Number and algebraProducts and factorsweb2.hunterspt-h.schools.nsw.edu.au/studentshared/MATHEMATICS/Year... · a 2a 3 3b 3 5c b 4p 3 3 3 2q c 40n2 4 5n d 8b2 4 40b2 e 6h 3 2k 3 (

3Number and algebra

Productsand factorsAlgebra is the branch of mathematics that uses symbols andformulas to describe number patterns and relationships inour natural and physical world. In 825 CE, the Persianmathematician al-Khwarizmi used the Arabic word ‘al-jabr’to describe the process of adding equal quantities to bothsides of an equation. When al-Khwarizmi’s book wastranslated into Latin and introduced to Europe, ‘al-jabr’became ‘algebra’ and the word was adopted as the name forthe whole subject.

n Chapter outlineProficiency strands

3-01 Adding and subtracting terms U F R C3-02 Multiplying and dividing terms U F R C3-03 Adding and subtracting algebraic

fractions U F R C3-04 Multiplying and dividing algebraic

fractions U F R C3-05 Expanding expressions U F R C3-06 Factorising expressions U F R C3-07 Expanding binomial products U F R C3-08 Perfect squares* U F R C3-09 Difference of two squares* U F R C3-10 Mixed expansions* F R C3-11 Factorising special binomial

products* U F R C3-12 Factorising quadratic expressions* U F R C3-13 Factorising quadratic expressions

of the form ax 2 þ bx þ c* U F R C3-14 Mixed factorisations* F R C3-15 Factorising algebraic fractions* U F R C

*STAGE 5.3

nWordbankbinomial An algebraic expression that consists of twoterms, for example, 4a þ 9, 3 � y, x2 � 4x

factorise To rewrite an expression with grouping symbols,by taking out the highest common factor; factorising is theopposite of expanding; for example, 9r2 þ 36r factorisedis 9r(r þ 4)

highest common factor (HCF) The largest term that isa factor of two or more terms, for example, the HCF of9r2 þ 36r is 9r

perfect square A square number or an algebraicexpression that represents one, for example, 64, (x þ 9) 2

quadratic expression An algebraic expression in whichthe highest power of the variable is 2, for example,2x2 þ 5x � 3 or x2 þ 2

Shut

ters

tock

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mac

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9780170193085

NEW CENTURY MATHS ADVANCEDfor the A u s t r a l i a n C u r r i c u l um9

n In this chapter you will:• simplify algebraic expressions involving the four operations• apply the distributive law to the expansion of algebraic expressions, including binomials, and

collect like terms where appropriate• add, subtract, multiply and divide simple algebraic fractions• factorise algebraic expressions• (STAGE 5.3) expand binomial products• (STAGE 5.3) factorise quadratic expressions• (STAGE 5.3) factorise and simplify expressions involving algebraic fractions

SkillCheck

1 Write an algebraic expression for each statement.

a The cost of p books at $Q each.b The selling price after a GST of $t is added to the marked price of $40.c The number of metres in h km.d The average of m, p, v and w.e The number of dollars in y cents.

2 If f ¼ 5, g ¼ �4 and r ¼ 2, then evaluate each expression.

a 2f þ g b 8 þ rf c f 2 � r2

d 3g2 � 5 e r(f � g) f 5r þ f3

3 If the perimeter of a rectangle with length l and width w has the formula P ¼ 2l þ 2w, findthe perimeter of a rectangle with length 4.7 cm and width 2.5 cm.

4 Simplify each expression.

a 5n þ 6n b 9h � 3h c 10pq � qp d x2 þ 4x2

e 4 3 w f 8 3 5x g 2g 3 3h h 3y 3 6y

i 10x 4 2 j 6b 4 b k 16xy8x

l 2n2

4n

5 Find the highest common factor (HCF) of each pair of numbers.

a 8 and 12 b 18 and 27 c 45 and 30

6 Evaluate each expression.

a 23 3

34 b 5

8�14 c 2

5 478 d 3

8þ45

Worksheet

StartUp assignment 3

MAT09NAWK10025

Puzzle sheet

Generalised arithmetic

MAT09NAPS10026

Skillsheet

Algebraic expressions

MAT09NASS10012

Puzzle sheet

Substitution puzzle

MAT09NAPS10027

Chapter 1 2 3 4 5 6 7 8 9 10 11 12 13

Products and factors

978017019308568

3-01 Adding and subtracting terms

Summary

Only like terms can be added or subtracted.

For example, 2x � 5y cannot be simplified because x and y represent different numbers.

Example 1

Simplify each expression.

a 5r þ 4s þ 7r � 2s b 8u2 � 4u þ 3u � 7u2

Solutiona 5r þ 4sþ 7r � 2s ¼ 5r þ 7r þ 4s� 2s

¼ 12r þ 2s

Group the like terms.

12r and 2s cannot be simplified further.

b 8u 2 � 4uþ 3u� 7u 2 ¼ 8u 2 � 7u 2 � 4uþ 3u

¼ u 2 � u

Example 2

Write a simplified algebraic expression for the perimeterof this rectangle. 2x

x + 3SolutionPerimeter ¼ xþ 3þ 2xþ xþ 3þ 2x

¼ xþ 2xþ xþ 2xþ 3þ 3

¼ 6xþ 6

Adding the lengths of the 4 sides.Group the like terms.

