Algebraic Chapter techniques - Teacher Superstore€¦ · Chapter Expanding binomial products...

Chapter

Expanding binomial productsPerfect squares and difference of perfect squaresFactorising algebraic expressionsFactorising the difference of two squaresFactorisation by groupingFactorising quadratic trinomials (Extending)Factorising trinomials of the form ax2 + bx + c (Extending)Simplifying algebraic fractions: multiplication and divisionSimplifying algebraic fractions: addition and subtractionFurther simpli� cation of algebraic fractions (Extending)Equations with algebraic fractions (Extending)

8A8B

8C8D8E8F

8G

8H

8I

8J

8K

N U M B E R A N D A L G E B R A

Patterns and algebraApply the distributive law to the expansion of algebraic expressions, including binomials, and collect like terms where appropriate

16x16 32x32

What you will learn

Australian curriculum

8 Algebraic techniques

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The distance (x units) of an object from the top of a building after it has been dropped (where air resistance is negligible) can be found using the formula x = ut + 1 2 at 2 where u is the initial velocity of the object, t the time since the object has been dropped and a the acceleration due to gravity, which is approximately equal to –9.8 m/s2. When an object is dropped it has an initial velocity of 0 m/s, so the distance the object has fallen becomes x = –4.9t 2. Using algebra, the distance from the building after t seconds can be found or the time taken to reach ground level could be calculated. If

_

the object is instead dropped from a hot air balloon ascending at 10 m/s, the object � rst travels in an upward direction. Its distance (x metres) above or below the height of the balloon from when the object is dropped can be found using x = 10t – 4.9t 2. Knowing the time taken for the object to reach the ground, we could again use algebra to � nd factors, such as the height of the balloon, the greatest height reached by the object and the time taken for the object to return to the height from which it was released.

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Free falling

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488 488 Chapter 8 Algebraic techniques

8A Expanding binomial productsA binomial is an expression with

two terms such as x + 5 or x2 + 3.

You will recall from Chapter 2 that

we looked at the product of a single

term with a binomial expression, e.g.

2(x – 3) or x(3x – 1). The product of

two binomial expressions can also be

expanded using the distributive law.

This involves multiplying every term

in one expression by every term in the

other expression.

Expanding the product of two binomial expressions can be applied toproblems involving the expansion of rectangular areas such as a farmer’spaddock.

Let’s start: Rectangular expansions

If (x + 1) and (x + 2) are the side lengths of a rectangle as shown, the total area can be found as an

expression in two different ways.

• Write an expression for the total area of the rectangle using length = (x + 2) and

width = (x + 1).

• Now find the area of each of the four parts of the rectangle and combine to give

an expression for the total area.

• Compare your two expressions above and complete this equation:

(x + 2)( ) = x2 + + .

• Can you explain a method for expanding the left-hand side to give the right-hand

side?

x

x

1

2

Keyideas

Expanding binomial products uses the distributive law.

(a + b)(c + d) = a(c + d) + b(c + d)

= ac + ad + bc + bd

Diagrammatically (a + b)(c + d) = ac + ad + bc + bd

a

ac bc

ad bdd

c

b

For example: (x + 1)(x + 5) = x2 + 5x + x + 5

= x2 + 6x + 5


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Number and Algebra 489 489

Example 1 Expanding binomial products

Expand the following.

(x + 3)(x + 5)a (x – 4)(x + 7)b

(2x – 1)(x – 6)c (5x – 2)(3x + 7)d

SOLUTION EXPLANATION

a (x + 3)(x + 5) = x2 + 5x + 3x + 15

= x2 + 8x + 15

Use the distributive law to expand the brackets

and then collect the like terms 5x and 3x.

b (x − 4)(x + 7) = x2 + 7x – 4x – 28

= x2 + 3x – 28

After expanding to get the four terms, collect

the like terms 7x and – 4x.

c (2x − 1)(x − 6) = 2x2 – 12x – x + 6

= 2x2 – 13x + 6

Remember 2x× x = 2x2 and –1 × (–6) = 6.

d (5x − 2)(3x + 7) = 15x2 + 35x – 6x – 14

= 15x2 + 29x – 14

Recall 5x× 3x = 5 × 3 × x× x = 15x2.

Exercise 8A

1 The given diagram shows the area (x + 2)(x + 3).

a Write down an expression for the area of each of the four regions inside the

rectangle.

b Copy and complete:

(x + 2)(x + 3) = + 3x + + 6

= + 5x +

x

x

2

3

2 The given diagram shows the area (2x + 3)(x + 1).

a Write down an expression for the area of each of the four regions

inside the rectangle.

b Copy and complete:

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

= + +

2x

x

1

3

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—1–3 3(½)


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8A3 Copy and complete these expansions.

(x + 1)(x + 5) = + 5x + + 5

= + 6x +

a (x – 3)(x + 2) = + – 3x –

= – x –

b

(3x – 2)(7x + 2) = + 6x – –

= – +

c (4x – 1)(3x – 4) = – – 3x +

= – 19x +

d UNDE

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4Example 1a Expand the following.

(x + 2)(x + 5)a (b + 3)(b + 4)b (t + 8)(t + 7)c(p + 6)(p + 6)d (x + 9)(x + 6)e (d + 15)(d + 4)f(a + 1)(a + 7)g (y + 10)(y + 2)h (m + 4)(m + 12)i

5Example 1b,c,d Expand the following.

(x + 3)(x – 4)a (x + 5)(x – 2)b (x + 4)(x – 8)c(x – 6)(x + 2)d (x – 1)(x + 10)e (x – 7)(x + 9)f(x – 2)(x + 7)g (x – 1)(x – 2)h (x – 4)(x – 5)i(4x + 3)(2x + 5)j (3x + 2)(2x + 1)k (3x + 1)(5x + 4)l(2x – 3)(3x + 5)m (8x – 3)(3x + 4)n (3x – 2)(2x + 1)o(5x + 2)(2x – 7)p (2x + 3)(3x – 2)q (4x + 1)(4x – 5)r(3x – 2)(6x – 5)s (5x – 2)(3x – 1)t (7x – 3)(3x – 4)u

6 Expand these binomial products.

(a + b)(a + c)a (a – b)(a + c)b (b – a)(a + c)c(x – y)(y – z)d (y – x)(z – y)e (1 – x)(1 + y)f(2x + y)(x – 2y)g (2a + b)(a – b)h (3x – y)(2x + y)i(2a – b)(3a + 2)j (4x – 3y)(3x – 4y)k (xy – yz)(z + 3x)l

FLUE

NCY

4–6(½)4–5(½) 4–6(½)

7 A room in a house with dimensions

4 m by 5 m is to be extended. Both

the length and width are to be

increased by x m.

a Find an expanded expression for

the area of the new room.

b If x = 3:

find the area of the new roomiby how much has the area

increased?

ii

PROB

LEM-SOLVING

8, 97 7, 8


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8A8 A picture frame 5 cm wide has a length which is twice the

width x cm.

a Find an expression for the total area of the frame and

picture.

b Find an expression in expanded form for the area of the

picture only.

2x cm5 cm

x cmPicture

9 The outside edge of a path around a rectangular swimming pool

is 15 m long and 10 m wide. The path is x metres wide.

a Find an expression for the area of the pool in expanded

form.

b Find the area of the pool if x = 2. 15 m

10 m

x m

Pool

PROB

LEM-SOLVING

10 Write the missing terms in these expansions.

(x + 2)(x + ) = x2 + 5x + 6a (x + )(x + 5) = x2 + 7x + 10b(x + 1)(x + ) = x2 + 7x +c (x + )(x + 9) = x2 + 11x +d(x + 3)(x – ) = x2 + x –e (x – 5)(x + ) = x2 – 2x –f(x + 1)( + 3) = 2x2 + +g ( – 4)(3x – 1) = 9x2 – +h(x + 2)( + ) = 7x2 + + 6i ( – )(2x – 1) = 6x2 – + 4j

11 Consider the binomial product (x + a)(x + b). Find the possible integer values of a and b if:

(x + a)(x + b) = x2 + 5x + 6a (x + a)(x + b) = x2 – 5x + 6b(x + a)(x + b) = x2 + x – 6c (x + a)(x + b) = x2 – x – 6d

REAS

ONING

10(½), 1110(½) 10(½), 11

Trinomial expansions

12 Using the distributive law (a + b)(c + d + e) = ac + ad + ae + bc + bd + be.

Use this knowledge to expand and simplify these products. Note: x× x2 = x3.

(x + 1)(x2 + x + 1)a (x – 2)(x2 – x + 3)b(2x – 1)(2x2 – x + 4)c (x2 – x + 1)(x + 3)d(5x2 – x + 2)(2x – 3)e (2x2 – x + 7)(4x – 7)f(x + a)(x2 – ax + a)g (x – a)(x2 – ax – a2)h(x + a)(x2 – ax + a2)i (x – a)(x2 + ax + a2)j

13 Now try to expand (x + 1)(x + 2)(x + 3).

ENRICH

MEN

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8B Perfect squares and difference of perfect squares

We know that 22 = 4, 152 = 225, x2 and (a + b)2 are all examples of perfect squares. To expand (a + b)2

we multiply (a + b) by (a + b) and use the distributive law:

(a + b)2 = (a + b)(a + b)

= a2 + ab + ba + b2

= a2 + 2ab + b2

A similar result is obtained for the

square of (a – b):

(a – b)2 = (a – b)(a – b)

= a2 – ab – ba + b2

= a2 – 2ab + b2

Another type of expansion involves

the case that deals with the product

of the sum and difference of the

same two terms. The result is the

difference of two perfect squares:

Binomial products can be used to calculate the most efficient way to cut theshapes required for a fabrication out of a metal sheet.

(a + b)(a – b) = a2 – ab + ba – b2

= a2 – b2 (since ab = ba, the two middle terms cancel each other out.)

Let’s start: Seeing the pattern

Using (a + b)(c + d) = ac + ad + bc + bd, expand and simplify the binomial products in the two sets

below.

Set A(x + 1)(x + 1) = x2 + x + x + 1

=

(x + 3)(x + 3) =

=

(x – 5)(x – 5) =

=

Set B(x + 1)(x – 1) = x2 – x + x – 1

=

(x – 3)(x + 3) =

=

(x – 5)(x + 5) =

=

• Describe what patterns you see in both sets of expansions above.

• Generalise your observations by completing the following expansions.

(a + b)(a + b) = a2 + + +

= a2 + +

(a – b)(a – b) =

=

A (a + b)(a – b) = a2 – + –

=

B


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Keyideas

32 = 9, a2, (2y)2, (x – 1)2 and (3 – 2y)2 are all examples of perfect squares.Expanding perfect squares

(a + b)2 = (a + b)(a + b)

= a2 + ab + ba + b2

= a2 + 2ab + b2

•

(a – b)2 = (a – b)(a – b)

= a2 – ab – ba + b2

= a2 – 2ab + b2

•

Difference of perfect squares (DOPS)

• (a + b)(a – b) = a2 – ab + ba – b2

= a2 – b2

• (a – b)(a + b) also expands to a2 – b2

• The result is a difference of two perfect squares.

Example 2 Expanding perfect squares

Expand each of the following.

(x – 2)2a (2x + 3)2b


a (x – 2)2 = (x – 2)(x – 2)

= x2 – 2x – 2x + 4

= x2 – 4x + 4

Alternative solution:

(x – 2)2 = x2 – 2 × x× 2 + 22

= x2 – 4x + 4

Write in expanded form.

Use the distributive law.

Collect like terms.

Expand using (a – b)2 = a2 – 2ab + b2 where

a = x and b = 2.

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

= 4x2 + 6x + 6x + 9

= 4x2 + 12x + 9


(2x + 3)2 = (2x)2 + 2 × 2x× 3 + 32

= 4x2 + 12x + 9

Write in expanded form.

Use the distributive law.

Collect like terms.

Expand using (a + b)2 = a2 + 2ab + b2 where

a = 2x and b = 3. Recall (2x)2 = 2x× 2x = 4x2.


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Example 3 Forming a difference of perfect squares

Expand and simplify the following.

(x + 2)(x – 2)a (3x – 2y)(3x + 2y)b


a (x + 2)(x – 2) = x2 – ��2x + ��2x – 4

= x2 – 4

Expand using the distributive law.

–2x + 2x = 0.


(x + 2)(x – 2) = (x)2 – (2)2

= x2 – 4

(a + b)(a – b) = a2 – b2. Here a = x and b = 2.

b (3x – 2y)(3x + 2y) = 9x2 + ��6xy – ��6xy – 4y2

= 9x2 – 4y2Expand using the distributive law.

6xy – 6xy = 0.


(3x – 2y)(3x + 2y) = (3x)2 – (2y)2

= 9x2 – 4y2(a + b)(a – b) = a2 – b2 with a = 3x and b = 2y

here.

