) and select another value of x to make solving for A easy.

Let x = 0:2x + 2 = A (x + 3) + B

2(0) + 2 = A (0 + 3) − 4

2 = 3A − 4

6 = 3A

We have A = 2 and B = − 4.

Therefore: 2x + 2(x + 3)(x + 3) = 2

x + 3 − 4(x + 3)2

Step 1: Use synthetic division to factorise the denominator.x2 − 7x + 2

x3 − 3x2 + 3x − 1 = x2 − 7x + 2(x−1)3

Step 2: Identify the type of partial fraction. In this case it has repeated linear factors.x2 − 7x + 2(x − 1)3

= Ax − 1 + B

(x − 1)2+ C

(x − 1)3

x2 − 7x + 2(x − 1)3

= A(x − 1)2

(x − 1)3+ B(x − 1)

(x − 1)3+ C

(x − 1)3

x2 − 7x + 2 = A(x − 1)2 + B (x − 1) + C (*)

Let x = 1:

x2 − 7x + 2 = A(x − 1)2 + B (x − 1) + C

12 − 7(1) + 2 = A(1 − 1)2 + B (1 − 1) + C

−4 = C

C = −4

Now substitute C = − 4 back into (*), expand the brackets and collect like terms.

x2 − 7x + 2 = A(x − 1)2 + B (x − 1) − 4

x2 − 7x + 2 = A(x2 − 2x + 1

)+ B (x − 1) − 4

x2 − 7x + 2 = Ax2 − 2Ax + A + Bx − B − 4

x2 − 7x + 2 = Ax2 − 2Ax + Bx + A − B − 4

x2 − 7x + 2 = Ax2 + (−2A + B)x + A − B − 4

x2 : 1 = A

Now set the coefficient in front of x on the LHS equal to the coefficient in front of x on the RHS.

x : −7 = −2A+B

−7 = −2(1) +B

−5 = B

B = −5

We have A = 1, B = − 5 and C = − 4.

Therefore: x2 − 7x + 2x3 − 3x2 + 3x − 1

= 1x − 1 − 5

(x − 1)2− 4

(x − 1)3

Step 1: Use synthetic division to factorise the denominator.−3x2 + 6x + 20

x3 − x2 − 8x + 12= −3x2 + 6x + 20

(x + 3)(x − 2)2

Step 2: Identify the type of partial fraction. In this case it has distinct linear factors and repeatedlinear factors.−3x2 + 6x + 20(x + 3)(x − 2)2

= A(x + 3) +

B(x − 2) +

C(x − 2)2

−3x2 + 6x + 20(x + 3)(x − 2)2

= A(x − 2)2

(x + 3)(x − 2)2+ B(x + 3)(x − 2)

(x + 3)(x − 2)2+ C(x + 3)

(x + 3)(x − 2)2

−3x2 + 6x + 20 = A(x − 2)2 +B (x + 3) (x − 2) + C (x + 3) (*)

Let x = 2:−3x2 + 6x + 20 = A(x − 2)2 +B (x + 3) (x − 2) + C (x + 3)

−3(2)2 + 6(2) + 20 = A(2 − 2)2 +B (2 + 3) (2 − 2) + C (2 + 3)

−12 + 12 + 20 = 5C

20 = 5C

Let x = − 3:−3x2 + 6x + 20 = A(x − 2)2 +B (x + 3) (x − 2) + C (x + 3)

−3(−3)2 + 6(−3) + 20 = A(−3 − 2)2 +B (−3 + 3) (−3 − 2) + C (−3 + 3)

−27 − 18 + 20 = 25A

−25 = 25A

A = −1

Now substitute A = − 1 and C = 4 back into (*) and select a value of x to simplify the calculation.

Let x = 0:−3x2 + 6x + 20 = A(x − 2)2 +B (x + 3) (x − 2) + C (x + 3)

−3(0)2 + 6(0) + 20 = −1(0 − 2)2 +B (0 + 3) (0 − 2) + 4 (0 + 3)

20 = −4− 6B + 12

12 = −6B

B = −2

We have A = − 1, B = − 2 and C = 4.

Therefore: −3x2 + 6x + 20x3 − x2 − 8x + 12

= −1x + 3 − 2

x − 2 + 4(x − 2)2

Step 1: Use synthetic division to factorise the denominator.x2 + 11x + 15x3 + 3x2 − 4

= x2 + 11x + 15(x − 1)(x + 2)2

Step 2: Identify the type of partial fraction. In this case it has distinct linear factors and repeatedlinear factors.x2 + 11x + 15(x − 1)(x + 2)2

= A(x − 1) +

B(x + 2) +

C(x + 2)2

Step 3: Obtain the fractions with a common denominator.x2 + 11x + 15(x − 1)(x + 2)2

= A(x + 2)2

(x − 1)(x + 2)2+ B(x − 1)(x + 2)

(x − 1)(x + 2)2+ C(x − 1)

(x − 1)(x + 2)2

x2 + 11x + 15 = A(x + 2)2 +B (x − 1) (x + 2) + C (x − 1) (*)

Let x = − 2:x2 + 11x + 15 = A(x + 2)2 +B (x − 1) (x + 2) + C (x − 1)

(−2)2 + 11(−2) + 15 = A(−2 + 2)2 +B (−2 − 1) (−2 + 2) + C (−2 − 1)

4 − 22 + 15 = −3C

−3 = −3C

Let x = 1:x2 + 11x + 15 = A(x + 2)2 +B (x − 1) (x + 2) + C (x − 1)

12 + 11(1) + 15 = A(1 + 2)2 +B (1 − 1) (1 + 2) + C (1 − 1)

27 = 9A

Now substitute A = 3 and C = 1 back into (*) and select a value of x to simplify the calculation.

