© Boardworks Ltd 2006 1 of 37 A2-Level Maths: Core 3 for Edexcel C3.1 Algebra and functions 1 This...

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A2-Level Maths: Core 3for Edexcel

C3.1 Algebra and functions 1

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Simplifying algebraic fractions

Adding and subtracting algebraic fractions

Multiplying and dividing algebraic fractions

Improper fractions and polynomial division

Examination-style questionCont

ents



Rational expressions

Remember, a rational number is any number that can be written in the form , where a and b are integers and b ≠ 0.

2

3

x 2

3 +1

2

x

x

3

2

2

+ 3 4

x

x x

ab

Numbers written in this form are often called fractions.

In algebra, a rational expression is an algebraic fraction that can be written in the form , where f(x) and g(x) are polynomials and g(x) ≠ 0.f

g( )( )xx

For example,

For which values of x are each of the above expressions undefined?

Rational expressions

That is, when x = –2. is undefined when x + 2 = 0.3

+ 2x

That is, when x = ±√2. is undefined when x2 – 2 = 0.2

3 +1

2

x

x

We can factorize this to give (x + 4)(x – 1) = 0.

is undefined when x2 + 3x – 4 = 0.3

2

2

+ 3 4

x

x x

So is undefined when x = –4 or x = 1.3

2

2

+ 3 4

x

x x

An algebraic fraction is undefined when the denominator is 0. So,

Simplifying fractions by cancelling

When the numerator and the denominator of a numerical fraction contain a common factor, the fraction can be simplified by cancelling.

For example, consider the fraction

This fraction can therefore be written in its simplest terms by dividing both the numerator and the denominator by 14.

The highest common factor of 28 and 42 is ___. 14

28

42

2

3

2=

3

Simplifying algebraic fractions by cancelling

Algebraic fractions can be cancelled in the same way.

For example,

2

3

6=

8

a

a

6= =

8

a a

a a a

3

4

3=

4a

When the numerator or the denominator contains more than one term, we have to factorize before cancelling. For example,

Simplifypq

p p 2

3

15 9

2

3=

15 9

pq

p p3

=3 (5 3 )

pq

p p 5 3

q

p

Simplifying algebraic fractions by cancelling

2

3 6=

2

b

b b

3( 2)=

( 2)

b

b b

3

b

Simplify x x

x

2

2

+ 2

1

Simplify2

3 6

2

b

b b

2

2

+ 2=

1

x x

x

( 1)( 2)=

( 1)( 1)

x x

x x

2

1

x

x

Using additive inverses

When manipulating algebraic fractions it is helpful to remember that an expression of the form a – b is the additive inverse of the expression b – a.

a – b = –(b – a)

and

b – a = –(a – b )

For example,

When cancelling, look for situations where a factor of –1 can be taken out of a pair of brackets.

2

2

3=

3

y

y

2

2

( 3)=

3

y

y

–1

i.e.

Using additive inversesSimplify

2

14 7

5 + 6

x

x x

2

14 7=

5 + 6

x

x x

7(2 )

( 2)( 3)

x

x x

7(2 )=

(2 )( 3)

x

x x

7=

3 x

7=

( 3)x

Simplifying complex fractionsSometimes the numerator or the denominator of an algebraic fraction contains another fraction. For example,

Simplify2

13 xx

x

This can be simplified by multiplying the numerator and the denominator by x.

2

13=xx

x

2

3

3 1x

x

Simplify2

3

1 aaa

Multiply the numerator and the denominator by 3a:

2

3

1=a

aa

2 2

3 6

3

a

a a

2

3 6=

4

a

a

Simplifying complex fractions

Simplify

To simplify this algebraic fraction we multiply the numerator and the denominator by the lowest common multiple of x, x2 and 3x. That is ___.

3x2

2

9 12

3 4

x

x x

3(3 + 4)=

(3 + 4)

x

x x

3=

x

23 4

43

=1x x

x

23 4

431

x x

x






ents



Adding and subtracting fractionsBefore looking at the addition and subtraction of algebraic fractions, let’s recall the method used for numerical fractions.

This is the lowest common multiple (LCM) of their denominators.