Exercise 8B

1 Complete these expansions.

(x + 3)(x + 3) = x2 + 3x + +

=

a (x + 5)(x + 5) = x2 + 5x + +

=

b

(x – 2)(x – 2) = x2 – 2x – +

=

c (x – 7)(x – 7) = x2 – 7x – +

=

d

2 a Substitute the given value of b into x2 + 2bx + b2 and simplify.

b = 3i b = 11ii b = 15iii

b Substitute the given value of b into x2 – 2bx + b2 and simplify.

b = 2i b = 9ii b = 30iii

3 Complete these expansions.

(x + 4)(x – 4) = x2 – 4x + –

=

a (x – 10)(x + 10) = x2 + 10x – –

=

b

(2x – 1)(2x + 1) = 4x2 + – –

=

c (3x + 4)(3x – 4) = 9x2 – + –

=

d

UNDE

RSTA

NDING

—1–3 1(½), 3(½)


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8B4Example 2a Expand each of the following perfect squares.

(x + 1)2a (x + 3)2b (x + 2)2c (x + 5)2d(x + 4)2e (x + 9)2f (x + 7)2g (x + 10)2h(x – 2)2i (x – 6)2j (x – 1)2k (x – 3)2l(x – 9)2m (x – 7)2n (x – 4)2o (x – 12)2p

5Example 2b Expand each of the following perfect squares.

(2x + 1)2a (2x + 5)2b (3x + 2)2c(3x + 1)2d (5x + 2)2e (4x + 3)2f(7 + 2x)2g (5 + 3x)2h (2x – 3)2i(3x – 1)2j (4x – 5)2k (2x – 9)2l(3x + 5y)2m (2x + 4y)2n (7x + 3y)2o(6x + 5y)2p (4x – 9y)2q (2x – 7y)2r(3x – 10y)2s (4x – 6y)2t (9x – 2y)2u

6 Expand each of the following perfect squares.

(3 – x)2a (5 – x)2b (1 – x)2c(6 – x)2d (11 – x)2e (4 – x)2f(7 – x)2g (12 – x)2h (8 – 2x)2i(2 – 3x)2j (9 – 2x)2k (10 – 4x)2l

7Example 3a Expand and simplify the following to form a difference of perfect squares.

(x + 1)(x – 1)a (x + 3)(x – 3)b (x + 8)(x – 8)c(x + 4)(x – 4)d (x + 12)(x – 12)e (x + 11)(x – 11)f(x – 9)(x + 9)g (x – 5)(x + 5)h (x – 6)(x + 6)i(5 – x)(5 + x)j (2 – x)(2 + x)k (7 – x)(7 + x)l

8Example 3b Expand and simplify the following.

(3x – 2)(3x + 2)a (5x – 4)(5x + 4)b (4x – 3)(4x + 3)c(7x – 3y)(7x + 3y)d (9x – 5y)(9x + 5y)e (11x – y)(11x + y)f(8x + 2y)(8x – 2y)g (10x – 9y)(10x + 9y)h (7x – 5y)(7x + 5y)i(6x – 11y)(6x + 11y)j (8x – 3y)(8x + 3y)k (9x – 4y)(9x + 4y)l

FLUE

NCY

4–8(½)4–7(½) 4–8(½)

9 Lara is x years old and her two best friends are (x – 2) and (x + 2) years old.

a Write an expression for:

the square of Lara’s ageithe product of the ages of Lara’s best friends (in expanded form).ii

b Are the answers from parts a i and ii equal? If not, by how much do they differ?

PROB

LEM-SOLVING

9, 109 9, 10


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8B10 A square piece of tin of side length 20 cm has four

squares of side length x cm removed from each corner.

The sides are folded up to form a tray. The centre

square forms the tray base.

a Write an expression for the side length of the base

of the tray.

b Write an expression for the base of the tray. Expand

your answer.

c Find the area of the tray base if x = 3.

d Find the volume of the tray if x = 3.

Traybase

x cm

x cm

20 cm

20 cm PROB

LEM-SOLVING

11 Four tennis courts are arranged as shown with a square storage area in the centre. Each court area

has the same dimensions a× b.

a Write an expression for the side length of the total area.

b Write an expression for the total area.

c Write an expression for the side length of the inside storage area.

d Write an expression for the area of the inside storage area.

e Subtract your answer to part d from your answer to part b to find the

area of the four courts.

a

b

f Find the area of one court. Does your answer confirm that your answer to part e is correct?

12 A square of side length x units has one side reduced by 1 unit and the other increased by 1 unit.

a Find an expanded expression for the area of the resulting rectangle.

b Is the area of the original square the same as the area of the resulting rectangle?

Explain why/why not?

REAS

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12, 1311 12, 13


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8B13 A square of side length b is removed from a square of side length a.

A B

C b

a

a Using subtraction write down an expression for the remaining area.

b Write expressions for the area of the regions:

Ai Bii Ciii

c Add all the expressions from part b to see if you get your answer from part a.

REAS

ONING

Extended expansions

14 Expand and simplify these expressions.

(x + 2)2 – 4a (2x – 1)2 – 4x2b(x + 3)(x – 3) + 6xc 1 – (x + 1)2dx2 – (x + 1)(x – 1)e (x + 1)2 – (x – 1)2f(3x – 2)(3x + 2) – (3x + 2)2g (5x – 1)2 – (5x + 1)(5x – 1)h(x + y)2 – (x – y)2 + (x + y)(x – y)i (2x – 3)2 + (2x + 3)2j(2 – x)2 – (2 + x)2k (3 – x)2 + (x – 3)2l2(3x – 4)2 – (3x – 4)(3x + 4)m 2(x + y)2 – (x – y)2n

The logical skills of algebra have applications in computer programming.

ENRICH

MEN

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8C Factorising algebraic expressionsThe process of factorisation is a key

step in the simplification of many

algebraic expressions and in the solution

of equations. It is the reverse process

of expansion and involves writing an

expression as a product of its factors.

expanding−−−−−−→

2(x – 3) = 2x – 6←−−−−−−factorising

Factorising is a key mathematical skill required in many diverseoccupations, such as in business, science, technology and engineering.

Let’s start: Which factorised form?

The product x(4x + 8) when expanded gives 4x2 + 8x.

• Write down three other products that when expanded give 4x2 + 8x. (Do not use fractions.)

• Which of your products uses the highest common factor of 4x2 and 8x? What is this highest common

factor?

Keyideas

When factorising expressions with common factors, take out the highest common factor (HCF).

The HCF could be:

• a number

For example: 2x + 10 = 2(x + 5)

• a pronumeral (or variable)

For example: x2 + 5x = x(x + 5)

• the product of numbers and pronumerals

For example: 2x2 + 10x = 2x(x + 5)

A factorised expression can be checked by using expansion.

For example: 2x(x + 5) = 2x2 + 10x.


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Example 4 Finding the HCF

Determine the HCF of the following.

6a and 8aba 3x2 and 6xyb


a 2a HCF of 6 and 8 is 2.

HCF of a and ab is a.

b 3x HCF of 3 and 6 is 3.

HCF of x2 and xy is x.

Example 5 Factorising expressions

Factorise the following.

40 – 16ba –8x2 – 12xb


a 40 – 16b = 8(5 – 2b) The HCF of 40 and 16b is 8. Place 8 in front

of the brackets and divide each term by 8.

b –8x2 – 12x = – 4x(2x + 3) The HCF of the terms is – 4x, including the

common negative. Place the factor in front of

the brackets and divide each term by – 4x.

Example 6 Taking out a binomial factor

Factorise the following.

3(x + y) + x(x + y)a (7 – 2x) – x(7 – 2x)b


a 3(x + y) + x(x + y)

= (x + y)(3 + x)

HCF = (x + y).

The second pair of brackets contains what

remains when 3(x + y) and x(x + y) are divided

by (x + y).

b (7 – 2x) – x(7 – 2x)

= 1(7 – 2x) – x(7 – 2x)

= (7 – 2x)(1 – x)

Insert 1 in front of the first bracket.

HCF = (7 – 2x).

The second bracket must contain 1 – x after

dividing (7 – 2x) and x(7 – 2x) by (7 – 2x).


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Exercise 8C

1 Write down the highest common factor (HCF) of these pairs of numbers.

8, 12a 10, 20b 5, 60c 24, 30d3, 5e 100, 75f 16, 24g 36, 72h

2 Write down the missing factor.

5 × = 5xa 7 × = 7xb 2a× = 2a2c5a× = 10a2d × 3y = –6y2e × 12x = –36x2f–3 × = 6xg –2x× = 20x2h × 7xy = –14x2yi

3 a Write down the missing factor in each part.

(x2 + 2x) = 6x2 + 12xi (2x + 4) = 6x2 + 12xii(x + 2) = 6x2 + 12xiii

b Which equation above uses the HCF of 6x2 and 12x?

c By looking at the terms left in the brackets, how do you know you have taken out the HCF?

UNDE

RSTA

NDING

—1–2(½), 3 3

4Example 4 Determine the HCF of the following.

6x and 14xya 12a and 18ab 10m and 4c 12y and 8d15t and 6se 15 and pf 9x and 24xyg 6n and 21mnh10y and 2yi 8x2 and 14xj 4x2y and 18xyk 5ab2 and 15a2bl

5Example 5a Factorise the following.

7x + 7a 3x + 3b 4x – 4c 5x – 5d4 + 8ye 10 + 5af 3 – 9bg 6 – 2xh12a + 3bi 6m + 6nj 10x – 8yk 4a – 20blx2 + 2xm a2 – 4an y2 – 7yo x – x2p3p2 + 3pq 8x – 8x2r 4b2 + 12bs 6y – 10y2t12a – 15a2u 9m + 18m2v 16xy – 48x2w 7ab – 28ab2x

6Example 5b Factorise the following by factoring out the negative sign as part of the HCF.

–8x – 4a – 4x – 2b –10x – 5yc –7a – 14bd–9x – 12e –6y – 8f –10x – 15yg – 4m – 20nh–3x2 – 18xi –8x2 – 12xj –16y2 – 6yk –5a2 – 10al–6x – 20x2m –6p – 15p2n –16b – 8b2o –9x – 27x2p

7Example 6 Factorise the following which involve a binomial common factor.

4(x + 3) + x(x + 3)a 3(x + 1) + x(x + 1)b 7(m – 3) + m(m – 3)cx(x – 7) + 2(x – 7)d 8(a + 4) – a(a + 4)e 5(x + 1) – x(x + 1)fy(y + 3) – 2(y + 3)g a(x + 2) – x(x + 2)h t(2t + 5) + 3(2t + 5)im(5m – 2) + 4(5m – 2)j y(4y – 1) – (4y – 1)k (7 – 3x) + x(7 – 3x)l

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4–8(½)4–7(½) 4–8(½)


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8C8 Factorise these mixed expressions.

6a + 30a 5x – 15b 8b + 18cx2 – 4xd y2 + 9ye a2 – 3afx2y – 4xy + xy2g 6ab – 10a2b + 8ab2h m(m + 5) + 2(m + 5)ix(x + 3) – 2(x + 3)j b(b – 2) + (b – 2)k x(2x + 1) – (2x + 1)ly(3 – 2y) – 5(3 – 2y)m (x + 4)2 + 5(x + 4)n (y + 1)2 – 4(y + 1)o

FLUE

NCY

9 Write down the perimeter of these shapes in factorised form.

2x

4

ax

6

b

5x

10

c

x

5

2

d

x + 1

1

e3

4x

f

10 The expression for the area of a rectangle is (4x2 + 8x) square units. Find an expression for its

width if the length is (x + 2) units.

11 The height, in metres, of a ball thrown in the air is given by

5t – t2, where t is the time in seconds.

a Write an expression for the ball’s height in factorised

form.

b Find the ball’s height at these times:

t = 0it = 2iit = 4iii

c How long does it take for the ball’s height to return to

0 metres? Use trial and error if required.

PROB

LEM-SOLVING

9(½), 10, 119 9, 10


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8C12 7 × 9 + 7 × 3 can be evaluated by firstly factorising to 7(9 + 3). This gives 7 × 12 = 84. Use a

similar technique to evaluate the following.

9 × 2 + 9 × 5a 6 × 3 + 6 × 9b –2 × 4 – 2 × 6c–5 × 8 – 5 × 6d 23 × 5 – 23 × 2e 63 × 11 – 63 × 8f

13 Common factors can also be removed from expressions with more than two terms.

For example: 2x2 + 6x + 10xy = 2x(x + 3 + 5y)

Factorise these expressions by taking out the HCF.

