Algebra ILearning OutcomesIn this chapter you will learn to:

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 Add, subtract, multiply and divide algebraic terms

 Factorise quadratic expressions

 Factorise certain cubic expressions

 Divide algebraic expressions using long division

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2 .1 expressIonsAlgebra has many uses, from the design of computer games to the modelling of weather patterns.

To be able to use algebra, we must first understand the rules involved in the basic operations of adding, subtracting, multiplying and dividing algebraic terms and expressions.

notation in Algebra

� In 5x, the variable is x.

� In 20y, the variable is y.

coefficient coefficient � In 5x, the coefficient is 5.

� In 20y, the coefficient is 20.

� In x, the coefficient is 1.

In 10x + 2, the constant is 2.

12(x) is an example of a term.

This is written as 12x.

x + 5y – 7 contains three terms:

� x (a variable) � 5y (a constant times a variable) � –7 (a constant)

.

5x + 2 is an expression. Other examples of expressions include x + 3y, 8y2 and 4pr3 – 7.

A polynomial in x has the form anxn + an – 1x

n – 1 + an – 2xn – 2 + ... +a2x

2 + a1x1 + a0, where all the

coefficients (a0, a1, a2, ..., an – 1, an) are constants and the powers are non-negative whole numbers.

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The degree (or order) of the polynomial is equal to the highest power of x.

� A polynomial of degree 1 is called linear. �A polynomial of degree 3 is called cubic.

� A polynomial of degree 2 is called quadratic. �A polynomial of degree 4 is called quartic.

For example, 5x6 + 2x2 – 3x + 7 is a polynomial in x of degree 6. It has four terms. The coefficient of x is –3 and the constant is 7.

Addition/subtraction When adding or subtracting algebraic terms, we must always remember the following rules:

Examples: 5x + 6y + 4x – 3y = 9x + 3yp + 2q + 3p – 4q = 4p – 2q

Examples: 20y2 + 8y2 = 28y2

xy – 3w2 + 5xy + 2w2 = 6xy – w2

MultiplyingUnlike when adding or subtracting, in algebra any term may be multiplied by any other term. When we multiply terms, we encounter another set of rules that are important to understand.

Example: (4x)(5y)

= (4)(x)(5)(y)

= (4)(5)(x)(y)... (Commutative)

= 20xy

Example: (4x3)(3x5)

= (4)(x)(x)(x)(3)(x)(x)(x)(x)(x)

= (4)(3)(x)(x)(x)(x)(x)(x)(x)(x)... (Commutative)

= 12x8

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pascal’s Triangle(a + b)5 would take a long time to expand and simplify using the methods employed above. However, by looking at the expansion of (a + b)n we can spot a pattern that we may be able to use for other expressions (n ∈ N or n = 0).

(a + b)0 = 1

(a + b)1 = a1 + b1

(a + b)2 = a2 + 2ab + b2

(a + b)3 = a3 + 3a2b + 3ab2 + b3

(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

(a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5

If we examine the powers, we notice that, for n ≥ 1, the power of each term in the expansion add up to the power of the original expression.

(a + b)0= 1

(a + b)1 = a1 + b1

(a + b)2 = a2 + 2a1b1 + b2

(a + b)3 = a3 + 3a2b1 + 3a1b2 + b3

(a + b)4 = a4 + 4a3b1 + 6a2b2 + 4a1b3 + b4

(a + b)5 = a5 + 5a4b1 + 10a3b2 + 10a2b3 + 5a1b4 + b5

Important Products

Worked example 2.1

Expand and simplify:

(i) 4(2a – b)(a – 5b) (ii) 23x – x(2x + 3)2 (iii) (4x – 5)3

Solution (i) 4(2a − b)(a − 5b)

4[(2a(a – 5b) – b(a – 5b)]

4[2a2 – 10ab – ab + 5b2]

4[2a2 – 11ab + 5b2]

8a 2 – 44ab + 20b2

(ii) 23x – x(2x + 3)2

23x – x[(2x)2 + 2(2x)(3) + (3)2]

23x – x[4x2 + 12x + 9]

23x – 4x3 – 12x2 – 9x

–4x3 – 12x2 + 14x

(iii) (4x – 5)3

(4x)3 – 3(4x)2(5) + 3(4x)(5)2 – (5)3

64x3 – 240x2 + 300x – 125

(ii)

(iii)

2.1ACTIVITY ACTIVITY ACTIVITY 2.12.12.12.1

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If we highlight (a + b)5:

(a + b)5 = a5b0 + 5a4b1 + 10a3b2 + 10a2b3 + 5a1b4 + a0b5

we notice that the first term is a5 and then the powers of a decrease by 1 each term and eventually reach zero. Also, the powers of the b start at zero and increase by 1 each term.

Examining the coefficients of each expansion, we notice another pattern.

The next row of coefficients can be calculated by adding the pairs of coefficients from the row above.

This pattern was first described by the French mathematician Blaise Pascal (1623–1662) and is named in his honour.

In general, using Pascal’s triangle in the expansion of (x + y)n:

� There are n + 1 terms.

� The coefficients of the first two terms are 1 and n.

� The sum of the powers of x and y in each term is n.

Worked example 2.2

Use Pascal’s triangle to expand (x – 2)3.

SolutionFrom Pascal’s triangle we can determine the initial coefficients as:

1 3 3 1

We next fill in the powers of each term.

The x term will start with a power of 3 and decrease to zero.

The (–2) term will start with a power of zero and increase to 3.