Exercise 3-01 Adding and subtracting terms1 Simplify each algebraic expression.

a 2fg þ 3fg � 4fg b 6uvw � uvw þ 4uvw c 3t þ 3 þ 5t

d 3x þ 2y þ 8y e 6g þ h � 4g f 2fg þ 3f � 4fg

g 7p2 � 3 � 2p2 h 4p � q � 3q i �8r þ 6r þ 3j �3a þ 4b � 7a k 4v þ 3 þ 2v þ 7 l 2y þ 6x þ y � 3x

m 7p þ 8q � p � q n 8 � 3w þ 7 � 2w o 2n � 3 � 12n þ 5p 8e � 3f � 2e � 4f q 2l2 þ 2l þ 3l2 � 6l r 9b � 12b2 þ 6b2 � 15b

2 Simplify 15a2 þ 12a þ 48 � 18a. Select the correct answer A, B, C or D.

A 24a þ 48 B 27a2 þ 30a C 15a2 � 6a þ 48 D 15a2 þ 30a � 48

Skillsheet

Algebra using diagrams

MAT09NASS10013

Each þ and � sign belongs tothe term that follows them.

See Example 1


9780170193085 69

3 Use the substitutions a ¼ 2, b ¼ 3 and c ¼ 4 to test whether each equation is correct orincorrect.

a 20c � 12c ¼ 8c b 9b þ b ¼ 9b2

c 7a þ 3b ¼ 10 þ a þ b d 3a2 þ 5a2 ¼ 8a2

e 4ab þ 3ab ¼ 7ab f 12c � c ¼ 12g 3ac � 5ac ¼ �2ac h 5b þ 2b2 ¼ 9b

i 2c � b þ 3b � c ¼ c þ 2b j 2a þ 2b ¼ 2(a þ b)

4 Write a simplified algebraic expression for the perimeter of each shape.

4d 4d

4d

4d

b

e

40h

9h

41h

2x2x

5y

c

f

2a

5c

9c

4a

4c

6a

2m 2m

3p

3pa

d

2p

p + 2

p + 2

2p − 1

5 Draw a triangle and write algebraic expressions for its side lengths so that it has a perimeter of7d þ 6.

6 Draw a rectangle and find algebraic expressions for its length and width so that it has aperimeter of 18a � 30.

3-02 Multiplying and dividing terms

Summary

When multiplying and dividing terms, multiply or divide the numbers first, then thevariables.

Example 3


a 2 3 5n 3 4m b 3d 3 4d c 2g 3 3h 3 (�5f)

Solutiona 2 3 5n 3 4m ¼ 2 3 5 3 4 3 n 3 m

¼ 40mn

Multiply the numbers and multiply the variables.

Remember to write the variables in alphabeticalorder.

See Example 2

Worksheet

Perimeter and area

MAT09NAWK10028

Homework sheet

Algebra 1

MAT09NAHS10009

Chapter 1 2 3 4 5 6 7 8 9 10 11 12 13


978017019308570

b 3d 3 4d ¼ 3 3 4 3 d 3 d

¼ 12d2 d 3 d ¼ d2

c 2g 3 3h 3 �5fð Þ ¼ 2 3 3 3 �5ð Þ3 g 3 h 3 f

¼ �30fgh Remember to write the variables inalphabetical order.

Example 4


a 15abc3ab

b a2b4ab

c 27cf 4 (�3fc) d 5p2 4 15p

Solutiona 15abc

3ab¼ 15

36 a6 bc6 a6 b

¼ 5c

Divide the numbers and divide thevariables.

b a2b4ab¼ aab

4ab

¼ 14

a6 a6 b6 a6 b

¼ a4

a2 ¼ a 3 a ¼ aa

Divide the numbers and divide thevariables.

c 27cf 4 ð�3fcÞ ¼ 27cf�3fc

¼ 27�36 c 6 f6 c 6 f

¼ �9

The variables divide by themselvesto give 1.

d 5p2 4 15p ¼ 5pp15p

¼ 515

p 6 p6 p

¼ 13 p

¼ p3

p2 ¼ pp

Exercise 3-02 Multiplying and dividing terms1 Simplify each expression.

a 5 3 8x b 5n 3 3 c b 3 5a

d t 3 6t e �4 3 3w f 2 3 (�3p)g 4d 3 3e h 6v 3 4w i 8q 3 8q

j (�5a) 3 (�4a) k 2y 3 3 3 4b l �6w 3 3w 3 2m (5n) 2 n (�3r)2 o (at)2

We can cross out the ab in thenumerator and denominator

because aa ¼ 1 and b

b¼ 1

See Example 3


9780170193085 71


a 12x3 b 15y

y c 36a4a

d 18mn 4 6m e 8vw 4 48vw f 64abc16b

g 18g�3g

h �25yz 4 5y i 9p�45pq

j �75xyz�25x

k 10r2 4 40r l �12hk2 4 2hk

3 Simplify9yx2

24xy2. Select the correct answer A, B, C or D.

A 3x2

8y2 B 3x3

8y3 C 3x8y

D 3y8x


a 2a 3 3b 3 5c b 4p 3 3 3 2q c 40n2 4 5n

d 8b2 4 40b2 e 6h 3 2k 3 (�3) f �2 3 5k 3 (�4k)

g 6d12 h 7r

49ri (�33m) 4 3

j �9w�45w

k 8ab 4 2bc l 2ef 3 (�3fe)

5 Use the substitutions p ¼ 5, q ¼ 4 and r ¼ 3 to test whether each equation is correct orincorrect.

a 3 3 4p ¼ 12p b q þ q ¼ q2

c 2r 3 3r ¼ 6r2 d 3q2 3 5q2 ¼ 15q2

e 12p3p¼ 4 f 7qr

qr ¼ 7

g 3r 3 5q ¼ 15qr h p 3 q 3 q ¼ 2pq

i 14r2

7r¼ 2r j 5pq

10p¼ q

2

6 Write a simplified algebraic expression for the area of each shape.

a

2d

3a

4p

6y

d

b

5m

5m

3a

2a

e

c

5h

k

c

3c4k

2kf

See Example 4

Worked solutions

Multiplying anddividing terms

MAT09NAWS10009

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7 Find an algebraic expression for the volume of each rectangular prism.