3a2 + 9a + 12a 5z2 – 10z + zyb x2 – 2xy + x2yc4by – 2b + 6b2d –12xy – 8yz – 20xyze 3ab + 4ab2 + 6a2bf

14 Sometimes we can choose to factor out a negative or a positive HCF. Both factorisations are

correct. For example:

–13x + 26 = –13(x – 2) (HCF is –13)

OR –13x + 26 = 13(–x + 2) (HCF is 13)

= 13(2 – x)

Factorise in two different ways: the first by factoring out a negative and the second by a

positive HCF.

– 4x + 12a –3x + 9b –8n + 8c –3b + 3d–5m + 5m2e –7x + 7x2f –5x + 5x2g – 4y + 22y2h–8n + 12n2i –8y + 20j –15mn + 10k –15x + 45l

REAS

ONING

13, 1412 12, 13

Factoring out a negative

15 Using the fact that a – b = –(b – a) you can factorise x(x – 2) – 5(2 – x) by following these steps.

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

= (x – 2)(x + 5)

Use this idea to factorise these expressions.

x(x – 4) + 3(4 – x)a x(x – 5) – 2(5 – x)b x(x – 3) – 3(3 – x)c3x(x – 4) + 5(4 – x)d 3(2x – 5) + x(5 – 2x)e 2x(x – 2) + (2 – x)f– 4(3 – x) – x(x – 3)g x(x – 5) + (10 – 2x)h x(x – 3) + (6 – 2x)i

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8D Factorising the difference of two squaresRecall that a difference of two perfect squares is formed when expanding the product of the sum and

difference of two terms. For example, (x + 2)(x – 2) = x2 – 4. Reversing this process means that a

difference of two perfect squares can be factorised into two binomial expressions of the form (a + b)

and (a – b).

Let’s start: Expanding to understand factorising

Complete the steps in these expansions then write the conclusion.

(x + 3)(x – 3) = x2 – 3x + –

= x2 –

∴ x2 – 9 = ( + )( – )

• (2x – 5)(2x + 5) = 4x2 + 10x – –

= –

∴ 4x2 – = ( + )( – )

•

(a + b)(a – b) = a2 – ab + –

= –

∴ a2 – = ( + )( – )

•

Keyideas

Factorising the difference of perfect squares (DOPS) uses the rule a2 – b2 = (a + b)(a – b).

• x2 – 16 = x2 – 42

= (x + 4)(x – 4)

• 9x2 – 100 = (3x)2 – 102

= (3x + 10)(3x – 10)

• 25 – 4y2 = 52 – (2y)2

= (5 + 2y)(5 – 2y)

First take out common factors where possible.

• 2x2 – 18 = 2(x2 – 9)

= 2(x + 3)(x – 3)

Example 7 Factorising DOPS

Factorise each of the following.

x2 – 4a 9a2 – 25b

81x2 – y2c 2b2 – 32d

(x + 1)2 – 4e


a x2 – 4 = x2 – 22

= (x + 2)(x – 2)

Write as a DOPS (4 is the same as 22).

Write in factorised form a2 –b2 = (a+b)(a–b).

Here a = x and b = 2.


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b 9a2 – 25 = (3a)2 – 52

= (3a + 5)(3a – 5)

Write as a DOPS. 9a2 is the same as (3a)2.

Write in factorised form.

c 81x2 – y2 = (9x)2 – y2

= (9x + y)(9x – y)

81x2 = (9x)2

Use a2 – b2 = (a + b)(a – b)

d 2b2 – 32 = 2(b2 – 16)

= 2(b2 – 42)

= 2(b + 4)(b – 4)

First, factor out the common factor of 2.

Write as a DOPS and then factorise.

e (x + 1)2 – 4 = (x + 1)2 – 22

= (x + 1 + 2)(x + 1 – 2)

= (x + 3)(x – 1)

Write as a DOPS. In a2 – b2 here, a is the

expression x + 1 and b = 2.

Write in factorised form and simplify.

Exercise 8D

1 Expand these binomial products to form a difference of perfect squares.

(x + 2)(x – 2)a (x – 7)(x + 7)b (2x – 1)(2x + 1)c(x + y)(x – y)d (3x – y)(3x + y)e (a + b)(a – b)f

2 Write the missing term. Assume it is a positive number.

( )2 = 9a ( )2 = 121b ( )2 = 81c ( )2 = 400d( )2 = 4x2e ( )2 = 9a2f ( )2 = 25b2g ( )2 = 49y2h

3 Complete these factorisations.

x2 – 16 = x2 – 42

= (x + 4)( – )

a x2 – 144 = x2 – ( )2

= ( + 12)(x – )

b

16x2 – 1 = ( )2 – ( )2

= (4x + )( – 1)

c 9a2 – 4b2 = ( )2 – ( )2

= (3a + )( – 2b)

dUN

DERS

TAND

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—1–3(½) 3(½)

4Example 7a Factorise each of the following.

x2 – 9a y2 – 25b y2 – 1c x2 – 64dx2 – 16e b2 – 49f a2 – 81g x2 – y2ha2 – b2i 16 – a2j 25 – x2k 1 – b2l36 – y2m 121 – b2n x2 – 400o 900 – y2p

FLUE

NCY

4–7(½)4–6(½) 4–7(½)


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8D5Example 7b,c Factorise each of the following.

4x2 – 25a 9x2 – 49b 25b2 – 4c 4m2 – 121d100y2 – 9e 81a2 – 4f 1 – 4x2g 25 – 64b2h16 – 9y2i 36x2 – y2j 4x2 – 25y2k 64a2 – 49b2l4p2 – 25q2m 81m2 – 4n2n 25a2 – 49b2o 100a2 – 9b2p

6Example 7d Factorise each of the following by first taking out the common factor.

3x2 – 108a 10a2 – 10b 6x2 – 24c4y2 – 64d 98 – 2x2e 32 – 8m2f5x2y2 – 5g 3 – 3x2y2h 63 – 7a2b2i

7Example 7e Factorise each of the following.

(x + 5)2 – 9a (x + 3)2 – 4b (x + 10)2 – 16c(x – 3)2 – 25d (x – 7)2 – 1e (x – 3)2 – 36f49 – (x + 3)2g 4 – (x + 2)2h 81 – (x + 8)2i

FLUE

NCY

8 The height above ground (in metres) of an object

thrown off the top of a building is given by 36 – 4t2

where t is in seconds.

a Factorise the expression for the height of the

object by firstly taking out the common factor.

b Find the height of the object:

initially (t = 0)iat 2 seconds (t = 2).ii

c How long does it take for

the object to hit the ground?

Use trial and error if you wish.

9 This ‘multisize’ square picture frame has side length 30 cm and can

hold a square picture with any side length less than 26 cm.

a If the side length of the picture is x cm, write an expression for:

the area of the pictureithe area of the frame (in factorised form).ii

b Use your result from part a ii to find the area of the frame if:

x = 20ithe area of the picture is 225 cm2.ii

30 cmx cm

PROB

LEM-SOLVING

8, 98 8, 9


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8D10 Initially it may not appear that an expression such as – 4 + 9x2 is a difference of perfect squares.

However, swapping the position of the two terms makes – 4 + 9x2 = 9x2 – 4, which can be

factorised to (3x + 2)(3x – 2). Use this idea to factorise these difference of perfect squares.

–9 + x2a –121 + 16x2b –25a2 + 4c –y2 + x2d–25a2 + 4b2e –36a2b2 + c2f –16x2 + y2z2g –900a2 + b2h

11 Olivia factorises 16x2 – 4 to get (4x + 2)(4x – 2) but the answer says 4(2x + 1)(2x – 1).

a What should Olivia do to get from her answer to the actual answer?

b What should Olivia have done initially to avoid this issue?

12 Find and explain the error in this working and correct it.

9 – (x – 1)2 = (3 + x – 1)(3 – x – 1)

= (2 + x)(2 – x)

REAS

ONING

10–1210 10, 11

Factorising with fractions and powers of 4

13 Some expressions with fractions or powers of 4 can be factorised in a similar way. Factorise these.

x2 – 14

a x2 – 425

b 25x2 – 916

c x2

9– 1d

a2

4– b

2

9e 5x2

9– 54

f 7a2

25– 28b2

9g a2

8– b

2

18h

x4 – y4i 2a4 – 2b4j 21a4 – 21b4k x4

3– y

4

3l

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8E Factorisation by grouping

When an expression contains four terms, such as x2 + 2x – x – 2,

it may be possible to factorise it into a product of two binomial

terms like (x – 1)(x + 2). In such situations the method of

grouping is often used.

Let’s start: Two methods – same result

The four-term expression x2 – 3x – 3 + x is written on the board.

Factorising by grouping is a bit like arrangingscattered objects into some sort of order.

Tommy chooses to rearrange the terms to give x2 – 3x + x – 3 then factorises by grouping.

Sharon chooses to rearrange the terms to give x2 + x – 3x – 3 then also factorises by grouping.

• Complete Tommy and Sharon’s factorisation working.

Tommy

x2 – 3x + x – 3 = x(x – 3) + 1( )

= (x – 3)( )

Sharon

x2 + x – 3x – 3 = x( ) – 3( )

= (x + 1)( )

• Discuss the differences in the methods. Is there any difference in their answers?

• Whose method do you prefer?

Keyideas

Factorisation by grouping is a method which is often used to factorise a four-term expression.

• Terms are grouped into pairs and factorised separately.

• The common binomial factor is then taken out to complete the

factorisation.

• Terms can be rearranged to assist in the search of a common factor.

x2 + 3x – 2x – 6

= x(x + 3) – 2(x + 3)

= (x + 3)(x – 2)

Example 8 Factorising by grouping

Use the method of grouping to factorise these expressions.

x2 + 2x + 3x + 6a x2 + 3x – 5x – 15b


a x2 + 2x + 3x + 6 = (x2 + 2x) + (3x + 6)

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

= (x + 2)(x + 3)

Group the first and second pair of terms.

Factorise each group.

Take the common factor (x + 2) out of both

groups.

b x2 + 3x – 5x – 15 = (x2 + 3x) + (–5x – 15)

= x(x + 3) – 5(x + 3)

= (x + 3)(x – 5)

Group the first and second pair of terms.

Factorise each group.

Take out the common factor (x + 3).


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Example 9 Rearranging an expression to factorise by grouping

Factorise 2x2 – 9 – 18x + x using grouping.


2x2 – 9 – 18x + x = 2x2 + x – 18x – 9

= x(2x + 1) – 9(2x + 1)

= (2x + 1)(x – 9)

Rearrange so that each group has a common

factor.

Factorise each group then take out (2x + 1).

Alternatively:

2x2 – 9 – 18x + x = 2x2 – 18x + x – 9

= 2x(x – 9) + 1(x – 9)

= (x – 9)(2x + 1)

Alternatively, you can group in another order

where each group has a comman factor. Then

factorise.

The answer will be the same.

Exercise 8E

1 Expand each expression.

2(x – 1)a 3(a + 4)b –5(1 – a)c–2(3 – x)d a(a + 5)e b(2 – b)fx(x – 4)g y(4 – y)h x(a + 1) + 2(a + 1)ia(x – 3) + 5(x – 3)j b(x – 2) – 3(x – 2)k c(1 – x) – 4(1 – x)l

2 Copy and then fill in the missing information.

2(x + 1) + x(x + 1) = (x + 1)( )a 3(x + 3) – x(x + 3) = (x + 3)( )b5(x + 5) – x(x + 5) = (x + 5)( )c x(x + 7) + 4(x + 7) = (x + 7)( )da(x – 3) + (x – 3) = (x – 3)( )e a(x + 4) – (x + 4) = (x + 4)( )f(x – 3) – a(x – 3) = (x – 3)( )g (4 – x) + 2a(4 – x) = (4 – x)( )h

3 Take out the common binomial term to factorise each expression.

x(x – 3) – 2(x – 3)a x(x + 4) + 3(x + 4)b x(x – 7) + 4(x – 7)c3(2x + 1) – x(2x + 1)d 4(3x – 2) – x(3x – 2)e 2x(2x + 3) – 3(2x + 3)f3x(5 – x) + 2(5 – x)g 2(x + 1) – 3x(x + 1)h x(x – 2) + (x – 2)i

UNDE

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NDING

—1–3(½) 3(½)

4Example 8 Use the method of grouping to factorise these expressions.

x2 + 3x + 2x + 6a x2 + 4x + 3x + 12b x2 + 7x + 2x + 14cx2 – 6x + 4x – 24d x2 – 4x + 6x – 24e x2 – 3x + 10x – 30fx2 + 2x – 18x – 36g x2 + 3x – 14x – 42h x2 + 4x – 18x – 72ix2 – 2x – xa + 2aj x2 – 3x – 3xc + 9ck x2 – 5x – 3xa + 15al

5 Use the method of grouping to factorise these expressions. The HCF for each pair includes

a pronumeral.