What is ?5 3

+6 4

Before we can add these two fractions we have to write them as equivalent fractions over a common denominator.

It is best to use the lowest common denominator of the two fractions.

Adding and subtracting fractionsThe LCM of 6 and 4 is ___.

So we write,5 3 10 + 9

+ =6 4 12

10 9

19=

12

12

We apply the same method to add or subtract algebraic fractions.

712=1

2 2

2 3 2 + 3+ =

x

x x x

Adding and subtracting fractionsWrite as a single fraction in its lowest terms.

2

2 3+

x x

The LCM of x2 and x is ___. x2

2 3x

Write as a single fraction in its lowest terms.3

y x

x y

2 23=

3 3

y x y x

x y xy

The LCM of 3x and y is ___. 3xy

y2 3x2


2 1+

+ 3 2 + 6

x

x x

The LCM of x + 3 and 2(x + 3) is ______.2(x + 3)

Start by factorizing where possible:2 1

++ 3 2( + 3)

x

x x

2 1+ =

+ 3 2( + 3)

x

x x

4 1+

2( + 3) 2( + 3)

x

x x

4 +1=

2( + 3)

x

x


2

2 + 83

+ 5

x

x

2

2 + 83 =

+ 5

x

x

2

2 2

3( + 5) 2 + 8

+ 5 + 5

x x

x x

2

2

3 +15 2 8=

+ 5

x x

x

Notice that this becomes – 8.

2

2

3 2 7=

+ 5

x x

x






ents



Multiplying and dividing fractionsBefore looking at the multiplication and division of algebraic fractions, let’s recall the methods used for numerical fractions.

What is ?3 12

×8 21

When multiplying two fractions, start by cancelling any common factors in the numerators and denominators:

3 12×

8 21

1

7

3

2Then multiply the numerators and multiply the denominators:

1 3× =

2 7

3

14

Multiplying and dividing fractionsTo divide by a fraction we multiply by its reciprocal.

What is ?7 14

÷9 15

This is equivalent to

7 15× =

9 14

1

2

5

3

1 5×

3 2

5=

6


We can apply the same methods to the multiplication and division of algebraic fractions. For example,

Simplify3 12 2

×2 + 4 4

x

x x

Start by factorizing where possible:

3 12 2× =

2 + 4 4

x

x x

3( 4) 2×

2( + 2) 4

x

x x

3=

+ 2x


Simplify2 4 3

×+ 2 2 4

x x

x x

2 4 3× =

+ 2 2 4

x x

x x

( 2)( 2) 3×

+ 2 2( 2)

x x x

x x

3=

2

x

Simplify5 15

÷2 p p

5 15÷ =

2 p p

5× =

2 15

p

p

1

63


Simplify2

14 7÷

+ 2 6a a a

2

14 7÷ =

+ 2 6a a a

214 6×

+ 2 7

a a

a

14 ( + 2)( 3)= ×

+ 2 7

a a

a

2

= 2(a – 3)






ents



Improper fractions and mixed numbers

Remember, a numerical fraction is called an improper fraction if the numerator is larger than the denominator.

Improper fractions are usually simplified by writing them as a whole number plus a proper fraction.

For example, the improper fraction can be converted to a mixed number as follows:296

29 24 + 5= =

6 6

24 5+ =

6 6

5

64

This is called a mixed number.

When 29 is divided by 6, 4 is the quotient and 5 is the remainder.

Improper algebraic fractionsAn algebraic fraction is called an improper fraction when the numerator is a polynomial of degree greater than, or equal to, the degree of the denominator.

For example,

are improper algebraic fractions.

3

2 5

x

x

24

( 4)( 2)

x

x x 3

4

x

x

and

Suppose we have an improper fraction . fg( )( )xx

Dividing f(x) by g(x) will give us a quotient q(x) and a remainder r(x), which gives us the identity:

f rq

g g

( ) ( )( )+

( ) ( )

x xx

x x

where the degree of f(x) ≥ the degree of g(x).

Writing improper fractions in proper form

We can think of the form as being the algebraic

equivalent of mixed number form. It is a polynomial plus a proper fraction.