Let x = 0:x2 + 11x + 15 = A(x + 2)2 +B (x − 1) (x + 2) + C (x − 1)

02 + 11(0) + 15 = 3(0 + 2)2 +B (0 − 1) (0 + 2) + 1 (0 − 1)

15 = 12− 2B − 1

4 = −2B

B = −2

We have A = 3, B = − 2 and C = 1.

Therefore: x2 + 11x + 15x3 + 3x2 − 4

= 3x − 1 − 2

x + 2 + 1(x + 2)2

Repeated factors exercise (page 21)

Step 1: Identify the type of partial fraction. In this case it has repeated factors.3x + 11(x + 1)2

= A(x + 1) +

B(x + 1)2

= A(x + 1)

(x + 1)2+ B

(x + 1)2

3x + 11 = A (x + 1) + B (*)

Let x = − 1:

3x + 11 = A (x + 1) + B

3 (−1) + 11 = A (−1 + 1) + B

−3 + 11 = B

Substitute B = 8 back into (*) and select another value of x.

Let x = 0:3x + 11 = A (x + 1) + B

3 (0) + 11 = A (0 + 1) + 8

11 = A + 8

Therefore: 3x + 11(x + 1)2

= 3x + 1 + 8

(x + 1)2

x2 + 8x + 16 = 2x + 11(x + 4)2

Step 2: Identify the type of partial fraction. In this case it has repeated factors.2x + 11(x + 4)2

= A(x + 4) +

B(x + 4)2

= A(x + 4)

(x + 4)2+ B

(x + 4)2

2x + 11 = A (x + 4) +B (*)

Let x = − 4:2x + 11 = A (x + 4) +B

2 (−4) + 11 = A (−4 + 4) +B

−8 + 11 = B

Substitute B = 3 back into (*) and select another value of x.

Let x = 0:2x + 11 = A (x + 4) +B

2 (0) + 11 = A (0 + 4) + 3

11 = 4A+ 3

Therefore: 2x + 11x2 + 8x + 16 = 2

x + 4 + 3(x + 4)2

Step 1: Identify the type of partial fraction. In this case it has repeated factors.3x + 5

(x + 2)(x + 2) =A

(x + 2) +B

(x + 2)2

(x + 2)(x + 2) =A(x + 2)

(x + 2)2+ B

(x + 2)2

3x + 5 = A (x + 2) +B (*)

Let x = − 2:3x + 5 = A (x + 2) +B

3 (−2) + 5 = A (−2 + 2) +B

−6 + 5 = B

B = −1

Substitute B = − 1 back into (*) and select another value of x.

Let x = 0:3x + 5 = A (x + 2) + B

3 (0) + 5 = A (0 + 2) − 1

5 = 2A − 1

Therefore: 3x + 5(x + 2)(x + 2) =

3x + 2 − 1

(x + 2)2

Step 1: Identify the type of partial fraction. In this case it has repeated factors.x2 − 2x + 3(x + 1)3

= A(x + 1) + B

(x + 1)2+ C

(x + 1)3

x2 − 2x + 3(x + 1)3

= A(x + 1)2

(x + 1)3+ B(x + 1)

(x + 1)3+ C

(x + 1)3

x2 − 2x + 3 = A(x + 1)2 + B (x + 1) + C (*)

Let x = − 1:

x2 − 2x + 3 = A(x + 1)2 + B (x + 1) + C

(−1)2 − 2 (−1) + 3 = A(−1 + 1)2 + B (−1 + 1) + C

1 + 2 + 3 = C

x2 − 2x + 3 = A(x + 1)2 + B (x + 1) + C

= A(x2 + 2x + 1

)+ B (x + 1) + 6

= Ax2 + 2Ax + A + Bx + B + 6

= Ax2 + (2A + B)x + A + B + 6

x2: 1 = A

Set the coefficient in front of x on the LHS equal to the coefficient in front of x on the RHS.

x: −2 = 2A + B

−2 = 2A + B

−2 = 2 (1) + B

B = −4

We have A = 1, B = − 4 and C = 6.

Therefore: x2 − 2x + 3(x + 1)3

= 1x + 1 − 4

(x + 1)2+ 6

(x + 1)3

Step 1: Factorise the denominator.2x2 + 7x + 8

x3 + 6x2 + 12x + 8= 2x2 + 7x + 8

(x + 2)3

Step 2: Identify the type of partial fraction. In this case it has repeated factors.2x2 + 7x + 8

(x + 2)3= A

(x + 2) + B(x + 2)2

+ C(x + 2)3

2x2 + 7x + 8(x + 2)3

= A(x + 2)2

(x + 2)3+ B(x + 2)

(x + 2)3+ C

(x + 2)3

2x2 + 7x + 8 = A(x + 2)2 + B (x + 2) + C (*)

Let x = − 2:

2x2 + 7x + 8 = A(x + 2)2 + B (x + 2) + C

2(−2)2 + 7 (−2) + 8 = A(−2 + 2)2 + B (−2 + 2) + C

8 − 14 + 8 = C

2x2 + 7x + 8 = A(x + 2)2 + B (x + 2) + C

= A(x2 + 4x + 4

)+ B (x + 2) + 2

= Ax2 + 4Ax + 4A + Bx + 2B + 2

= Ax2 + (4A + B)x + 4A + 2B + 2

x2: 2 = A

x: 7 = 4A + B

7 = 4A + B

7 = 4 (2) + B

B = −1

We have A = 2, B = − 1 and C = 2.