3ab + 5bc + 3ad + 5cda 4ab – 7ac + 4bd – 7cdb 2xy – 8xz + 3wy – 12wzc5rs – 10r + st – 2td 4x2 + 12xy – 3x – 9ye 2ab – a2 – 2bc + acf

FLUE

NCY

4–6(½)4–5(½) 4–6(½)


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8E6 Factorise these expressions. Remember to use a factor of 1 where necessary, for example,

x2 – ax + x – a = x(x – a) + 1(x – a).

x2 – bx + x – ba x2 – cx + x – cb x2 + bx + x + bcx2 + cx – x – cd x2 + ax – x – ae x2 – bx – x + bf

FLUE

NCY

7Example 9 Factorise these expressions by first rearranging the terms.

2x2 – 7 – 14x + xa 5x + 2x + x2 + 10b 2x2 – 3 – x + 6xc3x – 8x – 6x2 + 4d 11x – 5a – 55 + axe 12y + 2x – 8xy – 3f6m – n + 3mn – 2g 15p – 8r – 5pr + 24h 16x – 3y – 8xy + 6i

8 What expanded expression factorises to the following?

(x – a)(x + 4)a (x – c)(x – d)b (x + y)(2 – z)c (x – 1)(a + b)d(3x – b)(c – b)e (2x – y)(y + z)f (3a + b)(2b + 5c)g (m – 2x)(3y + z)h

9 Note that x2 + 5x + 6 = x2 + 2x + 3x + 6 which can be factorised by grouping. Use a similar

method to factorise the following.

x2 + 7x + 10a x2 + 8x + 15b x2 + 10x + 24cx2 – x – 6d x2 + 4x – 12e x2 – 11x + 18f

PROB

LEM-SOLVING

7, 97 7, 8

10 xa – 21 + 7a – 3x could be rearranged in two different ways before factorising.

Method 1

xa + 7a – 3x – 21 = a(x + 7) – 3( )=

Method 2

xa – 3x + 7a – 21 = x(a – 3) + 7( )=

a Copy and complete both methods for the above expression.

b Use different arrangements of the four terms to complete the factorisation of the following

in two ways. Show working using both methods.

xb – 6 – 3b + 2xi xy – 8 + 2y – 4xii 4m2 – 15n + 6m – 10mniii2m + 3n – mn – 6iv 4a – 6b2 + 3b – 8abv 3ab – 4c – b + 12acvi

11 Make up at least three of your own four-term expressions that factorise to a binomial

product. Describe the method that you used to make up each four-term expression.

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ONING

10, 1110 10

Grouping with more than four terms

12 Factorise by grouping.

2(a – 3) – x(a – 3) – c(a – 3)a b(2a + 1) + 5(2a + 1) – a(2a + 1)bx(a + 1) – 4(a + 1) – ba – bc 3(a – b) – b(a – b) – 2a2 + 2abdc(1 – a) – x + ax + 2 – 2ae a(x – 2) + 2bx – 4b – x + 2fa2 – 3ac – 2ab + 6bc + 3abc – 9bc2g 3x – 6xy – 5z + 10yz + y – 2y2h8z – 4y + 3x2 + xy – 12x – 2xzi –ab – 4cx + 3aby + 2abx + 2c – 6cyj

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8F Factorising quadratic trinomials EXTENDING

An expression that takes the form x2 + bx + c, where b and c are

constants, is an example of a monic quadratic trinomial which has

the coefficient of x2 equal to 1. To factorise a quadratic expression,

we need to use the distributive law in reverse. Consider the

expansion shown at right:

(x + 2)(x − 4) = x2 − 2x − 8

factorised form expanded formfactorising

expanding

If we examine the expansion above we can see how each term of the product is formed.

(x + 2)(x − 4) = x2 − 2x −8

Product of 2 and −4 is −8 (2 × (−4) = −8, the constant term)

Product of x and x is x2

(x + 2)(x − 4) = x2 − 2x − 8

x × (−4) = −4x

2 × x = 2x

Add −4x and 2x to give the middle term, −2x(−4 + 2 = −2, the coefficient of x)

Let’s start: So many choices

Mia says that since –2 × 3 = –6 then x2 + 5x – 6 must equal (x – 2)(x + 3).

• Expand (x – 2)(x + 3) to see if Mia is correct.

• What other pairs of numbers multiply to give –6?

• Which pair of numbers should Mia choose to correctly factorise x2 + 5x – 6?

• What advice can you give Mia when trying to factorise these types of trinomials?

Keyideas To factorise a quadratic trinomial of the form

x2 + bx + c, find two numbers which:

For example:

x2 – 3x – 10 = (x – 5)(x + 2) choose –5 and +2

since –5 × 2 = –10 and –5 + 2 = –3• multiply to give c and

• add to give b.

Check factorisation steps by expanding. check: (x – 5)(x + 2) = x2 + 2x – 5x – 10

= x2 – 3x – 10

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Example 10 Factorising quadratic trinomials

Factorise each of the following quadratic expressions.

x2 + 7x + 10a x2 + 2x – 8b x2 – 7x + 10c


a x2 + 7x + 10 = (x + 5)(x + 2) Factors of 10 include: (10, 1) and (5, 2).

The pair that adds to 7 is (5, 2).

b x2 + 2x – 8 = (x + 4)(x – 2) Factors of –8 are (–8, 1) or (8, –1) or (4, –2) or

(– 4, 2) and 4 + (–2) = 2 so choose (4, –2).

c x2 – 7x + 10 = (x – 2)(x – 5) Factors of 10 are: (10, 1) or (–10, –1) or (5, 2)

or (–5, –2).

To add to a negative (–7), both factors must

then be negative: –5 + (–2) = –7 so choose

(–5, –2).

Example 11 Factorising with a common factor

Factorise the quadratic expression 2x2 – 2x – 12.


2x2 – 2x – 12 = 2(x2 – x – 6)

= 2(x – 3)(x + 2)

First take out common factor of 2.

Factors of –6 are: (–6, 1) or (6, –1) or (–3, 2) or

(3, –2).

–3 + 2 = –1 so choose (–3, 2).

Exercise 8F

1 Expand these binomial products.

(x + 1)(x + 3)a (x + 2)(x + 7)b (x – 3)(x + 11)c(x – 5)(x + 6)d (x + 12)(x – 5)e (x + 13)(x – 4)f(x – 2)(x – 6)g (x – 20)(x – 11)h (x – 9)(x – 1)i

2 Decide what two numbers multiply to give the first number and add to give the second number.

6, 5a 10, 7b 12, 13c20, 9d –5, 4e –7, –6f–15, 2g –30, –1h 6, –5i18, –11j 40, –13k 100, –52l

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8F3Example 10a Factorise each of the following quadratic expressions.

x2 + 3x + 2a x2 + 4x + 3b x2 + 8x + 12cx2 + 10x + 9d x2 + 8x + 7e x2 + 15x + 14fx2 + 6x + 8g x2 + 7x + 12h x2 + 10x + 16ix2 + 8x + 15j x2 + 9x + 20k x2 + 11x + 24l

4Example 10b Factorise each of the following quadratic expressions.

x2 + 3x – 4a x2 + x – 2b x2 + 4x – 5cx2 + 5x – 14d x2 + 2x – 15e x2 + 8x – 20fx2 + 3x – 18g x2 + 7x – 18h x2 + x – 12i

5Example 10c Factorise each of the following quadratic expressions.

x2 – 6x + 5a x2 – 2x + 1b x2 – 5x + 4cx2 – 9x + 8d x2 – 4x + 4e x2 – 8x + 12fx2 – 11x + 18g x2 – 10x + 21h x2 – 5x + 6i

6 Factorise each of the following quadratic expressions.

x2 – 7x – 8a x2 – 3x – 4b x2 – 5x – 6cx2 – 6x – 16d x2 – 2x – 24e x2 – 2x – 15fx2 – x – 12g x2 – 11x – 12h x2 – 4x – 12i

7Example 11 Factorise each of the following quadratic expressions by first taking out a common factor.

2x2 + 10x + 8a 2x2 + 22x + 20b 3x2 + 18x + 24c2x2 + 14x – 60d 2x2 – 14x – 36e 4x2 – 8x + 4f2x2 + 2x – 12g 6x2 – 30x – 36h 5x2 – 30x + 40i3x2 – 33x + 90j 2x2 – 6x – 20k 3x2 – 3x – 36l

FLUE

NCY

3–7(½)3–6(½) 3–7(½)

8 Find the missing term in these trinomials if they are to factorise using integers. For example: the

missing term in x2 + + 10 could be 7x because x2 + 7x + 10 factorises to (x + 5)(x + 2) and 5

and 2 are integers. There may be more than one answer in each case.

x2 + + 5a x2 – + 9b x2 – – 12c x2 + – 12d

x2 + + 18e x2 – + 18f x2 – – 16g x2 + – 25h

9 A backyard, rectangular in area, has a length 2 metres more than its width (x metres). Inside the

rectangle are three square paved areas each of area 5 m2 as shown. The remaining area is lawn.

a Find an expression for:

the total backyard areaithe area of lawn in expanded formiithe area of lawn in factorised form.iii

b Find the area of lawn if:

x = 10ix = 7.ii

(x + 2) metres

x metres

PROB

LEM-SOLVING

8(½), 98 8(½), 9


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8F10 The expression x2 – 6x + 9 factorises to (x – 3)(x – 3) = (x – 3)2, which is a perfect square.

Factorise these perfect squares.

x2 + 8x + 16a x2 + 10x + 25b x2 + 30x + 225cx2 – 2x + 1d x2 – 14x + 49e x2 – 26x + 169f2x2 + 4x + 2g 5x2 – 30x + 45h –3x2 + 36x – 108i

11 Sometimes it is not possible to factorise quadratic trinomials using integers. Decide which of the

following cannot be factorised using integers.

x2 – x – 56a x2 + 5x – 4b x2 + 7x – 6cx2 + 3x – 108d x2 + 3x – 1e x2 + 12x – 53f

REAS

ONING

10(½), 1110(½) 10(½)

Completing the square

12 It is useful to be able to write a simple quadratic trinomial in the

form (x + b)2 + c. This involves adding (and subtracting) a special

number to form the first perfect square. This procedure is called

completing the square. Here is an example.

x2 − 6x − 8 = x2 − 6x + 9 − 9 − 8

= (x − 3)(x − 3) − 17 = (x − 3)2 − 17

62

(− )2 = 9

Complete the square for these trinomials.

x2 – 2x – 8a x2 + 4x – 1b x2 + 10x + 3cx2 – 16x – 3d x2 + 18x + 7e x2 – 32x – 11f

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8G Factorising trinomials of the form ax 2 + bx+ c EXTENDING

So far we have factorised quadratic trinomials where the coefficient of x2 is 1, such as x2 – 3x – 40.

These are called monic trinomials. We will now consider non-monic trinomials where the coefficient of

x is not equal to 1 and is also not a common factor to all three terms, such as in 6x2 + x – 15. The method

used in this section uses grouping which was discussed in section 8E.

Let’s start: How the grouping method works

Consider the trinomial 2x2 + 9x + 10.

• First write 2x2 + 9x + 10 = 2x2 + 4x + 5x + 10 then factorise

by grouping.

• Note that 9x was split to give 4x + 5x and the product of

2x2 and 10 is 20x2. Describe the link between the pair of

numbers {4, 5} and the pair of numbers {2, 10}.

• Why was 9x split to give 4x + 5x and not, say, 3x + 6x?

• Describe how the 13x should be split in 2x2 + 13x + 15 so

it can be factorised by grouping.

• Now try your method for 2x2 – 7x – 15.

Keyideas

To factorise a trinomial of the form ax2 + bx + c by grouping, find two numbers which sum to give

b and multiply to give a× c.

For example:

5x2 + 13x – 6

= 5x2 + 15x – 2x – 6

= 5x(x + 3) – 2(x + 3)

= (x + 3)(5x – 2)

a× c = 5 × (–6) = –30 so the two numbers are 15

and –2 since 15 + (–2) = 13 and 15 × (–2) = –30.

Mentally check your factors by expanding your answer.

(x + 3)(5x − 2)

5x2 −6

15x−2x 15x− 2x= 13x


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Example 12 Factorising trinomials of the form ax 2 + bx+ c

Factorise 2x2 + 7x + 3.


2x2 + 7x + 3 = 2x2 + x + 6x + 3

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

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

a× c = 2 × 3 = 6 then ask what factors of this

number (6) add to 7. The answer is 1 and 6, so

split 7x = x + 6x. Then factorise by grouping.

Example 13 Factorising trinomials with negative numbers

Factorise the quadratic trinomials.