∴ (x – 2)3 = (x + (–2))3 = 1x3 (–2)0 + 3x2(–2)1+ 3x1 (–2)2 + 1x0(–2)3 = x3 – 6x2+ 12x – 8

1

1 1

1 2 1

1 3 3 1

11 4 6 4

1

1 1

2 11

3 3 1

1

1

1

4 6 41

55 10 101

1

1 1

1 2 1

1 3 3 1

11 4 6 4

1

1 1

2 11

3 3 1

1

1

1

4 6 41

55 10 101

Worked example 2.3

Use Pascal’s triangle to expand (2p + 3)4.

SolutionFrom Pascal’s triangle we can determine the coefficients as:

1 4 6 4 1

The 2p term will start with a power of 4 and decrease to zero.

The 3 term will start with a power of zero and increase to 4.

∴ (2p + 3)4 = 1(2p)4 (3)0 + 4(2p)3 (3)1+ 6(2p)2 (3)2 + 4(2p)1 (3)3 + 1(2p)0(3)4

= 16p4 + 96p3 + 216p2 + 216p + 81

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Iexercise 2.1

1. If p = 3, q = – 4 and r = 7, find the value of:

(i) 2pq (iv) pqr – q2

(ii) (q + r)p (v) √_______________

q2 + rp + r + 4

______________ – q

___ p

(iii) q + p

______ 2r

2. Expand and simplify:

(i) 3x(x + 4) + 5(3x – 2)

(ii) 3(a2 – b) – (a – 3b)

(iii) 12x(3x2 + 2x + 1) + 5x(2x – 4)

(iv) –y(2y – 3x) – y(xy + 3x)

(v) b(b2 + 4b + c) – 4c(a2 + b)

3. Expand each of the following expressions. For each polynomial, state:

(a) The degree

(b) The value of the constant term

(c) The number of terms

(d) The coefficient of the x term

(i) (x + 2)(x – 3)

(ii) (2x – 5)(3x + 4)

(iii) –x(5x2 – 3)(2x2 – 3)

(iv) (x3 – 4)(2x2 – 8x + 3)

(v) (11x4 – 3x3 + x2 – 3x + 1)(4x – 7)

4. Multiply out the following expressions:

(i) (5x + 1)(7x + 1)

(ii) (x + 3)(4x – 3)

(iii) 5(3s – t)(2t – 1)

(iv) (x – y)(2x + 5y)(3x)

(v) (xy – y)(y – 6xy)(y)

5. Expand:

(i) (3p + 4)2 (iv) (a + b)(x + 5)2

(ii) (4y – x)2 (v) (y + x)3(2y – 3x)

(iii) 2(3a – 1)2 (vi) 3p(p + q)(p – q)

6. Expand:

(i) (3p – 4)(b2 + 4b + c)

(ii) (b2 + 4b + c)(4y – x)

(iii) (3a – 1)(a + 3b)(2a + b)

(iv) 5(x + y)(x + 4y)(y + x)

(v) (z + x)2(z – x)2

7. Expand:

(i) (p + q)2 (vi) (y – 3)3

(ii) (p – q)2 (vii) (x + 4)3

(iii) (p – q)(p + q) (viii) (2a + 5b)3

(iv) (x – 3)(x + 3) (ix) (p – 8q)3

(v) 2x(4x + 3)2 (x) (x + y)(x – y)y2

8. Expand:

(i) (2p – q)(2s + 4r) (v) 5(2a + 9b)3

(ii) (a – 3b)(a + 3b) (vi) (x2 – y2)(x – y)

(iii) (4m + 7n)3 (vii) [(a – b)b]3

(iv) (9x – 2y)3

9. Using Pascal’s triangle, expand the following polynomials:

(i) (a + 1)4 (iv) (2a + 3b)3

(ii) (b – 3)3 (v) (3y – 4x)4

(iii) (x + y)5 (vi) (3x – 2y)5

10. In a maths competition, a student gets 8 marks for a correct answer, and 3 marks are deducted if the answer is incorrect. One mark is awarded for any question not attempted. If there are 20 questions on the test, write an expression for a student’s total mark if the student got x questions correct from a total of y questions attempted.

11. Mark is y years old. In five years’ time, Aoife will be twice as old as Mark and Daniel will be two years younger than Aoife is now. Write an expression in y for the sum of their three ages now.

10.

11.

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12. Express as a polynomial the area and perimeter of each of the following shapes:

(i)

x + 2

(ii)

x + y

2x Ð 3y

13. An open-topped tank is in the shape of a cuboid. The length of the tank is (x + 7) cm. Its breadth is 3 cm less than its length and its height is 1 cm more than its breadth. Write an expression in terms of x for the volume and surface area of the tank.

14. Find the value of the coefficient of x3 in the expansion of (x + y)7.

15. Find the value of the third term in the expansion of (4a – 5b)8.

16. The middle term of the expansion of (ax + ay)n is 14,580x3y3. Find the value of a and the value of n, where a, n ∈ N.

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2.2 FAcTorIsIng Another important skill is that of factorising expressions.

� Highest Common Factor

x2 – 3x = x(x – 3) � Difference of Two Squares

4x2 – 25y2

= (2x)2 – (5y)2 (writing each term as a square)

= (2x – 5y)(2x + 5y) � Quadratic Trinomials

2x2 + 13x + 15 = (2x + 3)(x + 5)

FactorisingFactorising

productproductproduct

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Worked example 2.4

Factorise: (i) 49x2 – 225y2 (ii) 3x2 + 11x – 20 (iii) 8x2 – 18x + 9

Solution (i) 49x2 – 225y2

Write each term as a square.