2m

2m

2m

4y

5d

3t

6k2w

wa b c

3-03Adding and subtracting algebraicfractions

Summary

To add or subtract fractions, convert them (if needed) so that they have the samedenominator, then simply add or subtract the numerators.

Example 5


a r3þ

r3 b 7m

10 �3m10 c k

3þk2

d 3h4 þ

2h5 e 5m

6� 2m

3 f 34þ

7a10

Solution

a r3þ

r3 ¼

2r3

b 7m10 �

3m10 ¼

4m10

¼ 2m5

c k3þ

k2 ¼

2 3 k2 3 3þ

3 3 k3 3 2

¼ 2k6þ 3k

6

¼ 5k6

Common denominator ¼ 2 3 3 ¼ 6

Puzzle sheet

Algebraic fractions

MAT09NAPS00010

Animated example

Adding and subtractingfractions

MAT09NAAE00004

Video tutorial

Adding and subtractingalgebraic fractions

MAT09NAVT10004

9780170193085


73

d 3h4 þ

2h5 ¼

5 3 3h5 3 4 þ

4 3 2h4 3 5

¼ 15h20 þ

8h20

¼ 23h20

e 5m6� 2m

3 ¼15m18 �

12m18

¼ 3m18

¼ m6

or 5m6� 2m

3 ¼5m6� 4m

6¼ m

6

The lowest common denominator is 6.

f 34þ

7a10 ¼

1520þ

14a20

¼ 15þ 14a20

Exercise 3-03 Adding and subtracting algebraicfractions


a w4 þ

2w4 b 4k

8 þ7k8 c 7m

10 �2m10 d x

3þ2x3

e 4qþ

5q f 5

3d� 2

3dg 4r

w þrw h 5

c �2c

i 4t3 þ

s3 j 11y

2h� 7y

2hk 6

a�5a l 8

5eþ 7

5e

m pz þ

3z n 5u

8g� 3u

8go 4

9f� 1

9fp 7e

10�3e10

q 5a6þ 5a

6r 5

2d� 1

2ds 7

5kþ 13

5kt 12

4a� 8

4a


a x3þ

x4 b s

3�s7 c h

5þh3 d m

7 �m2

e w4 þ

w5 f 5t

4 �2t5 g 2p

5 þp3 h 5r

2 �5r3

i 3c2 �

c5 j 2d

11 �r3 k 3h

5 þ2a3 l 5

6þ 4w

5

m 34þ

2a7 n 7e

8 �2e3 o m

2 �n7 p 2k

5 þm6

See Example 5

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3-04Multiplying and dividing algebraicfractions

Summary

• To multiply fractions, cancel any common factors, then multiply the numerators anddenominators separately

• To divide by a fraction ab, multiply by its reciprocal b

a

Example 6

Simplify each product.

a 3d

34c b 4

k3

3k16

c 52r

38r11

Solution

a 3d

34c ¼

3 3 4d 3 c

¼ 12dc

b 4k

33k16¼

1 6 41 6 k

336 k1

164

¼ 34

c 52r

38r11 ¼

512r

38r 4

11

¼ 2011

¼ 1 911

Example 7

Simplify each quotient.

a 3h

44k

b xy5 4

3x25

Solution

a 3h

44k¼ 3

h3

k4

¼ 3k4h

b xy5 4

3x25 ¼

1 6 xy

1 6 53

255

36 x1

¼ 5y3

Puzzle sheet

Algebraic fractionspuzzle

MAT09NAPS10029

Puzzle sheet

Upside-down fractions

MAT09NAPS00018

9780170193085


75

Exercise 3-04 Multiplying and dividing algebraicfractions

1 Simplify each product.

a w3 3

12 b s

5 3t4 c 3

h3

56k

d 4m 3

3n e l

3 35f

f 1v 3

23v

g 2x 3

3x h a

b3

cd

i de 3

eg

j 4ad9 3

d16a

k 5p4 3

815p

l 4ak

33a5k

m 23 3

9u10 n u

3 33u o 3z

r 32r9dz

2 Simplify each quotient.

a r2 4

r5 b m

64

n3 c h

2 4h8

d q4 4

34 e 3y

5 42yd

f 4t9 4

3t5

g 3e 4

5e h 3

2a4

b6a

i 5mn 4

2m3n

j hk

4kh

k 8w3x

42w9x

l 3s4 4

6s11

m t3 4

3t5u

n 3e7g

4e

14go xh

5 43h15


a 3p2 4

p2

4 b 2w7 4

5w6

c 8y5 3

332y2

d 52g

42g e xy

z 35y f a

b3

cb

3b3a

g 5 435b

h 4t9 4 3 i 5mn

2d3

4dn 3

115mn

j 5p 47

10ptk s

3 45s2 3

3s7 l 7

h3

142h

m 5tyk

35ky

t n 6r 3

5r9 4

15yh

o 6cfq 4

5qf

410ca

Mental skills 3A Maths without calculators

Multiplying and dividing by 5, 15, 25 and 50It is easier to multiply or divide a number by 10 than by 5. So whenever we multiply ordivide a number by 5, we can double the 5 (to make 10) and then adjust the first number.

1 Study each example.

a To multiply by 5, halve the number, then multiply by 10.

18 3 5 ¼ 18 312 3 10 ðor 9 3 2 3 10Þ

¼ 9 3 10

¼ 90

See Example 6

See Example 7

Worked solutions

Multiplying and dividingalgebraic fractions

MAT09NAWS100010

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76

b To multiply by 50, halve the number, then multiply by 100.

26 3 50 ¼ 26 312 3 100 ðor 13 3 2 3 100Þ

¼ 13 3 100

¼ 1300c To multiply by 25, quarter the number, then multiply by 100.