If the degree of f(x) is n and the degree of g(x) is m then:

An improper algebraic fraction can be written in proper form by:

rewriting the numerator.

writing an appropriate identity to equate the coefficients.

using long division to divide the numerator by the denominator.

rq

g

( )( )+

( )

xx

x

The degree of the quotient q(x) will be equal to n – m. The degree of the quotient q(x) will be equal to n – m.

The degree of the remainder r(x) will be less than m.The degree of the remainder r(x) will be less than m.

Writing improper fractions in proper form

Write in the form A + .+ 3

1

x

x 1

B

x

Rewriting the numeratorA useful technique for writing improper fractions in proper form is to look for ways to rewrite the numerator so that part of it can be divided by the denominator. For example,

+ 3

1

x

x 1+ 4

=1

x

x

1 4= +

1 1

x

x x

4=1+

1x

Rewriting the numerator

2

2 2

3( +1) 2 3= +

+1 +1

x x

x x

2

2 3= 3 +

+1

x

x

We can write this in fraction form as:

2

2

3 + 2

+1

x x

x

2

2

3( +1)+ 2 3=

+1

x x

x

So when 3x2 + 2x is divided by x2 + 1 the quotient is 3 and the remainder is 2x – 3.

What is the quotient and the remainder when 3x2 + 2x is divided by x2 + 1?

Rewriting the numerator

2

2

( 3)+ 3=

3

x x x

x

2

2 2

( 3) 3= +

3 3

x x x

x x

We can write this in fraction form as:

3

2 3

x

x

3

2

3 + 3=

3

x x x

x

The remainder is 3x.

2

3= +

3

xx

x

Find the remainder when x3 is divided by x2 – 3.

Constructing an identityWhen the numerator cannot easily be manipulated to give an expression of the required form, we can write an identity using:

f rq

g g

( ) ( )( )+

( ) ( )

x xx

x x

What is x3 – 4x2 + 5 divided by x2 – 3?

where q(x) is the quotient and r(x) is the remainder when f(x) is divided by g(x).

Let the quotient be Ax + B. (It must be linear because the degree of the dividend minus the degree of the divisor is 1).

Let the remainder be Cx + D. (Its degree must be less than 2.)

Constructing an identityThis gives us the identity

3 2

2 2

4 + 5 ++ +

3 3

x x Cx DAx B

x x

Multiply through by x2 – 3:

3 2 24 + 5 ( + )( 3)+ +x x Ax B x Cx D 3 23 + 3 + +Ax Ax Bx B Cx D

Equate the coefficients:

x3: A = 1

x2:

x:

constants:

B = –4

C – 3A = 0 C = 3

D – 3B = 5 D + 12 = 5

D = –7

Constructing an identityWe can now substitute these values into the original identity to give:

3 2

2 2

4 + 5 3 74 +

3 3

x x xx

x x

Alternatively, use long division:

x3 – 4x2 + 0x + 5x2 – 3

x3 + 0x2 – 3x

– 4x2 + 3x + 5

x

– 4x2 +0x + 12

3x – 7

– 4

The quotient is x – 4 and the remainder is 3x – 7 so, as before:

3

2 2

2 3 74 +

3

4 + 5

3

x x

x

xx

x

Cont

ents






Examination-style question

Examination-style question

Examination-style questionGiven that

show that

f2

3 9( ) + { , 1, 2},

2 2x x x x x

x x x

f2 + 3

( ) .+1

x xx

x

2

3 9 3 9+ = +

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

x x x x x x

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

( 2)( +1) ( 2)( +1) ( 2)( +1)

x x x x

x x x x x x

3 2 2 3 3 + 9

=( 2)( +1)

x x x x

x x

3 2 5 + 6=

( 2)( +1)

x x x

x x

Examination-style questionNow divide x3 – x2 – 5x + 6 by x – 2:

x3 – x2 – 5x + 6x – 2

x3 – 2x2

x2 – 5x

+ x

x2 – 2x

–3x + 6

– 3

–3x + 6

0

x2

So,3 2 25 + 6 ( 2)( + 3)

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

x x x x x x

x x x x

x x

x

2 + 3=

+1as required.

© Boardworks Ltd 2006 1 of 37 A2-Level Maths: Core 3 for Edexcel C3.1 Algebra and functions 1 This...

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