(7x)2 – (15y)2

(7x – 15y)(7x + 15y)

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I (ii) 3x2 + 11x – 20

If the constant term is negative, then one factor of 20 is positive and the other is negative

4Ð

+ 53x

x 5Ð

+ 43x

x 5+

Ð 43x

x

3x × x = 3x2

5 × –4 = –20

3x × x = 3x2

4 × –5 = –20

3x × x = 3x2

–4 × 5 = –20 We now use the arrows:

3x × –4 = –12x

x × 5 = +5x –7x

We now use the arrows:

3x × –5 = –15x

x × 4 = +4x –11x

We must be on the right track, as we have the right number but the wrong sign.

We now use the arrows:

3x × 5 = +15x

x × –4 = – 4x +11x

So (3x – 4)(x + 5) are the correct factors.

(iii) We can also use the Guide Number Method to factorise quadratic trinomials.

8x2 – 18x + 9

Step 1 Multiply the coefficient of x2 by the constant.

8x2 – 18x + 9

8 × 9 = 72

Step 2 Find two factors of 72 that add to give the coefficient of the middle term, i.e. –18.

–12 and –6

Step 3 Use the answers from Step 2 to rewrite 8x2 – 18x + 9 as follows:

8x2 – 12x – 6x + 9

4x(2x – 3) – 3(2x – 3) (Factorise by grouping)

(4x – 3)(2x – 3) (Distributive property)

The factors of 8x2 – 18x + 9 are (4x – 3)(2x – 3).

sum and Difference of Two cubesWe may also be asked to factorise certain cubic expressions. ACTIVITY ACTIVITY ACTIVITY 2.22.22.22.2

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Worked example 2.5

Factorise:

(i) x3 + 125 (ii) 64p3 – 27q3 (iii) 1 + 216w3

Solution (i) x3 + 125

Write each term as a cube.

(x)3 + (5)3

= (x + 5)(x2 – 5x + 25)

(ii) 64p3 – 27q3

Write each term as a cube:

(4p)3 – (3q)3

= (4p – 3q)[(4p)2 + (4p)(3q) + (3q)2]

= (4p – 3q)(16p2 + 12pq + 9q2)

(iii) 1 + 216w2

Write each term as a cube.

(1)3 + (6w)3

= (1 + 6w) [ (1)2 – (1)(6w) + (6w)2 ] = (1 + 6w)(1 – 6w + 36w2)

(iii)

Worked example 2.6

Factorise fully:

(i) 9px2 + 24px – 9p

(ii) 5x3 – 625y3

Solution (i) 9px2 + 24px – 9p

First find the highest common factor:

3p[3x2 + 8x – 3]

Now factorise the quadratic:

3p(3x – 1)(x + 3)

(ii) 5x3 – 625y3

First, find the highest common factor:

5[x3 – 125y3]

Now factorise the expression inside the brackets:

5(x – 5y)(x2 + 5xy + 25y2)

exercise 2.2

Factorise fully:

1. 4ab2 – 12ab3

2. 7x2 + 9x + 2

3. 3y2 – 4y – 7

4. 5x2 + 12x + 4

5. x2 – 3x – 18

6. 3x2 + 10x – 8

7. 2y2 + 11y – 63

8. 7x2 + 2x – 57

9. 25a2 – 1

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5.

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I 10. 2x2 – 9x + 4

11. 5x2 + 52x + 96

12. 64a2 – 81b2

13. 2x2 – 23x – 12

14. 4x2 + 7x + 3

15. 10y2 + 27y + 17

16. 9x2 + 3x – 2

17. 9x2 – 21x + 10

18. 4y2 + 23y + 19

19. 6x2 + 37x + 45

20. 36p2 – 100q2

21. 12x2 + 11x – 56

22. 8x2 – 22x + 15

23. 10x2 – 4x – 6

24. 12x2 – 18x –12

25. 48y3 + 48y2 + 12y

26. 16x2 –100

27. 2x3 – 8x

28. 6pq2 + 11pq +4p

29. x4 – 36x2

Factorise fully the following expressions:

Factorise fully the following expressions:

30. x3 – 27

31. p3 + 8

32. x3 – y3

33. 64a3 – 1

34. 8a3 + 27b3

35. 125p3 + 512q3

36. 1,000x3 – 729

37. 343c3 + d3

38. 3x3 – 648

39. 128 + 16x3

40. 54a4 + 432ab3

41. (x + 2)2 – (x + 3)2

42. x2 + 2px + p2

43. a2c2 – b2

44. x4 – 25

45. a2b2 – 2ab + 1

46. x4 – y4

47. (x – y)2 – 9

48. 8x2 – 18xy + 9y2

49. a4 + a

50. a2 – (b + c)2

51. ab5 – ab2

52. x4 – 7x2 – 18

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35.

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2.3 AlgebrAIc FrAcTIons I: ADDITIon AnD subTrAcTIon

When adding or subtracting two algebraic fractions, we first find the lowest common denominator (LCD).