10x2 + 9x – 9a 6x2 – 17x + 12b


a 10x2 + 9x – 9 = 10x2 + 15x – 6x – 9

= 5x(2x + 3) – 3(2x + 3)

= (2x + 3)(5x – 3)

10 × (–9) = –90 so ask what factors of –90 add

to give 9. Choose 15 and –6. Then complete

the factorisation by grouping.

b 6x2 – 17x + 12 = 6x2 – 9x – 8x + 12

= 3x(2x – 3) – 4(2x – 3)

= (2x – 3)(3x – 4)

6 × 12 = 72 so ask what factors of 72 add to

give –17. Choose –9 and –8.

Complete a mental check.

(2x − 3)(3x − 4)−9x−8x −9x− 8x= −17x

Exercise 8G

1 List the two numbers which satisfy each part.

Multiply to give 6 and add to give 5a Multiply to give 12 and add to give 8bMultiply to give –10 and add to give 3c Multiply to give –24 and add to give 5dMultiply to give 18 and add to give –9e Multiply to give 35 and add to give –12fMultiply to give –30 and add to give –7g Multiply to give –28 and add to give –3h

UNDE

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8G2 Copy and complete.

2x2 + 7x + 5 = 2x2 + 2x + + 5

= 2x( ) + 5( )

= ( )( )

a 3x2 + 8x + 4 = 3x2 + 6x + + 4

= 3x( ) + 2( )

= ( )( )

b

2x2 – 7x + 6 = 2x2 – 3x – + 6

= x( ) – 2( )

= ( )( )

c 5x2 + 9x – 2 = 5x2 + 10x – – 2

= 5x( ) – 1( )

= ( )( )

d

4x2 + 11x + 6 = 4x2 + + 3x + 6

= (x + 2) + 3( )

= ( )( )

e 6x2 – 7x – 3 = 6x2 – 9x + – 3

= ( ) + 1( )

= ( )( )

f

UNDE

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3Example 12 Factorise these quadratic trinomials.

2x2 + 9x + 4a 3x2 + 7x + 2b 2x2 + 7x + 6c3x2 + 8x + 4d 5x2 + 12x + 4e 2x2 + 11x + 12f6x2 + 13x + 5g 4x2 + 5x + 1h 8x2 + 14x + 5i

4Example 13 Factorise these quadratic trinomials.

3x2 + 2x – 5a 5x2 + 6x – 8b 8x2 + 10x – 3c6x2 – 13x – 8d 10x2 – 3x – 4e 5x2 – 11x – 12f4x2 – 16x + 15g 2x2 – 15x + 18h 6x2 – 19x + 10i12x2 – 13x – 4j 4x2 – 12x + 9k 7x2 + 18x – 9l9x2 + 44x – 5m 3x2 – 14x + 16n 4x2 – 4x – 15o

FLUE

NCY

3–4(½)3–4(½) 3–4(½)

5 Factorise these quadratic trinomials.

10x2 + 27x + 11a 15x2 + 14x – 8b 20x2 – 36x + 9c18x2 – x – 5d 25x2 + 5x – 12e 32x2 – 12x – 5f27x2 + 6x – 8g 33x2 + 41x + 10h 54x2 – 39x – 5i12x2 – 32x + 21j 75x2 – 43x + 6k 90x2 + 33x – 8l

6 Factorise by firstly taking out a common factor.

30x2 – 14x – 4a 12x2 + 18x – 30b 27x2 – 54x + 15c21x2 – 77x + 42d 36x2 + 36x – 40e 50x2 – 35x – 60f

7 Factorise these trinomials.

–2x2 + 7x – 6a –5x2 – 3x + 8b –6x2 + 13x + 8c18 – 9x – 5x2d 16x – 4x2 – 15e 14x – 8x2 – 5f

PROB

LEM-SOLVING

5–6(½), 75(½) 5–6(½)


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8G8 When splitting the 3x in 2x2 + 3x – 20, you could write:

A 2x2 + 8x – 5x – 20 or B 2x2 – 5x + 8x – 20

Complete the factorisation using A.aComplete the factorisation using B.b

c Does the order matter when you split the 3x?

d Factorise these trinomials twice each. Factorise once by grouping then repeat but reverse

the order of the two middle terms in the first line of working.

3x2 + 5x – 12i 5x2 – 3x – 14ii 6x2 + 5x – 4iii

9 Make up five non-monic trinomials with the coefficient of x2 not equal to 1 which factorise using

the above method. Explain your method in finding these trinomials.

REAS

ONING

8, 98 8

The cross method

10 The cross method is another way to factorise trinomials of the form ax2 + bx + c. It involves

finding factors of ax2 and factors of c then choosing pairs of these factors that add to bx.

For example: Factorise 6x2 – x – 15.

Factors of 6x2 include (x, 6x) and (2x, 3x).

Factors of –15 include (15, –1), (–15, 1), (5, –3) and (–5, 3).

We arrange a chosen pair of factors vertically then cross-multiply and add to get –1x.

x × (−1) + 6x × 15 x × (−3) + 6x × 5 2x × (−5) + 3x × 3= −1x= 27x ≠ −1x= 89x ≠ −1x

15

−16x

x 5

. . . . . . . . . . .−36x

x 3

−53x (3x − 5)

(2x + 3)2x

You will need to continue until a particular combination works. The third cross-product gives a

sum of –1x so choose the factors (2x + 3) and (3x – 5) so:

6x2 – x – 15 = (2x + 3)(3x – 5)

Try this method on the trinomials from Questions 4 and 5.

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Progress quiz

138pt8A Expand the following.

(x + 4)(x + 2)a (a – 5)(a + 8)b(2x – 3)(x + 6)c (3a – b)(a – 2b)d

238pt8B Expand each of the following.

(y + 4)2a (x – 3)2b(2a – 3)2c (7k + 2m)2d

338pt8B Expand and simplify the following.

(x + 5)(x – 5)a (11x – 9y)(11x + 9y)b

438pt8C Factorise the following.

25a – 15a x2 – 7xb–12x2 – 16xc 2(a + 3) + a(a + 3)d7(8 + a) – a(8 + a)e k(k – 4) – (k – 4)f

538pt8D Factorise each of the following.

x2 – 81a 16a2 – 49b25x2 – y2c 2a2 – 50d12x2y2 – 12e (h + 3)2 – 64f

638pt8E Use the method of grouping to factorise these expressions.

x2 + 7x + 2x + 14a a2 + 5a – 4a – 20b x2 – hx + x – hc

738pt8E Use grouping to factorise these expressions by first rearranging.

2x2 – 9 – 6x + 3xa 3ap – 10 + 2p – 15ab

838pt8F Factorise each of the following quadratic expressions.

Extx2 + 6x + 8a a2 + 2a – 15bm2 – 11m + 30c 2k2 + 2k – 24d

938pt8G Factorise these quadratic trinomials.

Ext2k2 + 7k + 6a 2x2 + 11x + 5b3a2 + 10a – 8c 10m2 – 19m + 6d


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8H Simplifying algebraic fractions: multiplication and division

With a numerical fraction such as 69, the highest common factor of 6 and 9 is 3, which can be

cancelled 69=

1�3 × 21�3 × 3

= 23. For algebraic fractions the process is the same. If expressions are in a

factorised form, common factors can be easily identified and cancelled.

Let’s start: Correct cancelling

Consider this cancelling attempt:

5x + ��101

��202= 5x + 1

2

• Substitute x = 6 into the left-hand side to evaluate 5x + 1020

.

• Substitute x = 6 into the right-hand side to evaluate 5x + 12

.

• What do you notice about the two answers to the above? How can you explain this?

• Decide how you might correctly cancel the expression on the left-hand side. Show your steps and check

by substituting a value for x.

Keyideas

Simplify algebraic fractions by factorising and cancelling only common factors.

Incorrect

2x + �4 2

�21= 2x + 2

Correct

2x + 42

=1�2(x + 2)

�21

= x + 2

To multiply algebraic fractions:

• factorise expressions where possible

• cancel if possible

• multiply the numerators and the denominators.

To divide algebraic fractions:

• multiply by the reciprocal of the fraction following the division sign

• follow the rules for multiplication after converting to the reciprocal

- The reciprocal of abis ba.


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Example 14 Simplifying algebraic fractions

Simplify the following by cancelling.

3(x + 2)(x – 4)6(x – 4)

a 20 – 5x8 – 2x

b x2 – 4x + 2

c


a1�3(x + 2)��(x – 4)1

2�6��(x – 4)1= x + 2

2Cancel the common factors (x – 4) and 3.

b 20 – 5x8 – 2x

= 5��(4 – x)1

2��(4 – x)1

= 52

Factorise the numerator and denominator then

cancel common factor of (4 – x).

c x2 – 4x + 2

=1��(x + 2)(x – 2)

��(x + 2)1

= x – 2

Factorise the difference of squares in the

numerator then cancel the common factor.

Example 15 Multiplying and dividing algebraic fractions

Simplify the following.

3(x – 1)(x + 2)

× 4(x + 2)9(x – 1)(x – 7)

a (x – 3)(x + 4)x(x + 7)

÷ 3(x + 4)x + 7

bx2 – 425

×5x + 5

x2 – x – 2 Extc


a1�3��(x – 1)1

��(x + 2)1× 4��(x + 2)1

3�9��(x – 1)1(x – 7)

= 1 × 41 × 3(x – 7)

= 43(x – 7)

First, cancel any factors in the numerators with

a common factor in the denominators. Then

multiply the numerators and the denominators.

b (x – 3)(x + 4)x(x + 7)

÷ 3(x + 4)x + 7

= (x – 3)��(x + 4)1

x��(x + 7)1× ��(x + 7)1

3��(x + 4)1

= x – 33x

Multiply by the reciprocal of the fraction after the

division sign.

Cancel common factors and multiply remaining

numerators and denominators.


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c x2 – 425

×5x + 5

x2 – x – 2

=1��(x – 2)(x + 2)

��25 5 × �51��(x + 1)1

1��(x – 2)��(x + 1)1

= x + 25

First factorise all the algebraic expressions. Note

that x2 – 4 is a difference of perfect squares. Then

cancel as normal.

Exercise 8H

1 Simplify these fractions by cancelling.

515

a 2416

b 5x10

c 42x12

d

248x

e 918x

f 3(x + 1)6

g 22(x – 4)11

h

2 Factorise these by taking out common factors.

3x + 6a 20 – 40xb x2 – 7xc 6x2 + 24xd

3 Copy and complete.

2x – 48

= 2( )8

=4

a 12 – 18x2x – 3x2

= 6( )x( )

= 6x

b

x – 1x + 3

÷ 2(x – 1)(x + 3)(x + 2)

= x – 1x + 3

×

= x + 22

c

UNDE

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—1–3 3

4Example 14a Simplify the following by cancelling.

3(x + 2)4(x + 2)

a x(x – 3)3x(x – 3)

b 20(x + 7)5(x + 7)

c

(x + 5)(x – 5)(x + 5)

d 6(x – 1)(x + 3)9(x + 3)

e 8(x – 2)4(x – 2)(x + 4)

f

5Example 14b Simplify the following by factorising and then cancelling.

5x – 55

a 4x – 1210

b 2x – 43x – 6

c 12 – 4x6 – 2x

d

x2 – 3xx

e 4x2 + 10x5x

f 3x + 3y2x + 2y

g 4x – 8y3x – 6y

h

FLUE

NCY

5–9(½)4–8(½) 4–9(½)


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8H6Example 14c Simplify the following. These expressions involve difference of perfect squares.

x2 – 100x + 10

a x2 – 49x + 7

b x2 – 25x + 5

c

2(x – 20)

x2 – 400d

5(x – 6)

x2 – 36e

3x + 27

x2 – 81f

7Example 15a Simplify the following by cancelling.

2x(x – 4)4(x + 1)

× (x + 1)x

a (x + 2)(x – 3)x – 5

× x – 5x + 2

b

x – 3x + 2

× 3(x + 4)(x + 2)x + 4

c 2(x + 3)(x + 4)(x + 1)(x – 5)

× (x + 1)4(x + 3)

d

8Example 15b Simplify the following by cancelling.

x(x + 1)x + 3

÷ x + 1x + 3

a x + 3x + 2

÷ x + 32(x – 2)

b

x – 4(x + 3)(x + 1)

÷ x – 44(x + 3)

c 4xx + 2

÷ 8xx – 2

d

3(4x – 9)(x + 2)2(x + 6)

÷ 9(x + 4)(4x – 9)4(x + 2)(x + 6)

e 5(2x – 3)(x + 7)

÷ (x + 2)(2x – 3)x + 7

f

9Ext Simplify by firstly factorising.

x2 – x – 6x – 3

a x2 + 8x + 16x + 4

b x2 – 7x + 12x – 4

c

x – 2

x2 + x – 6d

x + 7

x2 + 5x – 14e

x – 9

x2 – 19x + 90f

FLUE

NCY

10Example 15c

Ext

These expressions involve a combination of trinomials, difference of perfect squares and simple

common factors. Simplify by firstly factorising where possible.

x2 + 5x + 6x + 5

÷x + 3

x2 – 25a

x2 + 6x + 8

x2 – 9÷ x + 4x – 3

b

x2 + x – 12

x2 + 8x + 16×

x2 – 16

x2 – 8x + 16c

x2 + 12x + 35

x2 – 25×x2 – 10x + 25

x2 + 9x + 14d

9x2 – 3x

6x – 45x2e

x2 – 4x

3x – x2f

3x2 – 21x + 36

2x2 – 32×2x + 106x – 18

g2x2 – 18x + 40

x2 – x – 12×

3x + 15

4x2 – 100h

PROB

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1010(½) 10


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8H

11 The expression 5 – 2x2x – 5

can be written in the form – 1(–5 + 2x)2x – 5

= – 1(2x – 5)2x – 5

, which can be

cancelled to –1.