Therefore: 2x2 + 7x + 8x3 + 6x2 + 12x + 8

= 2x + 2 − 1

(x + 2)2+ 2

(x + 2)3

Step 1: Factorise the denominator.6x2 + 5x + 2x3 − 3x − 2 = 6x2 + 5x + 2

(x − 2)(x + 1)2

Step 2: Identify the type of partial fraction. In this case it has a distinct linear factor and repeatedlinear factors.6x2 + 5x + 2

(x − 2)(x + 1)2= A

(x − 2) +B

(x + 1) +C

(x + 1)2

6x2 + 5x + 2(x − 2)(x + 1)2

= A(x + 1)2

(x − 2)(x + 1)2+ B(x − 2)(x + 1)

(x − 2)(x + 1)2+ C(x − 2)

(x − 2)(x + 1)2

6x2 + 5x + 2 = A(x + 1)2 + B (x − 2) (x + 1) + C (x − 2) (*)

Let x = − 1:

6x2 + 5x + 2 = A(x + 1)2 + B (x − 2) (x + 1) + C (x − 2)

6(−1)2 + 5 (−1) + 2 = A(−1 + 1)2 + B (−1 − 2) (−1 + 1) + C (−1 − 2)

6 − 5 + 2 = C (−3)

3 = −3C

C = −1

Let x = 2:6x2 + 5x + 2 = A(x + 1)2 + B (x − 2) (x + 1) + C (x − 2)

6(2)2 + 5 (2) + 2 = A(2 + 1)2 + B (2 − 2) (2 + 1) + C (2 − 2)

24 + 10 + 2 = A(3)2

36 = 9A

Now substitute A = 4 and C = − 1 back into (*), expand the brackets and collect like terms.

6x2 + 5x + 2 = A(x + 1)2 + B (x − 2) (x + 1) + C (x − 2)

= A(x2 + 2x + 1

(x2 − x − 2

)+ C (x − 2)

= 4(x2 + 2x + 1

(x2 − x − 2

) − 1 (x − 2)

= 4x2 + 8x + 4 + Bx2 − Bx − 2B − x + 2

= (4 + B) x2 + (7 − B) x + (6 − 2B)

x2: 6 = 4 + B so B = 2

We have A = 4, B = 2 and C = − 1.

Therefore: 6x2 + 5x + 2x3 − 3x − 2 = 4

x − 2 + 2x + 1 − 1

(x + 1)2

Irreducible quadratic factors practice (page 24)

Step 1: Use synthetic division to factorise the denominator.x2 − 4x + 14

x3 − x2 +2x − 8= x2 − 4x + 14

(x − 2)(x2 + x + 4)

Step 2: Identify the type of partial fraction. In this case it has a distinct linear factor and an irreduciblequadratic factor.

x2 − 4x + 14(x − 2)(x2 + x + 4)

= Ax − 2 + Bx+C

x2 + x + 4

x2 − 4x + 14(x − 2)(x2 + x + 4)

=A(x2 + x + 4)

(x − 2)(x2 + x + 4)+ (Bx+C)(x − 2)

(x − 2)(x2 + x + 4)

x2 − 4x + 14 = A(x2 + x + 4

)+ (Bx+ C) (x − 2) (*)

Let x = 2:x2 − 4x + 14 = A

(x2 + x + 4

)+ (Bx+C) (x − 2)

22 − 4(2) + 14 = A(22 + 2 + 4

)+ (B(2) + C) (2 − 2)

10 = 10A

Let x = 0:x2 − 4x + 14 = A

(x2 + x + 4

)+ (Bx+C) (x − 2)

02 − 4(0) + 14 = 1(02 + 0 + 4

)+ (B(0) + C) (0 − 2)

14 = 4 − 2C

10 = −2C

C = −5

Now substitute A = 1 and C = − 5 back into (*) and select another value of x to simplify thecalculation.

Let x = 1:x2 − 4x + 14 = A

(x2 + x + 4

)+ (Bx + C) (x − 2)

12 − 4(1) + 14 = 1(12 + 1 + 4

)+ (B(1) − 5) (1 − 2)

11 = 6 − B + 5

We have A = 1, B = 0 and C = − 5.

Therefore: x2 − 4x + 14x3 − x2 +2x − 8 = 1

x − 2 − 5x2 + x + 4

Step 1: Use synthetic division to factorise the denominator.10x2 + 16x + 9

2x3 + 7x2 + 5x + 6 = 10x2 + 16x + 9(x + 3)(2x2 + x + 2)

Step 2: Identify the type of partial fraction. In this case it has a distinct linear factor and an irreduciblequadratic factor.