44 3 25 ¼ 44 314 3 100 ðor 11 3 4 3 25Þ

¼ 11 3 100

¼ 1100d To multiply by 15, halve the number, then multiply by 30.

8 3 15 ¼ 8 312 3 30 ðor 4 3 2 3 15Þ

¼ 4 3 30

¼ 120e To divide by 5, divide by 10 and double the answer. We do this because there are two

5s in every 10.

140 4 5 ¼ 140 4 10 3 2

¼ 14 3 2

¼ 28f To divide by 50, divide by 100 and double the answer. This is because there are two

50s in every 100.

400 4 50 ¼ 400 4 100 3 2

¼ 4 3 2

¼ 8

g To divide by 25, divide by 100 and multiply the answer by 4. This is because there arefour 25s in every 100.

600 4 25 ¼ 600 4 100 3 4

¼ 6 3 4

¼ 24

h To divide by 15, divide by 30 and double the answer. This is because there are two 15sin every 30.

240 4 15 ¼ 240 4 30 3 2

¼ 8 3 2

¼ 16

2 Now evaluate each expression.

a 32 3 5 b 14 3 5 c 48 3 5 d 18 3 50e 52 3 50 f 36 3 25 g 28 3 5 h 12 3 25i 12 3 15 j 22 3 35 k 90 4 5 l 170 4 5m 230 4 5 n 1300 4 50 o 900 4 50 p 300 4 25q 1000 4 25 r 360 4 45 s 210 4 15 t 360 4 15

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3-05 Expanding expressions

Summary

The distributive law for expanding an expressionMultiply each term inside the brackets by the term outside

a(b þ c) ¼ ab þ ac

a(b � c) ¼ ab � ac

Example 8

Expand each expression.

a 4(5 þ d) b 2w(3w þ 10)

Solution

a 4(5 + d) = 4 × 5 + 4 × d = 20 + 4d

a(b þ c) ¼ ab þ ac

b 2w(3w + 10) = 2w × 3w + 2w × 10 = 6w2 + 20w

Example 9


a 3g(h � 2) b r(10 � 4r)

Solution

a 3g(h – 2) = 3g × h – 3g × 2 = 3gh – 6g

a(b � c) ¼ ab � ac

b r(10 – 4r) = r × 10 – r × 4r = 10r – 4r2

Skillsheet

Algebra using diagrams

MAT09NASS10013

Worksheet

Algebra review

MAT09NAWK10034

multiply each term inside thebrackets by the term outside

multiply each term inside thebrackets by the term outside

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Example 10


a �(y þ 4) b �3(2m � 7)

Solution

a –(y + 4) = –1(y + 4) = –1 × y + (–1) × 4 = –y + (–4) = –y – 4

�( ) is the same as �1 3 ( )

b –3(2m – 7) = –3 × 2m – (–3) × 7 = –6m – (–21) = –6m + 21

Example 11

Expand and simplify by collecting like terms.

a 4(3p þ 2) � 5p b 7(n � 3) þ n(n � 1)

Solution

a 4(3p + 2) – 5p = 4 × 3p + 4 × 2 – 5p = 12p + 8 – 5p = 7p + 8

Expanding

Collecting like terms to simplify.

b 7(n – 3) + n(n – 1) = 7 × n – 7 × 3 + n × n – n × 1 = 7n – 21 + n2 – n = n2 + 7n – n – 21 = n2 + 6n – 21

It’s neater to place n2 at the front.Collecting like terms to simplify.

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Exercise 3-05 Expanding expressions1 Expand each expression.

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

2 Expand each expression.

a 3(t � 2) b 7(5 � d) c 8(8g � 3)d w(w � 7) e a(a � 1) f 6h(3h � 1)g 4x(3x � 1) h 5x(2x � 4y) i 3a(4b � 7a)

3 Use the substitution x ¼ 5 to test whether each equation is correct or incorrect.

a 3(x þ 3) ¼ 3x þ 6 b 7(x � 2) ¼ 7x � 14 c x(4 � 2x) ¼ 4x � 2x2


a �(a � 5) b �(a þ 5) c �2(x þ 6)d �2(x � 6) e �(11 þ w) f �(11 � w)g �y(y þ 9) h �x(8 � z) i �4t(t � 8)

5 Expand �5u(8 þ 2u). Select the correct answer A, B, C or D.

A 3u � 3u2 B �40u þ 10u2 C �40u � 10u2 D �40u � 3u2

6 Expand and simplify by collecting like terms.

a 5(3m þ 2) þ 4m b 3(1 � 5e) þ 6e

c 4w � 2(5 þ 2w) d 8 � 5(2x � 7)e t(t þ 4) þ 3(t þ 4) f 4(3 þ h) þ h(7 � 2h)g 3x(2x þ 5) þ 4(2x þ 5) h v(2v þ 3) � 6(v þ 1)i 3(1 � 2w) � w(2 � w) j 2y(3y � 7) � 5(3y � 7)

3-06 Factorising expressions

Summary

The highest common factor (HCF) of two or more terms is the largest term that is a factorof all of the terms. To find the HCF of algebraic terms:

• find the HCF of the numbers• find the HCF of the variables• multiply them together

See Example 8

See Example 9

Worked solutions

Expanding expressions

MAT09NAWS10011

See Example 10

See Example 11

Skillsheet

Factorising usingdiagrams

MAT09NASS10015

Homework sheet

Algebra 2

MAT09NAHS10010

Puzzle sheet

Factorising puzzle

MAT09NAPS10030

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Example 12

Find the highest common factor (HCF) of each pair of terms.

a 20x2 and 15xy b 16 and 12b

SolutionFind the HCF of the numbers and the HCF of the variables. Their product is the HCF of theexpression.

a The HCF of 20 and 15 is 5.