Worked example 2.7

Write as a single fraction: 2x + 5 ______ 2 – x – 1 _____ 9 – 2

Solution 2x + 5 ______ 2 – x – 1 _____ 9 – 2 __ 1 =

9(2x + 5) – 2(x – 1) – 18(2) _______________________ 18

= 18x + 45 – 2x + 2 – 36 _____________________ 18 = 16x + 11 _________ 18

ACTIVITY ACTIVITY ACTIVITY 2.32.32.32.3ACTIVITY ACTIVITY ACTIVITY

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Worked example 2.8

(i) Write as a single fraction:

5 ______ 3x – 5 – 2 ______ 4x – 1

(ii) Write 1 _____ x – 2 – 1 _____ x + 2 as a single fraction in the form b _____ xa – b

, where a, b ∈ N.

Solution

(i) 5 ______ 3x – 5 – 2 ______ 4x – 1 = 5(4x – 1) – 2(3x – 5)

_________________ (3x – 5)(4x – 1)

= 20x – 5 – 6x + 10 ________________ (3x – 5)(4x – 1)

= 14x + 5 _____________ (3x – 5)(4x – 1)

(ii) 1 _____ x – 2 – 1 _____ x + 2 = 1(x + 2) – 1(x – 2)

________________ (x – 2)(x + 2)

= x + 2 – x + 2 _____________ x2 – 4

= 4 _____ x2 – 4

Simplify 5x – 10 _______ x2 – 4

.

SolutionWe factorise both numerator and denominator.

5x – 10 = 5(x – 2)

x2 – 4 = (x – 2)(x + 2)

∴ 5x – 10 _______ x2 – 4

= 5(x – 2) ____________ (x – 2)(x + 2) 1

= 5 _____ x + 2

Worked example 2.9 Worked example 2.10

Simplify 3a2 – 6a ________ 2 – a .

Solution 3a2 – 6a _______ 2 – a =

3a(a – 2) _________ –1 (a – 2)

1

1

= 3a ___ –1

= –3a

Worked example 2.11

(i) Express as a single fraction in its simplest form:

2 _____ x – 1 – 3x – 1 __________ x2 – 4x + 3

+ 4 _____ x – 3

(ii) Express as a single fraction in its simplest form:

3 _____ x – 2 + 2 _____ 2 – x

1

2

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ISolution (i) 2 _____ x – 1 – 3x – 1 __________

x2 – 4x + 3 + 4 _____ x – 3 = 2 _____ x – 1 – 3x – 1 ___________ (x – 1)(x – 3) + 4 _____ x – 3

= (2)(x – 3) – (3x – 1)(1) + 4(x – 1)

___________________________ (x – 1)(x – 3)

= 2x – 6 – 3x + 1 + 4x – 4 _____________________ (x – 1)(x – 3)

= 3x – 9 ___________ (x – 1)(x – 3)

= 3(x – 3) ___________ (x – 1)(x – 3) 1 1

= 3 _____ x – 1

exercise 2.3

Express as single fractions in their simplest form:

1. 4x – 1 ______ 4 + 2x – 5 ______ 2

2. 3x – 7 ______ 12 – 5x – 3 ______ 4

3. x – 1 _____ 5 + 2x – 4 ______ 3

4. 7 ______ 4x – 5 + 2

5. 1 ___ 5x – 2 ___ 7x

6. 8 _____ x + 1 + x

7. 3 _____ x + 5 – 1 __ x

8. 5 _____ x – 2 + 2 ______ 3x – 1

9. 1 ______ 3x + 5 + 3 ______ 2x – 1

10. x + y

_______ 3x + 3y

11. ab3 ___

a2b

12. x + y

_______ x2 – y2

13. ab + b2 _______

a2 – b2

14. p – q

_____ q – p

15. x2 – 2x ______ x – 2

16. 15x2 + 5x ______________ 15x2 + 20x + 5

17. x3 – 8 ___________ x2 + 2x + 4

18. 2x4 – 250x __________ 4x2 – 20x

19. x3 + y3

_______ x2 – y2

20. x4 – y4

______ x2 – y2

21. (p – q)3

_______ p2 – q2

22. x2 – 8x + 16 ___________

x2 – 16 – 16x ______

16 – x2

23. 2x _____ x – 1 – x _____ 1 – x

24. x + 4 ______ x2 –16

+ x – 5 ______ x2 – 25

25. 1 _____ x + 2

– 3 _____ x – 3

+ 4 ________ x2 – x – 6

26. 1 ___________ x2 + 3x + 2

+ 4 _____ x + 2

– 3 _____ x + 1

27. 3 _____ a – 1

– a + 1 ______ a2 – 1

+ a – 1 _____ a + 1

28. a _____ a + b

+ a _____ a – b

29. a + 2 _____ a – 2 + a – 3 _____ a + 3

30. p + 1

______ p – 1 + p + 2

______ p – 2

31. 1 ______ n + 1 + 1 __ n + 1 ______ n + 2

32. a + b ______ a2 – b2 – a – b ______

b2 – a2

33. a – b ____________ a2 – 2ab + b2 – a + b _____________

a2 + 2ab + b2

34. Write as a single fraction in the form a _____ xb – c

, where a, b and c ∈ N:

(i) 5 _____ x2 – 1

+ 1 _____ 1 – x2

(ii) 3x ___________ x2 + 3x – 18

– 18 ___________ 18 – 3x – x2

(ii) 3 _____ x – 2 + 2 _____ 2 – x

= 3 _____ x – 2 + 2 _____ 2 – x ⋅ –1 ___ –1

= 3 _____ x – 2 – 2 _____ x – 2

= 3 – 2 _____ x – 2

= 1 _____ x – 2

25. 1 3 + 4 31. 1 + 1 + 131.

32.

33.

34.