10x2 + 16x + 9(x + 3)(2x2 + x + 2) =

A(x + 3) +

Bx + C(2x2 + x + 2)

10x2 + 16x + 9(x + 3)(2x2 + x + 2)

=A(2x2 + x + 2)

(x + 3)(2x2 + x + 2)+ (Bx + C)(x + 3)

(x + 3)(2x2 + x + 2)

10x2 + 16x + 9 = A(2x2 + x + 2

)+ (Bx + C) (x + 3) (*)

Let x = − 3:10x2 + 16x + 9 = A

(2x2 + x + 2

)+ (Bx + C) (x + 3)

10(−3)2 + 16(−3) + 9 = A(2(−3)2 − 3 + 2

)+ (B(−3) + C) (−3 + 3)

51 = 17A

Let x = 0:10x2 + 16x + 9 = A

(2x2 + x + 2

)+ (Bx + C) (x + 3)

10(0)2 + 16(0) + 9 = 3(2(0)2 + 0 + 2

)+ (B(0) + C) (0 + 3)

9 = 6 + 3C

Let x = 1:10x2 + 16x + 9 = A

(2x2 + x + 2

)+ (Bx + C) (x + 3)

10(1)2 + 16(1) + 9 = 3(2(1)2 + 1 + 2

)+ (B(1) + 1) (1 + 3)

35 = 15 + 4B + 4

16 = 4B

We have A = 3, B = 4 and C = 1.

Therefore: 10x2 + 16x + 92x3 + 7x2 + 5x + 6 = 3

x + 3 + 4x + 12x2 + x + 2

Irreducible quadratic factors exercise (page 24)

Step 1: Use synthetic division to factorise the denominator.5x2 − 4x + 1

x3 − 2x2 − 2x + 1 = 5x2 − 4x + 1(x + 1)(x2 − 3x + 1)

5x2 − 4x + 1(x + 1)(x2 − 3x + 1)

= A(x + 1) + Bx + C

(x2 − 3x + 1)

5x2 − 4x + 1(x + 1)(x2 − 3x + 1)

=A(x2 − 3x + 1)

(x + 1)(x2 − 3x + 1)+ (Bx + C)(x + 1)

(x + 1)(x2 − 3x + 1)

5x2 − 4x + 1 = A(x2 − 3x + 1

)+ (Bx + C) (x + 1) (*)

Let x = − 1:

5x2 − 4x + 1 = A(x2 − 3x + 1

)+ (Bx + C) (x + 1)

5(−1)2 − 4 (−1) + 1 = A((−1)2 − 3 (−1) + 1

)+ (B (−1) + C) (−1 + 1)

10 = 5A

Let x = 0:5x2 − 4x + 1 = A

(x2 − 3x + 1

)+ (Bx + C) (x + 1)

5(0)2 − 4 (0) + 1 = 2((0)2 − 3 (0) + 1

)+ (B (0) + C) (0 + 1)

1 = 2 + C

C = −1

5x2 − 4x + 1 = A(x2 − 3x + 1

)+ (Bx + C) (x + 1)

5(1)2 − 4 (1) + 1 = 2((1)2 − 3 (1) + 1

)+ (B (1) − 1) (1 + 1)

2 = 2B − 4

We have A = 2, B = 3 and C = − 1.

Therefore: 5x2 − 4x + 1x3 − 2x2 − 2x + 1

= 2x + 1 + 3x − 1

x2 − 3x + 1

Step 1: Use synthetic division to factorise the denominator.x + 4

x3 − 4x2 + 7x − 6= x + 4

(x − 2)(x2 − 2x + 3)

x + 4(x − 2)(x2 − 2x + 3)

= A(x − 2) + Bx + C

(x2 − 2x + 3)

x + 4(x − 2)(x2 − 2x + 3)

=A(x2 − 2x + 3)

(x − 2)(x2 − 2x + 3)+ (Bx + C)(x − 2)

(x − 2)(x2 − 2x + 3)

x + 4 = A(x2 − 2x + 3

)+ (Bx + C) (x − 2) (*)

Let x = 2:x + 4 = A

(x2 − 2x + 3

)+ (Bx + C) (x − 2)

2 + 4 = A(22 − 2 (2) + 3

)+ (B (2) + C) (2 − 2)

6 = 3A

Let x = 0:x + 4 = A

(x2 − 2x + 3

)+ (Bx + C) (x − 2)

0 + 4 = 2(02 − 2 (0) + 3

)+ (B (0) + C) (0 − 2)

3 = 6 − 2C

Let x = 1:x + 4 = A

(x2 − 2x + 3

)+ (Bx + C) (x − 2)

1 + 4 = 2(12 − 2 (1) + 3

)+ (B (1) + 1) (1 − 2)

5 = 3 − B

B = −2

We have A = 2, B = − 2 and C = 1.

Therefore: x + 4x3 − 4x2 + 7x − 6

= 2x − 2 + −2x + 1

x2 − 2x + 3

Step 1: Use synthetic division to factorise the denominator.2x2 + 11x + 22x3 + x − 10 = 2x2 + 11x + 22

(x − 2)(x2 + 2x + 5)

2x2 + 11x + 22(x − 2)(x2 + 2x + 5)

= A(x − 2) + Bx + C

(x2 + 2x + 5)

2x2 + 11x + 22(x − 2)(x2 + 2x + 5)

=A(x2 + 2x + 5)

(x − 2)(x2 + 2x + 5)+ (Bx + C)(x − 2)

(x − 2)(x2 + 2x + 5)

2x2 + 11x + 22 = A(x2 + 2x + 5

)+ (Bx + C) (x − 2) (*)

Let x = 2:2x2 + 11x + 22 = A

(x2 + 2x + 5

)+ (Bx + C) (x − 2)

2(2)2 + 11 (2) + 22 = A(22 + 2 (2) + 5

)+ (B (2) + C) ((2) − 2)

52 = 13A

Let x = 0:2x2 + 11x + 22 = A

(x2 + 2x + 5

)+ (Bx + C) (x − 2)

2(0)2 + 11 (0) + 22 = 4(02 + 2 (0) + 5

)+ (B (0) + C) (0 − 2)

22 = 20 − 2C

C = −1

Let x = 1:2x2 + 11x + 22 = A

(x2 + 2x + 5

)+ (Bx + C) (x − 2)

2(1)2 + 11 (1) + 22 = 4(12 + 2 (1) + 5

)+ (B (1) − 1) (1 − 2)

35 = 33 − B

B = −2

We have A = 4, B = − 2 and C = − 1.