The HCF of x2 and xy is x, since it is the ‘largest’ common part of x2 and xy.[ The HCF of 20x2 and 15xy ¼ 5 3 x ¼ 5x

b The HCF of 16 and 12 is 4.There is no HCF of the variables because b is in 12b but not in 16.[ The HCF of 16 and 12b is 4.

Factorising algebraic expressionsWhen 4(2y þ 5) is expanded, the answer is 8y þ 20.Factorising is the reverse of expanding. Factorisation breaksan expression into factors. To factorise 8y þ 20, we takeout the greatest common divisor and insert brackets. Theresult is 4(2y þ 5). The factors are 4 and (2y þ 5).

Expand(remove brackets)

4(2y + 5) 8y + 20

Factorise(insert brackets)

Summary

Factorising an expression• Find the HCF of the terms and write it outside the brackets• Divide each term by the HCF and write the answers inside the brackets

ab þ ac ¼ a(b þ c)ab � ac ¼ a(b � c)

• To check that the factorised answer is correct, expand it

Skillsheet

HCF by factor trees

MAT09NASS10014

Skillsheet

Factorising usingdiagrams

MAT09NASS10015

9780170193085


81

Example 13

Factorise each expression.

a 8y þ 16 b 25b2 � 20ab c v(4 þ w) þ 2(4 þ w)

Solutiona The HCF of 8y and 16 is 8.

) 8yþ 16 ¼ 8 3 yþ 8 3 2

¼ 8 yþ 2ð ÞRewrite the expression using the HCF 8.

Write the HCF at the front of the brackets.

b The HCF of 25b2 and 20ab is 5b.

) 25b2 � 20ab ¼ 5b 3 5b� 5b 3 4a

¼ 5b 5b� 4að ÞRewrite the expression using the HCF 5b.

Write the HCF at the front of the brackets.

c The HCF of v(4 þ w) þ 2(4 þ w) is(4 þ w).) v 4þ wð Þ þ 2 4þ wð Þ ¼ 4þ wð Þ3 vþ 4þ wð Þ3 2

¼ 4þ wð Þ vþ 2ð Þ

Factorising with negative terms

Example 14


a �x2 þ 3x b �a � ab

SolutionWhen factorising expressions that begin with a negative term, we use the ‘negative’ HCF.

a The greatest ‘negative’ common divisor of �x2 and 3x is �x.

) �x2 þ 3x ¼ �xð Þ3 xþ �xð Þ3 �3ð Þ¼ �xð Þ xþ �3ð Þð Þ¼ �x x� 3ð Þ

(�x) 3 (�3) ¼ þ3x

b The greatest ‘negative’ common divisor of �a and ab is �a.) �a� ab ¼ �a 3 1þ �að Þ3 b

¼ �a 1þ bð Þ(�a) 3 b ¼ �ab

Video tutorial

Factorising expressions

MAT09NAVT10005

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82

Summary

Factorising with negative terms• Find the ‘negative’ HCF of the terms and write it outside the brackets• Divide each term by the HCF and write the answers inside the brackets�ab � ac ¼ �a(b þ c)�ab þ ac ¼ �a(b � c)

• To check that the factorised answer is correct, expand it

Exercise 3-06 Factorising expressions1 Find the HCF of the following terms.

a 12, 6y b 8a2, 24 c 20b, 15d 6, 9p2 e 24mn, m f c, c2

g 18pq, 12p2 h 24w2, 16w i 3(x � 5), x(x � 5)

2 Copy and complete each factorisation.

a 18p2 þ 24 ¼ 6(… þ…) b 9m2 þ 3m ¼ 3m(… þ…)c 20d � 30d2 ¼ 10d(… �…) d 6x2y � 8xy ¼ 2xy(… �…)e x(x � 2) þ 5(x � 2) ¼ (x � 2)(… þ…) f n(3 þ n) � 3(3 þ n) ¼ (3 þ n)(… �…)g h(h þ 4) � 2(h þ 4) ¼ (… þ…)(h � 2) h v(1 � w) þ (1 � w) ¼ (… þ…)(v þ 1)

3 Factorise each expression.

a 24x þ 30 b 36 þ 27a c 16g2 � 64 d xy þ y

e mn � 3n f 6p þ pq g x2 þ x h 2y � y2

i 3d2 þ 6d j 16r2 � 12r k 6t2 þ 27t l 36p2 � 108p

m 12x2y � 16x n 18p2 þ 16pr o 4m2n � 4mn2 p 14abc þ 21bc

q 28vw � 21v2w2 r 45rt þ 54r2t s 36pq2r � 144pr t 48x2y þ 64xy2

u 75g3h2 � 125gh


a a(a � 3) þ 6(a � 3) b t(8 þ t) � 3(8 þ t) c b(b þ 5) � 2(b þ 5)d x(2 � y) � 6(2 � y) e 5(a � 7) þ b(a � 7) f s(a þ 3) þ 4(a þ 3)g 3p(g þ 5) � 2(g þ 5) h 5(2 � 3m) þ 2n(2 � 3m) i r(8 þ r) þ (8 þ r)j (y � 6) � y(y � 6)

5 Factorise each expression using the ‘negative’ HCF.

a �4q � 8 b �9u þ 18 c 6 � 3g d �18a þ 12e �n � 1 f �n þ 1 g �y2 � 9y h �6t þ 10t2

i �3a2 � 6ab j �ap þ aq k �20e2 � 22e l �9m þ 3m2

6 Factorise �10kr þ 4rn. Select the correct answer A, B, C or D.

A �2r(5k � 4n) B �2r(5k � 2n) C �5r(2k � 2n) D �2r(5k þ 2n)


a 12 pþ 1

2 g b 7g þ 9g2 c 14 x2 � 1

4 xy

d 8a þ 12y þ 4 e 20xy � 10x2 þ 5y f t2m þ tm2 þ tm

g 7y(m þ 2) � p(m þ 2) h �10p2 � 10pq i 5a(x � y) þ 10a2xy � 10a2y2

See Example 12

See Example 13

See Example 14

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3-07 Expanding binomial products(x þ 5) and (x � 1) are called binomial expressions because each expression has exactly two terms(binomial ¼ ‘2 terms’). (x þ 5)(x � 1) is called a binomial product because it is a product(multiplication answer) of two binomial expressions.