Use this idea to simplify these algebraic fractions.

7 – 3x3x – 7

a 4x – 11 – 4x

b 8x + 16– 2 – x

c

x + 3– 9 – 3x

d 5 – 3x18x – 30

e x2 – 93 – x

f

12Ext Just like a2

2acan be cancelled to a

2, (a + 5)2

2(a + 5)cancels to a + 5

2. Use this idea to cancel these

fractions.

(a + 1)2

(a + 1)a 5(a – 3)2

(a – 3)b 7(x + 7)2

14(x + 7)c

3(x – 1)(x + 2)2

18(x – 1)(x + 2)d x2 + 6x + 9

2x + 6 Exte 11x – 22x2 – 4x + 4 Extf

REAS

ONING

11, 1211(½) 11

All in together

13Ext Use your knowledge of factorisation and the ideas in Questions 11 and 12 above to simplify

these algebraic fractions.

2x2 – 2x – 2416 – 4x

ax2 – 14x + 49

21 – 3xb

x – 16x + 64

64 – x2c

4 – x2

x2 + x – 6×

2x + 6x + 4x + 4

d

2x2 – 18

x2 – 6x + 9×

6 – 2x

x2 + 6x + 9e

x2 – 2x + 14 – 4x

÷1 – x2

3x2 + 6x + 3f

4x2 – 9

x2 – 5x÷

6 – 4x15 – 3x

gx2 – 4x + 4

8 – 4x×

–2

4 – x2h

(x + 2)2 – 4

(1 – x)2×x2 – 2x + 13x + 12

i2(x – 3)2 – 50

x2 – 11x + 24÷x2 – 43 – x

j

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8I Simplifying algebraic fractions: addition and subtractionThe process required for adding or subtracting

algebraic fractions is similar to that used for fractions

without pronumerals.

To simplify 23+ 45, for example, you would find the

lowest common multiple of the denominators (15)

then express each fraction using this denominator.

Adding the numerators completes the task.

Let’s start: Compare the working

Here is the working for the simplification of the sum

of a pair of numerical fractions and the sum of a pair

of algebraic fractions.

Although algebraic fractions, seem abstract, performingoperations on them and simplifying them is essentialto many calculations in real-life mathematical problems.

25+ 34= 820

+ 1520

= 2320

2x5

+ 3x4

= 8x20

+ 15x20

= 23x20

• What type of steps were taken to simplify the algebraic fractions that are the same as for the numerical

fractions?

• Write down the steps required to add (or subtract) algebraic fractions.

Keyideas

To add or subtract algebraic fractions:

• determine the lowest common denominator (LCD)

• express each fraction using the LCD

• add or subtract the numerators.

Example 16 Adding and subtracting with numerals in the denominators

Simplify:x4– 2x

5a 7x

3+ x6

b x + 32

+ x – 25

c


a x4– 2x

5= 5x20

– 8x20

= – 3x20

Determine the LCD of 4 and 5, i.e. 20. Express

each fraction as an equivalent fraction with a

denominator of 20. 2x× 4 = 8x. Then subtract

numerators.


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b 7x3

+ x6= 14x

6+ x6

= 15x6

= 5x2

Note the LCD of 3 and 6 is 6 not 3 × 6 = 18.

Simplify 156

to 52in the final step.

c x + 32

+ x – 25

= 5(x + 3)10

+ 2(x – 2)10

= 5x + 15 + 2x – 410

= 7x + 1110

The LCD of 2 and 5 is 10, write as equivalent

fractions with denominator 10.

Expand the brackets and simplify the numerator

by adding and collecting like terms.

Example 17 Adding and subtracting with algebraic terms in the denominators

Simplify:

2x–

52x

a2x+

3

x2b


a 2x– 52x

= 42x

– 52x

= – 12x

The LCD of x and 2x is 2x, so rewrite the

first fraction in an equivalent form with a

denominator also of 2x.

b2x+

3

x2=

2x

x2+

3

x2

=2x + 3

x2

The LCD of x and x2 is x2 so change the first

fraction so its denominator is also x2, then add

numerators.

Exercise 8I

1 Find the lowest common multiple of these pairs of numbers.

(6, 8)a (3, 5)b (11, 13)c (12, 18)d

2 Write equivalent fractions by stating the missing expression.

2x5

=10

a 7x3

=9

b x + 14

=(x + 1)12

c

3x + 511

=(3x + 5)22

d 4x=

2xe 30

x + 1=

3(x + 1)f

UNDE

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8I3 Copy and complete these simplifications.

x4+ 2x

3=

12+

12=

12a 5x

7– 2x

5=

35–

35=

35b

x + 12

+ 2x + 34

=(x + 1)4

+2x + 3

4=

+ 2x + 3

4=

4c

4 Write down the LCD for these pairs of fractions.

x3, 2x5

a 3x7

, x2

b – 5x4

, x8

c 2x3

, – 5x6

d 7x10

, – 3x5

e

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5Example 16a,b Simplify:

x7+ x2

a x3+ x15

b x4– x8

c x9+ x5

d

y7– y8

e a2+ a11

f b3– b9

g m3– m6

h

m6+ 3m

4i a

4+ 2a

7j 2x

5+ x10

k p9– 3p

7l

b2– 7b

9m 9y

8+ 2y

5n 4x

7– x5

o 3x4

– x3

p

6Example 16c Simplify:

x + 12

+ x + 35

a x + 33

+ x – 44

b a – 27

+ a – 58

c

y + 45

+ y – 36

d m – 48

+ m + 65

e x – 212

+ x – 38

f

2b – 36

+ b + 28

g 3x + 86

+ 2x – 43

h 2y – 57

+ 3y + 214

i

2t – 18

+ t – 216

j 4 – x3

+ 2 – x7

k 2m – 14

+ m – 36

l

7Example 17a Simplify:

3x+ 52x

a 73x

– 2x

b 74x

– 52x

c 43x

+ 29x

d

34x

– 25x

e 23x

+ 15x

f – 34x

– 7x

g – 53x

– 34x

h

FLUE

NCY

5–7(½)5–6(½) 5–7(½)

8Example 17b Simplify:3x+

2

x2a

5

x2+

4x

b7x+

3

x2c

4x–

5

x2d

3

x2–8x

e –4

x2+

1x

f3x–

7

2x2g –

23x

+3

x2h

PROB

LEM-SOLVING

8–10(½)8(½) 8–9(½)


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8I9 Simplify these mixed algebraic fractions.

2x+ x4

a – 5x

+ x2

b – 2x

– 4x3

c 32x

– 5x4

d

3x4

– 56x

e 13x

– x9

f – 25x

+ 3x2

g – 54x

– 3x10

h

10 Find the missing algebraic fraction. The fraction should be in simplest form.

x2+ = 5x

6a + x

4= 3x

8b 2x

5+ = 9x

10c

2x3

– = 7x15

d – x3= 5x

9e 2x

3– = 5x

12f

PROB

LEM-SOLVING

11 Find and describe the error in each set of working. Then find the correct answer.

4x5

– x3= 3x

2a x + 1

5+ x2= 2x + 1

10+ 5x10

= 7x + 110

b

5x3

+ x – 12

= 10x6

+ 3x – 16

= 13x – 16

c2x–

3

x2=

2

x2–

3

x2

=–1

x2

d

12 A student thinks that the LCD to use when simplifying x + 12

+ 2x – 14

is 8.

a Complete the simplification using a common denominator of 8.

b Now complete the simplification using the actual LCD of 4.

c How does your working for parts a and b compare? Which method is preferable and why?RE

ASON

ING

11, 1211 11, 12

More than two fractions!

13 Simplify by finding the LCD.2x5

– 3x2

– x3

a x4– 2x

3+ 5x

6b 5x

8– 5x

6+ 3x

4c

x + 14

+ 2x – 13

– x5

d 2x – 13

– 2x7

+ x – 36

e 1 – 2x5

– 3x8

+ 3x + 12

f

23x

+ 5x– 1x

g – 12x

+ 2x– 43x

h – 45x

– 12x

+ 34x

i

4

x2+ 32x

– 53x

j 5x–

3

2x2– 57x

k2

x2– 49x

–5

3x2l

2x+ x5– x3

m 3x2

– 12x

+ x3

n – 4x9

+ 25x

+ 2x5

o

ENRICH

MEN

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8J Further simplification of algebraic fractions EXTENDING

More complex addition and subtraction of algebraic fractions involves expressions like:

2x – 13

– x + 44

and 2x – 3

–5

(x – 3)2

In such examples, care needs to be taken at each step in the working to avoid common errors.

Let’s start: Three critical errors

The following simplification of algebraic fractions has three

critical errors. Can you find them?

2x + 13

– x + 22

= 2x + 16

– 3(x + 2)6

= 2x + 1 – 3x + 66

= x + 76

The correct answer is x – 46

.

Fix the solution to produce the correct answer.

Keyideas

When combining algebraic fractions which involve subtraction signs, recall that:

• the product of two numbers of opposite sign is a negative number

• the product of two negative numbers is a positive number.

For example:2(x – 1)

6–3(x + 2)

6=

2x – 2 – 3x –6

6

and5(1 – x)

8–2(x – 1)

8=

5 – 5x – 2x +2

8

A common denominator can be a product of two binomial terms.

For example: 2x + 3

+ 3x – 1

= 2(x – 1)(x + 3)(x – 1)

+ 3(x + 3)(x + 3)(x – 1)

= 2x – 2 + 3x + 9(x + 3)(x – 1)

= 5x + 7(x + 3)(x – 1)


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Example 18 Simplifying with more complex numerators

Simplify:

x – 13

– x + 45

a 2x – 36

– 3 – x5

b


a x – 13

– x + 45

= 5(x – 1)15

– 3(x + 4)15

= 5x – 5 – 3x – 1215

= 2x – 1715

The LCD of 3 and 5 is 15. Insert brackets

around each numerator when multiplying.

Note: –3(x + 4) = –3x – 12 not –3x + 12.

b 2x – 36

– 3 – x5

= 5(2x – 3)30

– 6(3 – x)30

= 10x – 15 – 18 + 6x30

= 16x – 3330

Determine the LCD and express as equivalent

fractions. Insert brackets.

Expand the brackets, recall –6 × (–x) = 6x and

then simplify the numerator.

Example 19 Simplifying with more complex denominators

Simplify:4

x + 1+

3x – 2

a3

(x – 1)2–

2x – 1

b


a 4x + 1

+ 3x – 2

= 4(x – 2)(x + 1)(x – 2)

+ 3(x + 1)(x + 1)(x – 2)

= 4x – 8 + 3x + 3(x + 1)(x – 2)

= 7x – 5(x + 1)(x – 2)

The lowest common multiple of (x + 1) and

(x – 2) is (x + 1)(x – 2). Rewrite each fraction

as an equivalent fraction with this denominator

then add numerators.

b3

(x – 1)2–

2x – 1

=3

(x – 1)2–2(x – 1)

(x – 1)2

=3 – 2x + 2

(x – 1)2

=5 – 2x

(x – 1)2

Just like the LCD of 32 and 3 is 32, the LCD

of (x – 1)2 and x – 1 is (x – 1)2.

Remember that –2(x – 1) = –2x + 2.