Therefore: 2x2 + 11x + 22x3 + x − 10

= 4x − 2 + −2x − 1

x2 + 2x + 5

Step 1: Use synthetic division to factorise the denominator.x2 + 9x

2x3 − 3x2 − x − 2= x2 + 9x

(x − 2)(2x2 + x + 1)

x2 + 9x(x − 2)(2x2 + x + 1)

= A(x − 2) + Bx + C

(2x2 + x + 1)

x2 + 9x(x − 2)(2x2 + x + 1)

=A(2x2 + x + 1)

(x − 2)(2x2 + x + 1)+ (Bx + C)(x − 2)

(x − 2)(2x2 + x + 1)

x2 + 9x = A(2x2 + x + 1

)+ (Bx + C) (x − 2) (*)

Let x = 2:x2 + 9x = A

(2x2 + x + 1

)+ (Bx + C) (x − 2)

22 + 9 (2) = A(2(2)2 + 2 + 1

)+ (B (2) + C) (2 − 2)

22 = 11A

Let x = 0:x2 + 9x = A

(2x2 + x + 1

)+ (Bx + C) (x − 2)

02 + 9 (0) = 2(2(0)2 + 0 + 1

)+ (B (0) + C) (0 − 2)

0 = 2 − 2C

Let x = 1:x2 + 9x = A

(2x2 + x + 1

)+ (Bx + C) (x − 2)

12 + 9 (1) = 2(2(1)2 + 1 + 1

)+ (B (1) + 1) (1 − 2)

10 = 7 − B

B = −3

We have A = 2, B = − 3 and C = 1.

Therefore: x2 + 9x2x3 − 3x2 − x − 2

= 2x − 2 + −3x + 1

2x2 + x + 1

Algebraic long division practice (page 29)

a) Set up the division correctly.

b) Divide x2 into 3x3 and type in the first term of the quotient.

c) Multiply 3x by the divisor and write the answer below the dividend.

d) Subtract to give a new last line in the dividend.

e) Divide x2 into −2x2 and type in the second term of the quotient.

f) Multiply −2 by the divisor and write the answer below new the dividend.

g) Subtract to give a new last line in the dividend.

h) Give the final solution.3x3 − 2x2 + 6

x2 + 4 = 3x − 2 + 14 − 12xx2 + 4

Therefore: x3 + 8x2 + 13x − 10x + 5 = x2 + 3x − 2

Therefore: x4 + 7x2 + 13x − 4x2 + 4x

= x2 − 4x + 23 − 79x + 4x2 + 4x

Therefore: x5 − x3

x5 + 1= 1 − x3 + 1

x5 + 1

Algebraic long division exercise (page 29)

Therefore: 3x6 − 4x3 + 9x3 + 4 = 3x3 − 16 + 73

x3 + 4

Therefore: x5 − 2x4 + 5x + 3x2 − 2

= x3 − 24x2 + 2x − 48 + 9x − 93x2 − 2

Therefore: x4 − 2x + 5x2 + 4

= x2 − 4 − 2x − 21x2 + 4

Therefore: x5 − 4x3

x − 4 = x4 + 4x3 + 12x2 + 48x + 192 + 768x − 4

Therefore: x3 + 3x2 + 7x2 − 2x

= x + 5 + 10x + 7x2 − 2x

Summary of partial fraction forms (page 29)

Q47: 1-e, 2-d, 3-g, 4-h, 5-c, 6-a, 7-b, 8-f

Reduce improper rational functions by division practice (page ??)

If we expand the denominator it becomes x2 + 3x + 2. It is then clear that we are working with animproper rational function as the degree of the numerator > degree of the denominator.

This gives x3

(x + 1)(x + 2) = x − 3 + 7x + 6(x + 1)(x + 2) .

Step 2: Now express 7x + 6(x + 1)(x + 2) in partial fractions. In this case it has a distinct linear factors.

7x + 6(x + 1)(x + 2) =

A(x + 1) + B

(x + 2)

(x + 1)(x + 2) =A(x + 2)

(x + 1)(x + 2) + B(x + 1)(x + 1)(x + 2)

7x + 6 = A (x + 2) + B (x + 1) (*)

Let x = − 1:7x + 6 = A (x + 2) + B (x + 1)

7(−1) + 6 = A (−1 + 2) + B (−1 + 1)

−1 = A

A = −1

Let x = − 2:7x + 6 = A (x + 2) + B (x + 1)

7(−2) + 6 = A (−2 + 2) + B (−2 + 1)

−8 = −B

Therefore: 7x + 6(x + 1)(x + 2) =

−1(x + 1) + 8

(x + 2)

(x + 1) (x + 2)= x − 3 +

7x + 6

(x + 1) (x + 2)

= x − 3 +

[ −1

(x + 1)+

(x + 2)

= x − 3 − 1

(x + 1)+

(x + 2)

It is clear that we are working with an improper rational function as the degree of the numerator >degree of the denominator.

This gives x3 + 3x2 + 4x + 3(x + 1)(x + 2) = x + 2x + 3

(x + 1)(x + 2) .

Step 2: Now express 2x + 3(x + 1)(x + 2) in partial fractions. In this case it has a distinct linear factors.