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2 .4 AlgebrAIc FrActIons II: MultIplIcAtIon And dIvIsIon

Worked example 2.12

Simplify 3x2 – 27 ____________ 4x2 – 15x + 9

× 4x – 3 _______ x2 + 3x

.

Solution 3x2 – 27 _____________ 4x2 – 15x + 9 × 4x – 3 _______

x2 + 3x

= 3(x2 – 9) ____________ (4x – 3)(x – 3) ⋅

(4x – 3) ________

x (x + 3)

= 3(x +3) (x – 3) (4x – 3)

____________________ (4x – 3)(x – 3)(x)(x + 3)

= 3 __ x

1 1 1

1 1 1

Worked example 2.13

Simplify x – 5 _____ x + 1 ÷ x2 – 25 ___________ x2 + 4x + 3

.

Solution

x – 5 _____ x + 1 ÷ x2 – 25 ___________ x2 + 4x + 3

= x – 5 _____ x + 1 × x2 + 4x + 3 ___________

x2 – 25

= (x – 5)

______ (x + 1) ⋅ (x + 3)(x + 1)

____________ (x + 5)(x – 5)

= x + 3 _____ x + 5

1 1

1 1

1

1

Worked example 2.14

Simplify 1 – 9 __

x2 ______

2 + 6 __ x .

SolutionNumerator: 1 – 9 __

x2 = x2 – 9 _____ x2

= (x – 3)(x + 3)

____________ x2

Denominator: 2 + 6 __ x = 2x + 6 ______ x

= 2(x + 3)

_______ x

∴ 1 – 9 __

x2 ______

2 + 6 __ x =

(x – 3)(x + 3) ____________

x2 ÷ 2(x + 3)

_______ x

= (x – 3)(x + 3)

____________ x2 × x _______ 2(x + 3)

= x – 3 _____ 2x

Alternative Method

Multiply the numerator and the denominator by x2.

= 1 – 9 __

x2 ______

2 + 6 __ x · x

2 __

x2

= x2 – 9 ________ 2x2 + 6x

= (x + 3) (x – 3)

____________ 2x (x + 3)

= x – 3 _____ 2x

Alternative Method

Mul

=

1 1

1

1

1

complex FractionsSometimes a fraction’s numerator and/or denominator may be a fraction as well. These types of fractions are called complex fractions.

For example: 1 + 1 _ 3

_____ 2 _ 5 + 3 _ 7

2.4ACTIVITY ACTIVITY ACTIVITY 2.42.42.42.4

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Simplify:

6x3 + 5x2 – 10x + 3 __________________ 2x –1

Solution 3x2 + 4x – 3

2x – 1 6x3 + 5x2 – 10x + 3 – (6x3 – 3x2) We multiply 2x –1 by 3x2 to get 6x3 –3x2 and subtract to get 8x2.

8x2 – 10x Bring down the next term. We divide 8x2 by 2x to get 4x.

– (8x2 – 4x) We multiply 2x –1 by 4x to get 8x2 –4x and subtract to get –6x.

–6x + 3 Bring down the next term. We divide –6x by 2x to get –3.

– (–6x + 3) We multiply 2x –1 by –3 to get –6x + 3 and subtract.

0 We get a remainder of 0.

∴ 6x3 + 5x2 – 10x + 3 __________________ 2x – 1 = 3x2 + 4x – 3

2.5 long DIvIsIon In AlgebrAAnother approach to dividing one algebraic expression by another is to use long division.

Arrange the divisor and dividend in descending powers of x.We divide 6x3 by 2x to get 3x2.

Worked example 2.15

exercise 2.4

1. 10b2 _____

5a2 × 25a3 _____

2b

2. 4x ___ 2y ÷ 8x3 ___

4y2

3. 2 ______ 2x – 1 ÷ 4 ______ 4x2 – 1

4. x2 + x – 2 __________

x2 + 2x – 3 × 2x + 6 ______ 4x – 4

5. 4x – 4 ______ x ÷ x2 – 1 _____ x3

6. y2 – 64

_______ y2 – 16

× 2y2 – 8y

_______ 2y – 16

7. 2x2 – x – 1 __________ 2x2 + x – 1

∙ 4x2 – 1 ______ x2 – 1

8. 8x2 – 34x – 9 ____________ 4x + 1 × 3x ________ 4x2 – 81

5.

6.

7.

8.

For Questions 1 to 19, express as single fractions in their simplest form.

9. 6x2 – 20x + 16 _____________ 4x2 – 16x + 16

÷ 9x2 – 16 ____________ 2x2 + 2x – 12

10. 4x + 3 _____ x2 – 49

________ 16x2 – 9 ______ x – 7

11. x – y

____ xy _____

5 __ xy

12. x – 5 ____ x + 5

_______ 1 _____ x2 – 25

13. x ___ x+1 +1

_______ 3 ___ x+1

14. 1 – 1 __ x ______ 2 – 2 __

x2

15. x __________ x2 – 6x + 9

__________ 5 ___________ (3 – x)(x – 3)

16. a – 1 _ b

_____ 1 – 1 _ b

17. p + q

_____ 1 ______

1 __ p + 1 _ q

18. x + 3 ____ y – x

______ x

2 + 3x _____ x2 – y2

19. x2 – y2

______ x2 _____________

x2 + 2xy + y2

____________ x

20. If u = x – 1 __ x and v = x2 – 1 __ x2 ,

show that u2(u2 + 4) = v2.