Expanding binomial products using an area diagram

Example 15

Expand each binomial product using an area diagram.

a (a þ 2)(a þ 5) b (n þ 4)(n � 3)

Solutiona Draw an area diagram (rectangle) with length

(a þ 2) and width (a þ 5) and divide thediagram into 4 smaller rectangles.

2

5

a

a

a + 5

a + 2

Find the area of each smaller rectangle.2

5 5a 10

2aa a2

a

a + 5

a + 2

Expand (a þ 2)(a þ 5) by adding the areas ofthe 4 rectangles.

aþ 2ð Þ aþ 5ð Þ ¼ a2 þ 2aþ 5aþ 10

¼ a2 þ 7aþ 10

Expanding

Simplifying by collecting like terms.

b Draw an area diagram with length (n þ 4)and width (n � 3), then increase the width ton by adding 3. This creates 4 rectangles.

3

4

n – 3n

nn + 4

Worksheet

Area diagrams

MAT09NAWK10031

Worksheet

Binomial products

MAT09NAWK10032

Homework sheet

Algebra 3

MAT09NAHS10011

Puzzle sheet

Expanding brackets

MAT09NAPS00009

Technology

GeoGebra: Binomialexpansion

MAT09NATC00003

Technology worksheet

Excel worksheet:Expanding binomials

MAT09NACT00021

Technology worksheet

Excel spreadsheet:Expanding binomials

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The 2 shaded upper rectangles together give theproduct (n þ 4)(n � 3). To find its area, wesubtract the areas of the 2 lower rectangles fromthe area of the whole rectangle.

3

4

3n

n

n + 4

12

(n + 4)(n – 3)n – 3n

nþ 4ð Þ n� 3ð Þ ¼ n nþ 4ð Þ � 3n� 12

¼ n2 þ 4n� 3n� 12

¼ n2 þ n� 12

Whole rectangle � 2 lower rectanglesExpandingSimplifying by collecting like terms.

Expanding binomial products algebraicallyWhen expanding a binomial product algebraically, each term in the first binomial is multiplied byeach term in the second binomial to give four terms, which are collected and simplified.

Example 16

Expand each binomial product.

a (x þ 5)(x þ 9) b (k þ 3)(k � 7)

Solutiona xþ 5ð Þ xþ 9ð Þ ¼ x xþ 9ð Þ þ 5 xþ 9ð Þ

¼ x2 þ 9xþ 5xþ 45

¼ x2 þ 14xþ 45

Each term in (x þ 5) is multiplied by (x þ 9).

Expanding to make 4 terms.

Adding 9x and 5x.

b k þ 3ð Þ k � 7ð Þ ¼ k k � 7ð Þ þ 3 k � 7ð Þ¼ k2 � 7k þ 3k � 21

¼ k2 � 4k � 21

Each term in (k þ 3) is multiplied by (k � 7).

Expanding to make 4 terms.

Adding �7k and 3k.

Summary

The distributive law for expanding a binomial productMultiply each term in the first binomial by eachterm in the second binomial.(a þ b)(c þ d) ¼ ac þ ad þ bc þ bd

accc + d

d

aa + b

b

ad bd

bc

Video tutorial

Expanding binomialproducts

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One way of remembering which pairs of terms to multiplytogether in a binomial product is called the FOIL method,as shown below.

• F means multiply the first terms: k 3 k ¼ k2

• O means multiply the outside terms: k 3 (�7) ¼ �7k

• I means multiply the inside terms: 3 3 k ¼ 3k

• L means multiply the last terms: 3 3 (�7) ¼ �21

( + 3)( – 7) = 2 – 7 + 3 – 21

= 2 – 4 – 21

FO

IL

k

k k

kkkk

Example 17

Expand each binomial product.

a (x � 6)(4x þ 2) b (3t � 1)(2t � 5)

Solutiona x� 6ð Þ 4xþ 2ð Þ ¼ x 4xþ 2ð Þ � 6 4xþ 2ð Þ

¼ 4x2 þ 2x� 24x� 12

¼ 4x2 � 22x� 12

Expanding

Simplifying

b 3t � 1ð Þ 2t � 5ð Þ ¼ 3t 2t � 5ð Þ � 1 2t � 5ð Þ¼ 6t2 � 15t � 2t þ 5

¼ 6t2 � 17t þ 5

ExpandingSimplifying

Exercise 3-07 Expanding binomial products1 Expand each binomial product using the given area diagram.

a (x þ 10)(x þ 8)

10

8

x

x

b (a þ 3)(a þ 6)

3

6

a

a

c (d þ 2)(2d þ 5)

2d

5

2d

d (4h þ 1)(3h þ 7)

7

3h

14h

Video tutorial


MAT09NAVT10006

Video tutorial


MAT09NAVT00004

See Example 15

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2 Expand each binomial product using the given area diagram. (Find the shaded area.)

a (w þ 5)(w � 1)

5

1

w

ww – 1

b (x þ 9)(x � 6)

9

6

x – 6

x

x

c (t � 8)(t þ 4)

8

4

t – 8t

t

d (n � 5)(n � 7)

n – 5

n – 7

7

5n

n

3 Draw an area diagram for each binomial product and use it to expand the product.

a (2y þ 5)(y þ 3) b (r þ 5)(3r þ 2) c (5k þ 3)(4k þ 1)d (w � 5)(w þ 3) e (v þ 4)(3v � 1) f (p � 3)(p � 5)