2x + 3(x + 1)(x + 2) =

A(x + 1) +

B(x + 2)

(x + 1)(x + 2) =A(x + 2)

(x + 1)(x + 2) + B(x + 1)(x + 1)(x + 2)

2x + 3 = A (x + 2) + B (x + 1) (*)

Let x = − 1:

2x + 3 = A (x + 2) + B (x + 1)

2(−1) + 3 = A (−1 + 2) + B (−1 + 1)

Let x = − 2:

2x + 3 = A (x + 2) + B (x + 1)

2(−2) + 3 = A (−2 + 2) + B (−2 + 1)

−1 = −B

x3 + 3x2 + 4x + 3

(x + 1) (x + 2)= x +

2x + 3

(x + 1) (x + 2)

= x +1

(x + 1)+

(x + 2)

Reduce improper rational functions by division exercise (page 34)

If we expand the denominator it becomes x2 − 2x + 1. It is then clear that we are working with animproper rational function as the degree of the numerator = degree of the denominator. We musttherefore use algebraic long division before writing it as the sum of partial fractions.

This gives 5x2 − 6x − 1(x − 1)2

= 5 + 4x − 6(x − 1)2

Step 2: Now express 4x − 6(x − 1)2

in partial fractions. In this case it has repeated linear factors.

4x − 6(x − 1)2

= Ax − 1 + B

(x − 1)2

Step 3: Obtain the fractions with a common denominator.4x − 6(x − 1)2

= A(x − 1)

(x − 1)2+ B

(x − 1)2

4x − 6 = A (x − 1) + B (*)

Let x = 1:4x − 6 = A (x − 1) + B

4 (1) − 6 = A (1 − 1) + B

B = −2

Now substitute B = − 2 back into (*) and select a value of x to simplify the calculation.

Let x = 0:4x − 6 = A (x − 1) + B

4 (0) − 6 = A (0 − 1) − 2

−6 = A (−1) − 2

Therefore: 4x − 6(x − 1)2

= 4x − 1 − 2

(x − 1)2

Step 6: Substitute partial fractions back into the expression.5x2 − 6x − 1

(x − 1)2= 5 + 4

x − 1 − 2(x − 1)2

If we expand the denominator it becomes x2 − 3x + 2. It is then clear that we are working with animproper rational function as the degree of the numerator > degree of the denominator. We musttherefore use algebraic long division before writing it as the sum of partial fractions.

This gives 2x3 − 5x2 − x + 3(x − 1)(x − 2) = 2x + 1 + −2x + 1

(x − 1)(x − 2) .

Step 2: Now express −2x + 1(x − 1)(x − 2) in partial fractions. In this case it has a distinct linear factors.

−2x + 1(x − 1)(x − 2) =

A(x − 1) + B

(x − 2)

(x − 1)(x − 2) =A(x − 2)

(x − 1)(x − 2) + B(x − 1)(x − 1)(x − 2)

−2x + 1 = A (x − 2) + B (x − 1) (*)

Let x = 1:−2x + 1 = A (x − 2) + B (x − 1)

−2 (1) + 1 = A (1 − 2) + B (1 − 1)

−1 = A (−1)

Let x = 2:−2x + 1 = A (x − 2) + B (x − 1)

−2 (2) + 1 = A (2 − 2) + B (2 − 1)

−3 = B (1)

B = −3

Therefore: −2x + 1(x − 1)(x − 2) =

1x − 1 − 3

x − 2

Step 6: Substitute partial fractions back into the expression.2x3 − 5x2 − x + 3

(x − 1)(x − 2) = 2x + 1 + 1x − 1 − 3

x − 2

It is clear that we are working with an improper rational function as the degree of the numerator =degree of the denominator. We must therefore use algebraic long division before writing it as thesum of partial fractions.

This gives −7x2 − 63x − 110−x2 − 8x − 12 = 7 + −7x − 26

−x2 − 8x − 12 .

Step 2: Now express −7x − 26−x2 − 8x − 12

in partial fractions by first factorising the denominator.−7x − 26

−x2 − 8x − 12= −7x − 26

(−x − 2)(x + 6)

Step 3: Now express −7x − 26(−x − 2)(x + 6) in partial fractions. In this case it has a distinct linear factors.

−7x − 26(−x − 2)(x + 6) =

A−x − 2 + B

Step 4: Obtain the fractions with a common denominator.−7x − 26

(−x − 2)(x + 6) =A(x + 6)

(−x − 2)(x + 6) + B(−x − 2)(−x − 2)(x + 6)

−7x − 26 = A (x + 6) + B (−x − 2) (*)

Let x = − 2:−7x − 26 = A (x + 6) + B (−x − 2)

−7 (−2) − 26 = A (−2 + 6) + B (− (−2) − 2)

−12 = 4A

A = −3

Let x = − 6:−7x − 26 = A (x + 6) + B (−x − 2)

−7 (−6) − 26 = A (−6 + 6) + B (− (−6) − 2)

16 = 4B

Therefore: −7x − 26−x2 − 8x − 12

= −3−x − 2 + 4

Step 7: Substitute partial fractions back into the expression.−7x2 − 63x − 110−x2 − 8x − 12

= 7 + −3−x − 2 + 4

It is clear that we are working with an improper rational function as the degree of the numerator >degree of the denominator. We must therefore use algebraic long division before writing it as thesum of partial fractions.