21. If x = a3 – b3 ________

a3 + ab2 and y = a3 – ab2

________ a2b + b3 ,

show that x __ y = b(a2 + ab + b2)

______________ a2(a + b)

.

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Worked example 2.16

Divide x3 – 4x2 + 3 by x – 1.

SolutionAs the x term is missing, we write 0x into the expression.

x2 – 3x – 3

x – 1 x3 – 4x2 + 0x + 3

– (x3 – x2)

–3x2 + 0x

– (–3x2 + 3x)

–3x + 3

– (–3x + 3)

0

∴ (x3 – 4x2 + 3) ÷ (x – 1) = x2 – 3x – 3

Long division is also useful as it can be used to find the other factors of an expression.

Consider the following question: ‘Find the prime factors of 30.’

Starting with the smallest prime factor of 30, which is 2, we have:

30 ÷ 2 = 15

Continuing:

15 ÷ 3 = 5

So the prime factors of 15 are 3 and 5.

∴ 30 = 2(3)(5). Its prime factors are 2, 3 and 5.

Long division in algebra is very similar to finding the prime factors of a positive whole number.

ACTIVITY ACTIVITY ACTIVITY 2.52.52.52.5

Worked example 2.17

Show that x – 4 is a factor of 12x3 – 55x2 + 29x – 4 and find the other two factors.

Solution 12x2 – 7x + 1

x – 4

12x3 – 55x2 + 29x – 4

– (12x3 – 48x2)

– 7x2 + 29x

– (–7x2 + 28x)

x – 4

– (x – 4)

0

There is no remainder, so (x – 4) is a factor.

∴ 12x3 – 55x2 + 29x – 4 = (x – 4) (12x2 – 7x + 1)

= (x – 4) (4x – 1)(3x – 1)

∴ The other two factors are (4x – 1) and (3x – 1).

There is no remainder, so (

∴

= (

∴

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Iexercise 2.5

Simplify each of the following:

1. (x3 + 2x2 – 7x – 2) ÷ (x – 2)

2. (3x3 + 13x2 – 18x – 40) ÷ (3x + 4)

3. (6x3 – 29x2 + 21x – 4) ÷ (3x – 1)

4. (14x3 + 33x2 – 5x) ÷ (7x – 1)

5. (36x3 – 28x + 8) ÷ (3x – 2)

6. (15x4 – 11x3 – 77x2 + 31x + 42) ÷ (5x + 3)

7. Divide x3 – 13x – 12 by x + 1. Hence, find the other factors.

8. Find the quotient when: 4x4 – 8x3 – 13x2 + 2x + 3 is divided by 2x – 1.

9. Divide x5 – x4 – 6x3 – 8x2 + 8x + 48 by x2 – x – 6. Hence, fully factorise x5 – x4 – 6x3 – 8x2 + 8x + 48.

10. Show that x – 1 is a factor of 2x4 – 14x3 + 22x2 – 10x.

11. Find all four factors of 36x4 – 289x2 + 400, if 4x2 – 25 divides evenly into the quartic expression.

12. Show that 2x + 1 is not a factor of 6x3 – 13x2 – 19x + 12. Give a reason for your answer.

13. Find all four factors of 72x4 + 18x3 – 29x2 – 3x + 2, if 6x2 + x – 2 divides evenly into the quartic expression.

14. Investigate if g(x) = 3x + 4 and h(x) = 5x + 4 are factors of the function f(x) = –6x4 – 29x3 – 16x2 + 16x.

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11.

12.

13.

14.

revision exercises

1. (a) Factorise the following:

(i) x2 – x – 90

(ii) 4x2 + 4x + 1

(iii) 10x2 – x – 2

(iv) 9x2 – 12xy + 4y2

(v) 14x2 – 15x + 4

(b) Simplify:

(i) 3a – 9b ________ 6a – 18b

(ii) 8x – 10 _________ 16x2 – 25

(iii) a3 – b3

_______ 6a – 6b

(iv) by + b – y – 1

_____________ b3 – 1

(v) x2 – x __________ x2 – 4x + 3

(c) Expand:

(i) (3x + 7)2

(ii) (2x – 1)3

(iii) (4x – 5)3

(iv) (4p + 3)3(p – 2)

(v) 2a(3a + 4)2(2a + 5)3

(d) Simplify ( 1 __ x – 1 _____

x + h ) ÷ h.

2. (a) Factorise the following:

(i) x2 – 100

(ii) 4x2 – 81

(iii) 25x2 – 49y2

(iv) 121a2 – 144b2

(v) (5x + 6y)2 – (x + y)2

(b) Write as a single fraction in its simplest form:

(i) 3 _____ x – 3 + 5 _____ x + 4

(ii) 7 ______ 2y + 1 – 6 ______ 2y – 1

(iii) 2 _____ x + 1 – x _____ x – 1

(c) Using Pascal’s triangle, expand the following polynomials:

(i) (2a + 1)5

(ii) (4b – 7c)4

(iii) (6x – 5)6

(d) Show that 2 _____ x – 4 + 2(9 – 2x)

________ 4 – x

simplifies to a constant for

all x ∈ R, x ≠ 4.