4 Expand each binomial product algebraically.

a (n þ 6)(n þ 3) b (t þ 6)(t þ 5) c (p þ 10)(p � 10)d (x � 8)(x þ 3) e (b � 2)(9 þ b) f (u � 8)(u � 7)g (15 � r)(r þ 1) h (a � 10)(a � 9) i (5 � c)(3 � c)j (t � 1)(t þ 2) k (y � 4)(y þ 10) l (n � 9)(11 þ n)

5 Expand (b þ 7)2. Select the correct answer A, B, C or D.

A b2 þ 49 B b2 þ 49b C b2 þ 7b þ 49 D b2 þ 14b þ 49

6 Expand each binomial product.

a (2x þ 5)(x þ 3) b (3e þ 2)(4e þ 5) c (10 þ 3p)(p � 1)d (7d � 2) (7d � 2) e (2f � 2)(3f þ 5) f (4m � 5)(5 þ 3m)g (3 � 4h)(2 þ 5h) h (4p � 5)(2p � 7) i (2m � 3)(4 � 5m)j (3t þ 5)(2t � 1) k (5y � 5)(5y þ 5) l (6 � 2a)(2a � 6)

7 A rectangular barbecue plate has a length of 100 cm and a width of 75 cm. The length andwidth are both increased by x cm.a Write an expression for the new length of the plate.

b Write an expression for the new width of the plate.

c Hence find a simplified expression for the new area of the plate.

d By how much has the area of the plate increased?

e If x ¼ 0.1, find the increase in the area of the plate.

See Example 16

See Example 17

Worked solutions


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8 A family room in a house is to be extended. The room is 4 metres long and 3 metres wide. Thelength is to be increased by x metres and the width by y metres.a Write down expressions for the new length and width.

b Write down a binomial expression for the new area of the room.

c Expand and simplify your expression for the area.

d By how much has the area of the room increased?

Investigation: Expanding perfect squares

There is a special pattern when you expand a binomial by itself, for example,(y þ 5)(y þ 5).This is called the square of a binomial or a perfect square.1 a Expand and simplify the perfect square (y þ 5)2 if (y þ 5)2 ¼ (y þ 5)(y þ 5).

b How many terms are there in your answer?c The terms in the binomial (y þ 5) are y and 5. The first term of your answer in part a is

y2, which is the square of y. How are the other terms in the answer related to y and 5?2 Expand and simplify each perfect square.

a (k þ 3)2 b (m þ 7)2 c (p þ 2)2

3 For each of the expansions in question 2:a How many terms are there?b Describe how each is related to the two terms of the perfect square.c Compare and discuss your results with other students.

4 Expand and simplify each perfect square.a (t � 1)2 b (g � 6)2 c (d � 5)2

5 For each of the expansions in question 4:a How many terms are there?b Describe how each is related to the two terms of the perfect square.c Compare and discuss your results with other students.

6 Describe the expansion of the perfect square (first term þ second term)2 by copying andcompleting this statement:The square of a binomial is equal to the square of the first term plus double the product of

the two terms plus _______

7 a Find a formula for (a þ b)2.b Check this formula by expanding (a þ b)2 using an area diagram.c Find a formula for (a � b)2.d Check this formula by expanding (a � b)2 using an area diagram.

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3-08 Perfect squares16, 49, v2 and (y þ 5)2 are called perfect squares because they are square numbers. The rule forexpanding the perfect square of a binomial is:

Summary

(a þ b)2 ¼ a2 þ 2ab þ b2

(a � b)2 ¼ a2 � 2ab þ b2

Proof:

aþ bð Þ2 ¼ aþ bð Þ aþ bð Þ¼ a aþ bð Þ þ b aþ bð Þ¼ a2 þ abþ baþ b2

¼ a2 þ 2abþ b2

a� bð Þ2 ¼ a� bð Þ a� bð Þ¼ a a� bð Þ � b a� bð Þ¼ a2 � ab� baþ b2

¼ a2 � 2abþ b2

Example 18

Copy and complete the expansion of each perfect square.

a (x þ 4)2 ¼ x2 þ ____ þ 16 b (y � 6)2 ¼ y2 � 12y þ ____

c (5g þ 9)2 ¼ ____ þ 45g þ 81 d (3d � 5)2 ¼ 9d2 � ____ þ 25

Solutiona In the expansion,

¼ 2 3 x 3 4

¼ 8x

) xþ 4ð Þ2 ¼ x2 þ 8xþ 16

Doubling the product of the two terms.

b In the expansion,

¼ 62

¼ 36

) y� 6ð Þ2 ¼ y2 � 12yþ 36

The second term squared.

c ¼ 5gð Þ2

¼ 25g2

) 5g þ 9ð Þ2 ¼ 25g2 þ 45g þ 81

The first term squared.

d ¼ 2 3 3d 3 5

¼ 30d

) 3d � 5ð Þ2 ¼ 9d2 þ 30d þ 25

Doubling the product of the two terms.