This gives 3x4+12x3+16x2 + 7x + 1x3 + 4x2 + 5x + 2

= 3x + x2 + x + 1x3 + 4x2 + 5x + 2

Step 2: Now express x2 + x + 1x3 + 4x2 + 5x + 2

in partial fractions by first factorising the denominator.

x2 + x + 1x3 + 4x2 + 5x + 2

= x2 + x + 1(x + 2)(x + 1)2

Step 3: Now express x2 + x + 1(x + 2)(x + 1)2

in partial fractions. In this case it has a distinct linear factor and

repeated linear factors.x2 + x + 1

(x + 2)(x + 1)2= A

(x + 2) + B(x + 1) + C

(x + 1)2

x2 + x + 1(x + 2)(x + 1)2

= A(x + 1)2

(x + 2)(x + 1)2+ B(x + 2)(x + 1)

(x + 2)(x + 1)2+ C(x + 2)

(x + 2)(x + 1)2

x2 + x + 1 = A(x + 1)2 + B (x + 2) (x + 1) + C (x + 2) (*)

Let x = − 2:

x2 + x + 1 = A(x + 1)2 + B (x + 2) (x + 1) + C (x + 2)

(−2)2 + (−2) + 1 = A(−2 + 1)2 + B (−2 + 2) (−2 + 1) + C (−2 + 2)

4 − 2 + 1 = A(−1)2

Let x = − 1:

x2 + x + 1 = A(x + 1)2 + B (x + 2) (x + 1) + C (x + 2)

(−1)2 + (−1) + 1 = A(−1 + 1)2 + B (−1 + 2) (−1 + 1) + C (−1 + 2)

1 = C (1)

Now substitute A = 3 and C = 1 back into (*).

Let x = 0:x2 + x + 1 = A(x + 1)2 + B (x + 2) (x + 1) + C (x + 2)

(0)2 + (0) + 1 = 3(0 + 1)2 + B (0 + 2) (0 + 1) + 1 (0 + 2)

1 = 3 + B (2) (1) + 2

B = −2

We have A = 3, B = − 2 and C = 1.

Therefore: x2 + x + 1x3 + 4x2 + 5x + 2

= 3x + 2 − 2

x + 1 + 1(x + 1)2

Step 7: Substitute partial fractions back into the expression.3x4+12x3+16x2 + 7x + 1

x3 + 4x2 + 5x + 2 = 3x + 3x + 2 − 2

x + 1 + 1(x + 1)2

This gives x5 − 2x4 + x3 + x2 − 10x + 5x3 − 2x2 − x + 2

= x2 + 2 + 3x2 − 8x + 1x3 − 2x2 − x + 2

Step 2: Now express 3x2 − 8x + 1x3 − 2x2 − x + 2

3x2 − 8x + 1x3 − 2x2 − x + 2

= 3x2 − 8x + 1(x + 1)(x − 1)(x − 2)

Step 3: Now express 3x2 − 8x + 1(x + 1)(x − 1)(x − 2) in partial fractions. In this case it has a distinct linear factors.

3x2 − 8x + 1(x + 1)(x − 1)(x − 2) =

A(x + 1) + B

(x − 1) + C(x − 2)

Step 4: Obtain the fractions with a common denominator.3x2 − 8x + 1

(x + 1)(x − 1)(x − 2) =A(x − 1)(x − 2)

(x + 1)(x − 1)(x − 2) + B(x + 1)(x − 2)(x + 1)(x − 1)(x − 2) + C(x + 1)(x − 1)

(x + 1)(x − 1)(x − 2)

3x2 − 8x + 1 = A (x − 1) (x − 2) + B (x + 1) (x − 2) + C (x + 1) (x − 1) (*)

Let x = − 1:

3x2 − 8x + 1 = A (x − 1) (x − 2) + B (x + 1) (x − 2) + C (x + 1) (x − 1)

3(−1)2 − 8 (−1) + 1 = A (−1 − 1) (−1 − 2) + B (−1 + 1) (−1 − 2) + C (−1 + 1) (−1 − 1)

3 + 8 + 1 = A (−2) (−3)

12 = 6A

Let x = 1:3x2 − 8x + 1 = A (x − 1) (x − 2) + B (x + 1) (x − 2) + C (x + 1) (x − 1)

3(1)2 − 8 (1) + 1 = A (1 − 1) (1 − 2) + B (1 + 1) (1 − 2) + C (1 + 1) (1 − 1)

3 − 8 + 1 = B (2) (−1)

−4 = −2B

Let x = 2:3x2 − 8x + 1 = A (x − 1) (x − 2) + B (x + 1) (x − 2) + C (x + 1) (x − 1)

3(2)2 − 8 (2) + 1 = A (2 − 1) (2 − 2) + B (2 + 1) (2 − 2) + C (2 + 1) (2 − 1)

12 − 16 + 1 = C (3) (1)

−3 = 3C

C = −1

Therefore: 3x2 − 8x + 1x3 − 2x2 − x + 2

= 2x + 1 + 2

x − 1 − 1x − 2

Step 7: Substitute partial fractions back into the expression.x5 − 2x4 + x3 + x2 − 10x + 5

x3 − 2x2 − x + 2 = x2 + 2 + 2x + 1 + 2

x − 1 − 1x − 2

If we expand the denominator it becomes x3 − 6x2 + 12x − 8. It is then clear that we are workingwith an improper rational function as the degree of the numerator = degree of the denominator. Wemust therefore use algebraic long division before writing it as the sum of partial fractions.