3. (a) Factorise:

(i) x3 + 27

(ii) a3 + 8

(iii) b3 + 1000

(iv) 8x3 + 125

(v) 125x3 + 27y3

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(b) Simplify the following:

(i) 2x2 + 9x + 4 _____________ 2x2 + 11x + 5

(iv) 2x – 2y

_______ 3y – 3x

(ii) a2 – b2

______ a3 – b3 (v) x – 3 _____

9 – x2

(iii) a2 – b2

______ a4 – b4

(c) Write as single fractions in their lowest terms:

(i) 5 _____ x – 2 + 1 _____ 2 – x (iii) 2 _____ b – a

+ 2 _____ a – b

(ii) 7 ______ 2y – 1 + 5 ______ 1 – 2y (iv) 9 ______ 2x – 1 – 4 ______ 1 – 2x

4. (a) Factorise:

(i) y3 – 1 (iv) x3 – 216

(ii) 8y3 – 1 (v) 1000a3 – 343b3

(iii) 27a3 – 8b3

(b) Express as single fractions in their lowest terms:

(i) 6x _____ x2 – 9

– 1 _____ x + 3 (ii) 4 ______ 3y – 2 – 2y _______

9y2 – 4

(c) The sides of a triangle are (m2 + n2), (m2 – n2) and 2mn in length.

(i) Prove that the triangle is right-angled.

2mn

m2 Ð n2

m2 + n2

(ii) Deduce the three lengths of the sides if m = 5 and n = 2.

5. (a) Factorise fully:

(i) 3x2 – 75 (iv) x4 – 81

(ii) 9x3 – 25x (v) ax2 – bx2 – ay2 + by2

(iii) x4 – y4

(b) Simplify:

(i) ax + ay – cx – cy

________________ ax + ay + cx + cy

(ii) 9y3 – y ___________

3y2 + 8y – 3

(iii) (a + b)2 – c2

___________ a2 – (b + c)2

(c) Write 2x _____ x + 3 + 3x _____ x – 3 – 5x2 + 9 _______ x2 – 9

as a fraction in the form k _____ x + t

where k ∈ Z.

6. (a) Simplify each of the following:

(i) (4x3 – 5x2 – 2x + 3) ÷ (x – 1)

(ii) (36x3 – 18x2 – 10x + 4) ÷ (3x – 1)

(iii) (24x4 – 22x3 – 7x2 + 4x + 1) ÷ (3x + 1)

(iv) (2x4 + 13x3 + 19x2 – 10x – 24)÷ (2x + 3)

(iv) (36x5 + 117x4 + 95x3 + 5x2 – 11x – 2) ÷ (3x – 1)

(b) (i) Show that 8x + 1 is a factor of 48x3 + 62x2 – 33x – 5 and find the other factors.

(ii) Show that x – 2 is not a factor of 8x3 + 22x2 – 7x – 3.

(c) Write down the dividend, divisor and quotient when 6x5 – 40x4 – 26x3 + 356x2 – 312x – 144 is divided by 3x2 + 10x + 3.

7. (a) Rearrange the order of the terms in these expressions and then factorise them:

(i) 51 + x2 + 20x

(ii) –x2 + 169

(iii) ax + by + ay + bx

(iv) a2 + b2 – 2ab

(v) a2 + b2 – c2 – 2ab

(b) (i) Write 1 + x __ y as a single fraction.

(ii) Hence, simplify ( 1 + x __ y ) ( y2

______ y2 – x2 ) .

(c) (i) Show that x – 1 is not a factor of 4x4 – 20x3 – 7x2 + 32x + 15. Give a reason for your answer.

(ii) Show that 4x2 – 1 is a factor of 12x4 + 4x3 – 59x2 – x + 14 and find the other factors.

(iii) Show that x3 – x2 is a factor of 4x5 – 7x3 + 3x2 and find the other factors.

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I 8. (a) Factorise fully the following expressions:

(i) x3 + y3 + 3x + 3y

(ii) x2 – y2 + 5x + 5y

(iii) x2 – 2xy + y2 – 4z2

(iv) x3 – y3 + x2 – y2

(v) a2 – (b + c)2

(b) Simplify the following:

(i) x2 – 2xy + y2 – z2

_______________ x2 – y2 – 2yz – z2 (iii) 9 ________

4x2 – 16 × 2x – 4 ______ 27

(ii) x2 – 25 ______

x2 – 64 × x

2 – 8x ______ x – 5

(c) Car A leaves a town and drives north at a speed of 40 km/hr. Car B leaves 3 hours later and travels north along the same route at a speed of 60 km/hr. After a certain distance, Car B catches up with Car A.

Let x be the time in hours that Car A travels before Car B catches up with it. Express for each car, in terms of x, the distance travelled.

(d) Write 8 _______ 6x – 15 – 5 _______ 4x – 10

in the form a ________ b(2x – 5)

where a, b ∈ Z.

9. (a) Factorise the following:

(i) x2 ____ 100 –

y2

___ 49 (iv) a2b2 – 64

(ii) x3 + 1 __ x3 (v) x2 – y2 + 2x + 1

(iii) x4 – 16

(b) Simplify:

(i) 15x2 ____

8y2 ÷ 5x3 ___ 2 (ii)

1 __ 9 (2x2 – 50) ____________

1 ___ 45 (4x2 – 20x)

(c) The area of a rectangle can be expressed as 2x2 + 3x – 20. The length of the rectangle is x + 4.

(i) Find the breadth of the rectangle in terms of x.

This rectangle is used as a base for a rectangular box. The volume of this box can be expressed as 6x3 + 7x2 – 63x + 20.

(ii) Find the height of this rectangular box in terms of x.

(d) (i) Write as single fractions:

y – 1 __ y and 2 + 2 _____ y – 1

(ii) Simplify ( y – 1 __ y ) ( 2 + 2 _____ y – 1 ) . 10. (a) (i) Factorise x3 + y3.