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Exercise 8J

1 Expand the following.

–2(x + 3)a –5(x + 1)b –7(2 + 3x)c–3(x – 1)d –10(3 – 2x)e –16(1 – 4x)f

2 Write the LCD for these pairs of fractions.

13, 59

a 316

, 18

b 3x,

5

x2c – 5

2x, 32

d

3x – 1

, 2x + 1

e 7x – 2

, 3x + 3

f 42x – 1

, – 1x – 4

g5

(x + 1)2, 4x + 1

h

UNDE

RSTA

NDING

—1, 2 2(½)


x + 34

– x + 23

a x – 13

– x + 35

b x – 43

– x + 16

c

3 – x5

– x + 42

d 5x – 14

– 2 + x8

e 3x + 214

– x + 44

f

1 + 3x4

– 2x + 36

g 2 – x5

– 3x + 13

h 2x – 36

– 4 + x15

i

4Example 18b Simplify:

x + 53

– x – 12

a x – 45

– x – 67

b 3x – 74

– x – 12

c

5x – 97

– 2 – x3

d 3x + 24

– 5 – x10

e 9 – 4x6

– 2 – x8

f

4x + 33

– 5 – 2x9

g 2x – 14

– 1 – 3x14

h 3x – 28

– 4x – 37

i


3x – 1

+ 4x + 1

a 5x + 4

+ 2x – 3

b 3x – 2

+ 4x + 3

c

3x – 4

+ 2x + 7

d 7x + 2

– 3x + 3

e 3x + 4

– 2x – 6

f

– 1x + 5

+ 2x + 1

g – 2x – 3

– 4x – 2

h 3x – 5

– 5x – 6

i

FLUE

NCY

3–5(½)3–5(½) 3–5(½)

6Example 19b Simplify:4

(x + 1)2–

3x + 1

a2

(x + 3)2–

4x + 3

b3

x – 2+

4

(x – 2)2c

–2x – 5

+8

(x – 5)2d

–1x – 6

+3

(x – 6)2e

2

(x – 4)2–

3x – 4

f

5

(2x + 1)2+

22x + 1

g9

(3x + 2)2–

43x + 2

h4

(1 – 4x)2–

51 – 4x

i

PROB

LEM-SOLVING

6–7(½)6(½) 6–7(½)


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8J7 Simplify:

3x

(x – 1)2+

2x – 1

a3x + 23x

+712

b2x – 1

4+

2 – 3x10x

c

2xx – 5

–x

x + 1d

34 – x

–2xx – 1

e5x + 1

(x – 3)2+

xx – 3

f

3x – 7

(x – 2)2–

5x – 2

g–7x

2x + 1+

3xx + 2

hx

x + 1–

5x + 1

(x + 1)2i

PROB

LEM-SOLVING

8 One of the most common errors made when subtracting

algebraic fractions is hidden in this working shown on the

right:

a What is the error and in which step is it made?

b By correcting the error how does the answer change?

7x2

– x – 25

= 35x10

– 2(x – 2)10

= 35x – 2x – 410

= 33x – 410

9 Simplify:

1(x + 3)(x + 4)

+ 2(x + 4)(x + 5)

a 3(x + 1)(x + 2)

– 5(x + 1)(x + 4)

b

4(x – 1)(x – 3)

– 6(x – 1)(2 – x)

c 5x(x + 1)(x – 5)

– 2x – 5

d

3x – 4

+ 8x(x – 4)(3 – 2x)

e 3x(x + 4)(2x – 1)

– x(x + 4)(3x + 2)

f

10 Use the fact that a – b = –1(b – a) to help simplify these.

31 – x

– 2x – 1

a 4x5 – x

+ 3x – 5

b 27x – 3

– 73 – 7x

c

14 – 3x

+ 2x3x – 4

d – 3x5 – 3x

– 53x – 5

e 4x – 6

+ 46 – x

fRE

ASON

ING

9, 10(½)8 8, 9(½)

Factorise first

11 Factorising a denominator before further simplification is a useful step. Simplify these by

firstly factorising the denominators if possible.

3x + 2

+5

2x + 4a

73x – 3

–2

x – 1b

38x – 4

–5

1 – 2xc

4

x2 – 9–

3x + 3

d

52x + 4

+2

x2 – 4e

103x – 4

–7

9x2 – 16f

7

x2 + 7x + 12+

2

x2 – 2x – 15g

3

(x + 1)2 – 4–

2

x2 + 6x + 9h

3

x2 – 7x + 10–

210 – 5x

i1

x2 + x–

1

x2 – xj

ENRICH

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8K Equations with algebraic fractions EXTENDING

For equations with more than one fraction it is often best to try to simplify the equation by dealing with

all the denominators at once. This involves finding and multiplying both sides by the lowest common

denominator.

Let’s start: Why use the LCD?

For this equation follow each instruction.

x + 13

+ x4= 1

• Multiply every term in the equation by 3. What effect does this have on the fractions on the left-hand

side?

• Starting with the original equation, multiply every term in the equation by 4. What effect does this have

on the fractions on the left-hand side?

• Starting with the original equation, multiply every term in the equation by 12 and simplify.

Which instruction above does the best job in simplifying the algebraic fractions? Why?

Keyideas

For equations with more than one fraction multiply both sides by the lowest commondenominator (LCD).• Multiply every term on both sides, not just the fractions.

• Simplify the fractions and solve the equation using the methods learnt earlier.

Alternatively, express each fraction using the same denominator then simplify by adding or

subtracting and solve.

Example 20 Solving equations involving algebraic fractions

Solve each of the following equations.2x3

+ x2= 7a x – 2

5– x – 1

3= 1b

52x

– 43x

= 2c 3x + 1

= 2x + 4

d


a 2x3

+ x2= 7

2x�31

× �6 2 + x�21

× �6 3 = 7 × 6

4x + 3x = 42

7x = 42

x = 6

Multiply each term by the LCD (LCD of 3 and 2

is 6) and cancel.

Simplify and solve for x.


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OR 2x3

+ x2= 7

4x6

+ 3x6

= 7

7x6

= 7

7x = 42

x = 6

Alternatively, write each term on the left-hand

side using the LCD = 6.

Simplify by adding the numerators and solve the

remaining equation.

b x – 25

– x – 13

= 1

��15 3(x – 2)�51

– ��15 5(x – 1)�31

= 1 × 15

3(x – 2) – 5(x – 1) = 15

3x – 6 – 5x + 5 = 15

–2x – 1 = 15

–2x = 16

x = –8

Multiply each term on both sides by 15 (LCD of

3 and 5 is 15) and cancel.

Expand the brackets and simplify by combining

like terms. Note: –5(x – 1) = –5x + 5 not –5x – 5.

(Alternatively, write each term using the

LCD = 15 then combine the numerators and

solve. 3(x – 2)15

– 5(x – 1)15

= 1)

c 52x

– 43x

= 2

5��2x1

×��6x 3 – 4��3x1

×��6x2 = 2 × 6x

15 – 8 = 12x

7 = 12x

x = 712

LCD of 2x and 3x is 6x.

Multiply each term by 6x. Cancel and simplify.

Solve for x leaving the answer in fraction form.

(Alternative solution: 156x

– 86x

= 2)

d 3x + 1

= 2x + 4

3��(x + 1)(x + 4)��(x + 1)

= 2(x + 1)��(x + 4)��(x + 4)

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

3x + 12 = 2x + 2

x + 12 = 2

OR x = –10

3x + 1

= 2x + 4

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

3x + 12 = 2x + 2

x + 12 = 2

x = –10

Multiply each term by the common denominator

(x + 1)(x + 4).

Expand the brackets.

Subtract 2x from both sides to gather x terms on

one side then subtract 12 from both sides.

Since each side is a single fraction you can

‘cross-multiply’:3

x + 12

x + 4

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Exercise 8K

1 Write down the lowest common denominator of all the fractions in these equations.x3– 2x

5= 1a x

2– 3x

4= 3b x + 1

3– x6= 5c

2x – 17

+ 5x + 24

= 6d x3+ x2= 15

e x – 14

– 3x8

= 18

f

2 Simplify the fractions by cancelling.12x3

a 21x7

b 4(x + 3)2

c

5(2x + 5)5

d 15x5x

e –7(x + 1)(x + 2)7(x + 1)

f

36(x – 7)(x – 1)9(x – 7)

g 18(3 – 2x)(1 – x)9(3 – 2x)

h –8(2 – 3x)(2x – 1)–8(2 – 3x)

i

UNDE

RSTA

NDING

—1–2(½) 2(½)

3Example 20a Solve each of the following equations.x2+ x5= 7a x

2+ x3= 10b y

3+ y4= 14c

x2– 3x

5= –1d 5m

3– m2

= 1e 3a5

– a3= 2f

3x4

– 5x2

= 14g 8a3

– 2a5

= 34h 7b2

+ b4= 15i

4Example 20b Solve each of the following equations.x – 12

+ x + 23

= 11a b + 32

+ b – 43

= 1b n + 23

+ n – 22

= 1c

a + 15

– a + 16

= 2d x + 52

– x – 14

= 3e x + 32

– x + 13

= 2f

m + 43

– m – 44

= 3g 2a – 82

+ a + 76

= 1h 2y – 14

– y – 26

= –1i

5 Solve each of the following equations.x + 12

= x3

a x – 23

= x2

b n + 34

= n – 12

c

a + 23

= a + 12

d 3 + y2

= 2 – y3

e 2m + 44

= m + 63

f

6Example 20c Solve each of the following equations.34x

– 12x

= 4a 23x

– 12x

= 2b 42m

– 25m

= 3c

12x

– 14x

= 9d 12b

+ 1b= 2e 1

2y+ 13y

= 4f

13x

+ 12x

= 2g 23x

– 1x= 2h 7

2a– 23a

= 1i

FLUE

NCY

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8K7Example 20d Solve each of the following equations.

3x + 1

= 1x + 2

a 2x + 3

= 3x + 2

b 2x + 5

= 3x – 2

c

1x – 3

= 12x + 1

d 2x – 1

= 12x + 1

e 1x – 2

= 23x + 2

f

FLUE

NCY

8 Half of a number (x) plus one-third of twice the same number is equal to 4.

a Write an equation describing the situation.

b Solve the equation to find the number.

9 Use your combined knowledge of all the methods learnt earlier to solve these equations with

algebraic fractions.2x + 31 – x

= 4a 5x + 2x + 2

= 3b 3x – 2x – 1

= 2c

2x3

+ x – 14

= 2x – 1d3

x2– 2x= 5x

e1 – 3x

x2+ 32x

= 4x

f

x – 12

+ 3x – 24

= 2x3

g 4x + 13

– x – 36

= x + 56

h 1x + 2

– 2x – 3

= 5(x + 2)(x – 3)

i

10 Molly and Billy each have the same number

of computer games (x computer games each).

Hazel takes one-third of Molly’s computer

games and a quarter of Billy’s computer

games to give her a total of 77 computer

games.

a Write an equation describing the total

number of computer games for Hazel.

b Solve the equation to find how many

computer games Molly and Billy

each had.

PROB

LEM-SOLVING

9(½), 108 8, 9(½)

11 A common error when solving equations with

algebraic fractions is made in this working. Find the

error and explain how to avoid it.

3x – 14

+ 2x = x3

(LCD = 12)

12(3x – 1)4

+ 2x = 12x3

3(3x – 1) + 2x = 4x

9x – 3 + 2x = 4x

7x = 3

x = 37

REAS

ONING

12, 1311, 12 11, 12


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8K12 Another common error is made in this working. Find

and explain how to avoid this error.x2– 2x – 1

3= 1 (LCD = 6)

6x2

– 6(2x – 1)3

= 6

3x – 2(2x – 1) = 6

3x – 4x – 2 = 6

–x = 8

x = –8

13 Some equations with decimals can by solved by firstly multiplying by a power of 10. Here

is an example.

0.8x – 1.2 = 2.5

8x – 12 = 25 Multiply both sides by 10 to remove all decimals.

8x = 37

x = 378

Solve these decimal equations using the same idea. For parts d–f you will need to multiply

by 100.

0.4x + 1.4 = 3.2a 0.3x – 1.3 = 0.4b 0.5 – 0.2x = 0.2c1.31x – 1.8 = 2.13d 0.24x + 0.1 = 3.7e 2 – 3.25x = 8.5f

REAS

ONING

Literal equations

14 Solve each of the following equations for x in terms of the other pronumerals.

Hint: you may need to use factorisation.xa– x2a

= ba axb

– cx2

= db x – ab

= xc

c

x + ab

= d + ec

d ax + b4

= x + c3

e x + a3b

+ x – a2b

= 1f

2a – ba

+ ax= ag 1

a– 1x= 1c

h ax= bc

i

ax+ b = c

xj ax – b

x – b= ck cx + b

x + a= dl

2a + xa

= bm 1x – a

= 1ax + b

n ab– aa + x

= 1o

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InvestigationExpanding quadratics using areasConsider the expansion of the quadratic (x + 3)(x + 6). This can be represented by finding the area of the

rectangle shown.

Total area = A1 + A2 + A3 + A4

= x2 + 6x + 3x + 18

Therefore:

(x + 3)(x + 6) = x2 + 9x + 18A1

A3

A2

A4

x

x 6

3

Expanding with positive signs

a Draw a diagram and calculate the area to determine the expansion of the following quadratics.

(x + 4)(x + 5)i (x + 7)(x + 8)ii

(x + 3)2iii (x + 5)2iv

b Using the same technique establish the rule for expanding (a + b)2.