This gives x3 − 5x2 + 9x − 11(x − 2)3

= 1 + x2 − 3x − 3(x − 2)3

Step 2: Now express x2 − 3x − 3(x − 2)3

in partial fractions. In this case it has repeated linear factors.

x2 − 3x − 3(x − 2)3

= Ax − 2 + B

(x − 2)2+ C

(x − 2)3

x2 − 3x − 3(x − 2)3

= A(x − 2)2

(x − 2)3+ B(x − 2)

(x − 2)3+ C

(x − 2)3

x2 − 3x − 3 = A(x − 2)2 + B (x − 2) + C (*)

Let x = 2:x2 − 3x − 3 = A(x − 2)2 + B (x − 2) + C

(2)2 − 3 (2) − 3 = A(2 − 2)2 + B (2 − 2) + C

C = −5

Now substitute C = − 5 back into (*), expand the brackets and collect like terms.

x2 − 3x − 3 = A(x − 2)2 + B (x − 2) + C

= A(x2 − 4x − 4

)+ B (x − 2) − 5

= Ax2 − 4Ax − 4A + Bx − 2B − 5

= Ax2 + (−4A + B) x − 4A − 2B − 5

−3 = − 4A + B so B = 1 since we know A = 1.

We have A = 1, B = 1 and C = − 5.

Therefore: x2 − 3x − 3(x − 2)3

= 1x − 2 + 1

(x − 2)2− 5

(x − 2)3

Step 6: Substitute partial fractions back into the expression.x3 − 5x2 + 9x − 11

(x − 2)3= 1 + 1

x − 2 + 1(x − 2)2

− 5(x − 2)3

This gives 2x4 + 9x3 + 22x2 + 27x + 152x3 + 7x2 + 5x + 6 = x + 1 + 10x2 + 16x + 9

2x3 + 7x2 + 5x + 6 .

Step 2: Now express 10x2 + 16x + 92x3 + 7x2 + 5x + 6

10x2 + 16x + 92x3 + 7x2 + 5x + 6

= 10x2 + 16x + 9(x + 3)(2x2 + x + 2)

Step 3: Now express 10x2 + 16x + 9(x + 3)(2x2 + x + 2)

in partial fractions. In this case it has a distinct linear factorand an irreducible quadratic factor.

10x2 + 16x + 9(x + 3)(2x2 + x + 2)

= A(x + 3) + Bx + C

(2x2 + x + 2)

10x2 + 16x + 9(x + 3)(2x2 + x + 2) =

A(2x2 + x + 2)(x + 3)(2x2 + x + 2) + (Bx + C)(x + 3)

(x + 3)(2x2 + x + 2)

10x2 + 16x + 9 = A(2x2 + x + 2

)+ (Bx + C) (x + 3) (*)

Let x = − 3:

10x2 + 16x + 9 = A(2x2 + x + 2

)+ (Bx + C) (x + 3)

10(−3)2 + 16 (−3) + 9 = A(2(−3)2 + (−3) + 2

)+ (B (−3) + C) (−3 + 3)

90 − 48 + 9 = A (18 − 3 + 2)

51 = 17A

Let x = 0:10x2 + 16x + 9 = A

(2x2 + x + 2

)+ (Bx + C) (x + 3)

10(0)2 + 16 (0) + 9 = 3(2(0)2 + (0) + 2

)+ (B (0) + C) (0 + 3)

9 = 6 + 3C

Let x = 1:10x2 + 16x + 9 = A

(2x2 + x + 2

)+ (Bx + C) (x + 3)

10(1)2 + 16 (1) + 9 = 3(2(1)2 + (1) + 2

)+ (B (1) + 1) (1 + 3)

35 = 19 + 4B

16 = 4B

We have A = 3, B = 4 and C = 1.

Therefore: 10x2 + 16x + 92x3 + 7x2 + 5x + 6

= 3x + 3 + 4x + 1

2x2 + x + 2

Step 7: Substitute partial fractions back into the expression.2x4 + 9x3 + 22x2 + 27x + 15

2x3 + 7x2 + 5x + 6 = x + 1 + 3x + 3 + 4x + 1

2x2 + x + 2

End of topic 1 test (page 37)

Q57: 3x − 1 + 2

x − 2

Q58: − 1x + 3 − 5

Q59: − 3x − 2 − 1

Q60: − 1x + 1 + 3

x + 2 + 2x − 3

Q61: 9x − 1 + 13

x − 2 + 1x + 1

Q62: 2x − 2 − 3

(x − 2)2

Q63: 2x + 3 + 2

(x + 3)2

Q64: 3x − 1 − 2

x + 2 − 3(x + 2)2

Q65: 2x − 2 + 3

(x − 2)2+ 4

(x − 2)3

Q66: 3x + 1 + x + 2

x2 + 4x + 7

Q67: − 72(x − 1) + 7x + 9

2(x2 − 2x + 3)

Q68: x − 3 + 7x + 6x2 + 3x + 2

Q69: x − 3 − 1x + 1 + 8

Q70: 2x2 + 4x + 6 − 5x − 8x2 − 3x + 2

Q71: 2x2 + 4x + 6 − 3x − 1 − 2

x − 2

Q72: 4x + 4x − 11x2 − 4x + 4

Q73: 4x + 4x − 2 − 3

(x − 2)2

Q74: −3x+ 5− x2−x+7x3−2x2+2x−5

Q75: −3x + 5 − 1x − 1 − 2

x2 − 3x + 5

SCHOLAR Study Guide CfE Advanced Higher Mathematics Course ... · TOPIC 1. PARTIAL FRACTIONS 3 1.1...

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