(ii) Hence, write x3 + y3 + 3xy (x + y) as a perfect cube (i.e. as (expression)3).

(iii) Hence, factorise x3 + y3 + z3 + 3xy(x + y).

(b) If x = 5b2 ______

a2 – ab and y = 10ab ______

b2 – a2 ,

show that x __ y = –b(b + a)

_________ 2a2 .

(c) Simplify z _____ z – 1 + z _____ z + 1

______________ z _____ z – 1 – z _____ z + 1

.

11. (a) (i) Show that 3t – 10 _______ 3t – 2 – 8 ______ 2 – 3t

simplifies to a constant, t ≠ 2 __ 3 .

(ii) Simplify 8x3 + 27 ________ 4x2 – 9

.

(iii) Simplify x2 – x – 6 ________ x2 – 4

.

(b) Factorise:

(i) a2 + 2ab + b2

(ii) a2 + b2 – c2 + 2ab

(c) Show that

x2 ___________ (x – y) (x – z) +

y2

___________ (y – z) (y – x) + yz ___________ (z – x) (z – y)

simplifies to x _____ (x – z) .

12. (a) Write 1 _____ x + 2 – 1 ______ 2x – 1 as a single fraction and hence solve:

1 _____ x + 2 – 1 ______ 2x – 1 = 1 ___________ 2x2 + 3x – 2

(b) Express in the form k _____ x + 3 :

4 _____ x – 2 + 5 _____ x + 3 – 20 _________ x2 + x – 6

(c) Factorise 8x3 – 27y3 + 4x2 – 9y2.

10.

11.

12.

ACTIVE MATHS – ACTIVITIES 5

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Algebra I 222Activity 2.1

(a) (x + y)0 = 1 (x + y)1 = x + y (x + y)2 = x2 + 2xy + y2

(x + y)3 = x3 + 3x2y + 3xy2 + y3

(x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4

(i) Look at the powers of x in each expansion. What pattern can be observed?

(ii) Look at the powers of y in each expansion. What pattern can be observed?

(iii) Calculate the sum of the powers of x and y in each term. What do you notice?

(iv) Count the number of terms for each expansion. What do you notice?

(b) Using the triangular grid below, fill in the values of the coefficients of the terms for each expansion, up to (x + y)7.

1

1 1

1 12

The first three have been done for you.

(x + y)0 = 1

(x + y)1 = 1x + 1y

(x + y)2 = 1x2 + 2xy + 1y2

ACTIVE MATHS – ACTIVITIES

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(c) Use Pascal’s triangle to expand:

(i) (x + y)7 =

(ii) (2x – y)5 =

Activity 2.2

(a) Simplify (a + b)(a2 − ab + b2) =

Simplify (a − b)(a2 + ab + b2) =

What do you notice?

(b) Below are the factors of various expressions. Expand and simplify for each part:

(i) (x – 3y)(x2 + 3xy + 9y2) =

(ii) (y + 2x)(y2 – 2xy + 4x2) =

(iii) (3a – 5b)(9a2 + 15ab + 25b2) =

(iv) (x – 3y)(x2 + 3xy + 9y2) =

(v) (p – q)(p4 + p3q + p2q2 + pq3 + q4) =

Activity 2.3

Express as a single fraction in its simplest form:

(i) 2 _____ x – 4 – 7 ______ 2x – 3

LCD =

(2) – (7)

__________________ (x – 4)(2x – 3)

Finish from here:

(ii) 3 _____ x + 2 + 4 _____ x – 4 – 1 _________ x2 – 2x – 8

3 _____ x + 2 + 4 _____ x – 4 – 1 ______________

( ) ( )

(3) + (4) – (1)

____________________________

( ) ( )

Finish from here:

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Activity 2.4

(i) Simplify 4 __ 5 – 3 __ 4

_____ 2 __ 3 – 1 __ 2

.

Method 1

Numerator: 4 __ 5 – 3 __ 4 = (4) – (3)

__________________ 20 =

Denominator: 2 __ 3 – 1 __ 2 = (2) – (1)

__________________ =

4 __ 5 – 3 __ 4

_____ 2 __ 3 – 1 __ 2

= ______ 20

______

______

= ______ 20 × ______ = ______ =

Method 2

LCD of 5, 4, 3 and 2 =

Multiply above and below by the LCD.

( 4 __ 5 ) – ( 3 __ 4 )

___________________ ( 2 __ 3 ) – ( 1 __ 2 )

= –

_____________

–

= ______

(ii) Simplify 3 – 2 __ x ______ 9 – 4 __

x2 .

Method 1

Numerator: 3 – 2 __ x = Denominator: 9 – 4 __ x2 =

3 – 2 __ x ______ 9 – 4 __

x2 =

Method 2

LCD of x and x2 =

(3) – ( 2 __ x )

__________________ (9) – ( 4 __

x2 ) =

– _____________

– =

ACTIVE MATHS – ACTIVITIES

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Activity 2.5

(a) (i) Factorise 8x2 + 10x – 3.

Factors: ( )( )

(ii) Divide 8x2 + 10x – 3 by either factor. Use long division.

(iii) What do you notice?

(b) (i) Expand (x + 3)(x + 2)(2x – 4) =

(ii) Divide the resultant polynomial by any of the three factors. Use long division.

(iii) What do you notice?

(c) Given that f(x) = x2 + x – 20 divides into g(x) = 2x4 + 5x3 – 39x2 – 62x + 40, factorise g(x) fully.