Expanding with negative signs

Consider the expansion of (x – 4)(x – 7).

Area required = total area – (A2 + A3 + A4)

= x2 – [(A2 + A4) + (A3 + A4) – A4]

= x2 – (7x + 4x – 28)

= x2 – 11x + 28

Therefore:

(x – 4)(x – 7) = x2 – 11x + 28

A1

A3

A2

A4

x

x

4

This area is counted twicewhen we add 7x + 4x.

7

a Draw a diagram and calculate the area to determine the expansion of the following quadratics.

(x – 3)(x – 5)i

(x – 6)(x – 4)ii

(x – 4)2iii

(x – 2)2iv

b Using the same technique, establish the rule for expanding (a – b)2.

Difference of perfect squares

Using a diagram to represent (a – b)(a + b), determine the appropriate area and establish a rule for the

expansion of (a – b)(a + b).


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Numerical applications of perfect squaresThe expansion and factorisation of perfect squares and difference of perfect squares can be applied to

the mental calculation of some numerical problems.

Evaluating a perfect square

The perfect square 322 can be evaluated using (a + b)2 = a2 + 2ab + b2.

322 = (30 + 2)2 (Let a = 30, b = 2)

= 302 + 2(30)(2) + 22

= 900 + 120 + 4

= 1024

a Use the same technique to evaluate these perfect squares.

222i 212ii 332iii 512iv

1.22v 3.22vi 6.12vii 9.012viii

Similarly, the perfect square 292 can be evaluated using (a – b)2 = a2 – 2ab + b2.

292 = (30 – 1)2 (Let a = 30, b = 1)

= 302 – 2(30)(1) + 12

= 900 – 60 + 1

= 841

b Use the same technique to evaluate these perfect squares.

192i 392ii 982iii 872iv

1.92v 4.72vi 8.82vii 3.962viii

Evaluating the difference of perfect squares

The difference of perfect squares 142 – 92 can be evaluated using a2 – b2 = (a + b)(a – b).

142 – 92 = (14 + 9)(14 – 9) (Let a = 14, b = 9)

= 23 × 5

= 115

a Use the same technique to evaluate these difference of perfect squares.

132 – 82i 252 – 232ii 422 – 412iii 852 – 832iv

1.42 – 1.32v 4.92 – 4.72vi 10012 – 10002vii 2.012 – 1.992viii

The expansion (a + b)(a – b) = a2 – b2 can also be used to evaluate some products. Here is

an example:

31 × 29 = (30 + 1)(30 – 1) (Let a = 30, b = 1)

= 302 – 12

= 900 – 1

= 899

b Use the same technique to evaluate these products.

21 × 19i 32 × 28ii 63 × 57iii 105 × 95iv

2.1 × 1.9v 7.4 × 6.6vi 520 × 480vii 915 × 885viii


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Problems and challengesUp for a challenge? If you get stuck on a question,

check out the 'Working with Unfamiliar Questions' poster

at the end of the book to help you.

1 a The difference between the squares of two consecutive numbers is 97.

What are the two numbers?

b The difference between the squares of two consecutive odd numbers is 136.


c The difference between the squares of two consecutive multiples of 3 is 81.


2 a If x2 + y2 = 6 and (x + y)2 = 36, find the value of xy.

b If x + y = 10 and xy = 2, find the value of 1x+ 1y.

3 Find the values of the different digits a, b, c and d if the four digit number abcd× 4 = dcba.

4 a Find the quadratic rule that relates the width n to the number of matches in the pattern below.

n = 1 n = 2 n = 3

b Draw a possible pattern for these rules.

n2 + 3i

n(n – 1)ii

5 Factorise n2 – 1 and use the factorised form to explain why when n is prime and greater than 3,

n2 – 1 is:

divisible by 4i

divisible by 3ii

thus divisible by 12.iii

6 Prove that this expression is equal to 1.

2x2 – 8

5x2 – 5÷ x – 2

5x – 5÷2x2 – 10x – 28

x2 – 6x – 7

7 Prove that 4x2 – 4x + 1 ≥ 0 for all x.

8 In a race over 4 km Ryan ran at a constant speed. Sophie, however, ran the first 2 km

at a speed 1 km/h more than Ryan and ran the second 2 km at a speed 1 km/h less

than Ryan. Who won the race?


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Solving equations with algebraic fractions (Ext)

Find lowest common denominator(LCD) and multiply every termby the LCD.e.g. LCD of 2x and 3x is 6x LCD of (x − 1) and (x + 3) is (x − 1)(x + 3) LCD of 2x and x2 is 2x2

Factorisation

The reverse process of expansion.Always remove the highest commonfactor first.e.g. 2ab + 8b HCF = 2b = 2b (a + 4)e.g. 3(x + 2) + y (x + 2) HCF = x + 2 = (x + 2)(3 + y )

By grouping

If there are four terms, we may be able to group into two binomial terms.e.g. 5x + 10 − 3x2 − 6x

Monic quadratic trinomials (Ext)

These are of the form:x2 + bx + c

Require two numbers that multiplyto c and add to be.g. x2 − 7x −18 = (x − 9)(x + 2)since −9 × 2 = −18and −9 + 2 = −7

Trinomials of the form ax 2 + bx + c (Ext)

Can be factorised using grouping alsoe.g. 3x 2 + 7x − 6, 3 × −6 = −18 = 3x 2 + 9x − 2x − 6, Use 9 and –2= 3x (x + 3) − 2(x + 3) since= (x + 3)(3x − 2) 9 × (−2) = −18 and 9 + (−2) = 7

Special cases

DOPS (difference of perfect squares)

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

e.g. (x −− −

−

−

−

−

5)(x + 5) = x2 + 5x 5x 25 = x2 − 25Perfect squares

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

(a b)2 = a2 2ab + b2

e.g. (2x + 3)2 = (2x + 3)(2x + 3) = 4x2 + 6x + 6x + 9 = 4x2 + 12x + 9

Expansion

The process of removing brackets.

e.g. 2(x + 5) = 2x + 10

(x + 3)(2x − 4) = x(2x − 4) + 3(2x − 4) = 2x 2 − 4x + 6x − 12 = 2x 2 + 2x − 12

2

5

x

Algebraic techniques

2xx − 7

35x

7(x − 1)2

Algebraic fractions

These involve algebraic expressionsin the numerator and/or denominator.

e.g. , ,

DOPS

÷

×=

=

Multiply/divide

To divide, multiply by the reciprocal.Factorise all expressions, canceland then multiply.

3x2 − 9

152x + 6

3(x − 3)(x + 3)

2(x + 3)15

e.g.

1

5

1

1

25(x − 3)

Add/subtract

Must find lowest commondenominator (LCD) beforeapplying operation.

=

3x4

9x + 4x − 412

= +9x 12

4(x − 1) 12

= 13x − 412

e.g. + x −13

a2 − b2 = (a – b)(a + b)e.g. x2 − 16 = x2 − 42

e.g. 4x2 − 9 = (2x )2 − 32

= (x – 4)(x + 4)

= (2x − 3)(2x + 3)

= 5(x + 2) − 3x (x + 2)= (x + 2)(5 − 3x )


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Multiple-choice questions

138pt8A (2x – 1)(x + 5) in expanded and simplified form is:

2x2 + 9x – 5A x2 + 11x – 5B 4x2 – 5C3x2 – 2x + 5D 2x2 + 4x – 5E

238pt8B (3a + 2b)2 is equivalent to:

9a2 + 6ab + 4b2A 9a2 + 4b2B 3a2 + 6ab + 2b2C3a2 + 12ab + 2b2D 9a2 + 12ab + 4b2E

338pt8D 16x2 – 49 in factorised form is:

(4x – 7)2A (16x – 49)(16x + 49)B (2x – 7)(8x + 7)C(4x – 7)(4x + 7)D 4(4x2 – 49)E

438pt8E The factorised form of x2 + 3x – 2x – 6 is:

(x – 3)(x + 2)A x – 2(x + 3)2B (x + 3)(x – 2)Cx – 2(x + 3)D x(x + 3) – 2E

538pt8F If (x – 2) is a factor of x2 + 5x – 14 the other factor is:

ExtxA x + 7B x – 7C x – 16D x + 5E

638pt8G The factorised form of 3x2 + 10x – 8 is:

AExt

(3x + 1)(x – 8)

B (x – 4)(3x + 2)

C (3x + 2)(x + 5)

D (3x – 2)(x + 4)

E (x + 1)(3x – 8)

738pt8H The simplified form of 3x + 6(x + 5)(x + 1)

× x + 5x + 2

is:

3x + 1

A15

2x2 + 5B x + 3C 5

x + 2D 3(x + 5)E

838pt8I x + 25

+ 2x – 13

written as a single fraction is:

11x + 115

A 11x + 98

B 3x + 18

C 11x + 715

D 13x + 115

E

938pt8J The LCD of 3x + 12x

and 4x + 1

is:

Ext 8xA 2x(x + 1)B (x + 1)(3x + 1)C8x(3x + 1)D x(x + 1)E

1038pt8K The solution to 31 – x

= 42x + 3

is:

Ext x = – 12

A x = – 911

B x = 17

C x = 2D x = – 45

E


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Short-answer questions

138pt8A Expand the following binomial products.

(x – 3)(x + 4)a (x – 7)(x – 2)b(2x – 3)(3x + 2)c 3(x – 1)(3x + 4)d

238pt8B Expand the following.

(x + 3)2a (x – 4)2b (3x – 2)2c(x – 5)(x + 5)d (7 – x)(7 + x)e (11x – 4)(11x + 4)f

338pt8C Write the following in fully factorised form by removing the highest common factor.

4a + 12ba 6x – 9x2b –5x2y – 10xyc3(x – 7) + x(x – 7)d x(2x + 1) – (2x + 1)e (x – 2)2 – 4(x – 2)f

438pt8D Factorise the following DOPS.

x2 – 100a 3x2 – 48b 25x2 – y2c49 – 9x2d (x – 3)2 – 81e 1 – x2f

538pt8E Factorise the following by grouping.

a x2 – 3x + 6x – 18

b 4x2 + 10x – 2x – 5

c 3x – 8b + 2bx – 12

638pt8F/G Factorise the following trinomials.

Extx2 + 8x + 15a x2 – 3x – 18b x2 – 7x + 6c 3x2 + 15x – 42d2x2 + 16x + 32e 5x2 + 17x + 6f 4x2 – 4x – 3g 6x2 – 17x + 12h

738pt8H Simplify the following.

3x + 123

a 2x – 163x – 24

b x2 – 95(x + 3)

c

838pt8H Simplify the following algebraic fractions by first factorising and cancelling where possible.

32x

× x6

a x(x – 4)8(x + 1)

× 4(x + 1)x

b x2 + 3x3x + 6

× x + 2x + 3

c

2x5x + 20

÷ xx + 4

d4x2 – 9

10x2÷10x – 15

xe Ext

x2 + 5x + 6x + 3

÷x2 – 44x – 8

f

938pt8I/J Simplify the following by first finding the lowest common denominator.

x4+ 2x

3a Ext

x – 16

– x + 38

b 34x

+ 12x

c

7x–

2

x2d Ext

3x + 1

+ 5x + 2

e Ext7

(x – 4)2– 2x – 4

f

1038pt8K Solve the following equations involving fractions.

Extx4+ 2x

5= 13a 4

x– 23x

= 20b

x + 32

+ x – 43

= 6c 41 – x

= 5x + 4

d


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Extended-response questions

1 A pig pen for a small farm is being redesigned. It is originally a square of side length x m.

a In the planning the length is initially kept as x m and the width altered such that the area of the

pen is (x2 + 3x) square metres. What is the new width?

b Instead, it is determined that the original length will be increased by 1 metre and the original

width will be decreased by 1 metre.

What effect does this have on the perimeter of the pig pen compared with the original size?iDetermine an expression for the new area of the pig pen in expanded form. How does this

compare to the original area?

ii

c The final set of dimensions requires an extra 8 m

of fencing to go around the pen compared with

the original pen. If the length of the pen has been

increased by 7 m, then the width of the pen must

decrease. Find:

the change that has been made to the width of

the pen

i

the new area enclosed by the peniiwhat happens when x = 3.iii

2 The security tower for a palace is on a small square

piece of land 20 m by 20 m with a moat of width

x metres the whole way around it as shown.

a State the area of the piece of land.

20 m

Land

x m

b i Give expressions for the length and the width of

the combined moat and land.

ii Find an expression, in expanded form, for the

entire area occupied by the moat and the land.

c If the tower occupies an area of (x + 10)2 m2, what fraction of the total area in part b ii is this?d Use your answers to parts a and b to give an expression for the area occupied by the moat alone,

in factorised form.

e Use trial and error to find the value of x such that the area of the moat alone is 500 m2.


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