Stage 5.3

Puzzle sheet

Factorominoes

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Example 19

Expand each perfect square.

a (n � 5)2 b (k þ 7)2 c (3y � 8)2

Solutiona n� 5ð Þ2 ¼ n2 � 2 3 n 3 5þ 52

¼ n2 � 10nþ 25

1st term squared � double product þ2nd term squared

b k þ 7ð Þ2 ¼ k2 þ 2 3 k 3 7þ 72

¼ k2 þ 14k þ 49

c 3y� 8ð Þ2 ¼ 3yð Þ2�2 3 3y 3 8þ 82

¼ 9y2 � 48yþ 64

Exercise 3-08 Perfect squares1 Copy and complete the expansion of each perfect square.

a (x þ 10)2 ¼ x2 þ _____ þ 100 b (m � 8)2 ¼ _____ � 16m þ 64

c (p � t)2 ¼ p2 � 2pt þ _____ d (h þ 4)2 ¼ h2 _____ þ 16

e (k � 9)2 ¼ k2 _____ þ 81 f (8 þ 5f )2 ¼ 64_____ þ 25f 2

g (2d þ 3)2 ¼ _____ þ _____ þ 9 h (6a þ 1)2 ¼ _____ þ 12a þ _____

2 Expand each perfect square.

a (m þ 9)2 b (u þ 3)2 c (y � 6)2

d (8 þ k)2 e (5 � h)2 f (7 þ k)2

g (f þ 20)2 h (q � 11)2 i (10 þ t)2

j (x � w)2 k (a þ g)2 l (2m � 3)2

m (5x � 6)2 n (9a þ 2)2 o (3e � 4)2

p (5 þ 7b)2 q (4 � 5p)2 r (11 � 2c)2

s (10g þ 3)2 t (3k þ 11)2 u (5 þ 2v)2


a (7h þ 2k)2 b (8a � 3y)2 c (xy þ z)2

d 1þ 1y

� �2e t � 1

t

� �2f 3

wþ w� �2

4 Use expansion to evaluate each square number without using a calculator.

a 212 ¼ (20 þ 1)2 b 452 ¼ (40 þ 5)2 c 292 ¼ (30 � 1)2

d 592 ¼ (60 � 1)2 e 1022 ¼ (100 þ 2)2 f 982 ¼ (100 � 2)2

Stage 5.3

See Example 18

See Example 19

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5 A 10 cm square sheet of cardboard has a square of length x cm cut from each corner. It is thenfolded to form a square-based prism.

xx

x

xx

xx

x

10

10

x

b

b

a Why is the length of the square base (b) equal to (10 � 2x) cm?

b Find the area of the square base.

c Find the area of one side face of the prism.

d Hence show that the surface area of the prism is (100 � 4x2) cm2.

Investigation: Squaring a number ending in 5

Study this mental short cut for squaring a number ending in 5.

• To evaluate 352, calculate 3 3 4 ¼ 12, add ‘25’ to the end: 352 ¼ 1225.• To evaluate 752, calculate 7 3 8 ¼ 56, add ‘25’ to the end: 752 ¼ 5625.• To evaluate 1052, calculate 10 3 11 ¼ 110, add ‘25’ to the end: 1052 ¼ 11 025.

Let n stand for the tens digit of the number ending in 5 being squared.Expand (10n þ 5)2 and investigate why the above method works.

Mental skills 3B Maths without calculators

Multiplying by 9, 11, 99 and 101

We can use expansion when multiplying by a number near 10 or near 100.

1 Study each example.

a 25 3 11 ¼ 25 3 10þ 1ð Þ¼ 25 3 10þ 25 3 1

¼ 250þ 25

¼ 275

b 14 3 9 ¼ 14 3 10� 1ð Þ¼ 14 3 10� 14 3 1

¼ 140� 14

¼ 126

Stage 5.3

Worked solutions

Perfect squares

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3-09 Difference of two squaresThe rule for expanding (a þ b)(a � b) is:

Summary

(a þ b)(a � b) ¼ a2 � b2

The answer is called the difference of two squares.

c 32 3 12 ¼ 32 3 10þ 2ð Þ¼ 32 3 10þ 32 3 2

¼ 320þ 64

¼ 384

d 7 3 99 ¼ 7 3 100� 1ð Þ¼ 7 3 100� 7 3 1

¼ 700� 7

¼ 693

e 27 3 101 ¼ 27 3 100þ 1ð Þ¼ 27 3 100þ 27 3 1

¼ 2700þ 27

¼ 2727

f 18 3 8 ¼ 18 3 10� 2ð Þ¼ 18 3 10� 18 3 2

¼ 180� 36

¼ 144

2 Now evaluate each product.

a 16 3 11 b 33 3 11 c 29 3 9 d 45 3 9e 62 3 11 f 7 3 101 g 18 3 101 h 36 3 99i 19 3 8 j 45 3 12 k 21 3 102 l 6 3 98m 32 3 9 m 7 3 99 o 39 3 101 p 71 3 12

Investigation: Expanding sums by differences

There is also a special pattern when you multiply the sum of two terms by the difference ofthose two terms, for example, (x þ 9)(x � 9).1 a Expand and simplify (x þ 9)(x � 9).

b How many terms are there in your answer?c How are the terms in the answer related to x and 9?

2 Expand and simplify each ‘sum by difference’.a (k þ 2)(k � 2) b (m þ 5)(m � 5) c (p � 6)(p þ 6)

3 Describe the expansion of (first term þ second term)(first term � second term) by copyingand completing this statement:The product of a ‘sum by difference’ is equal to the square of the first term ___________

4 a Find a formula for (a þ b)(a � b).b Check this formula by expanding (a þ b)(a � b) using an area diagram.

Stage 5.3

Worksheet

Special products

MAT09NAWK00034

Puzzle sheet

Enough time

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Proof:

a� bð Þ aþ bð Þ ¼ a aþ bð Þ � b aþ bð Þ¼ a2 þ ab� ba� b2

¼ a2 � b2 þ ab� ba ¼ 0

When the sum of two terms is multiplied by their difference, the answer is the square of the firstterm minus the square of the second term (the difference of two squares).

Example 20


a (d þ 3)(d � 3) b (2 þ r)(2 � r) c (7x þ 2)(7x � 2) d (4k � 5p)(4k þ 5p)

Solutiona d þ 3ð Þ d � 3ð Þ ¼ d2 � 32

¼ d2 � 9

b 2þ rð Þ 2� rð Þ ¼ 22 � r2

¼ 4� r2

c 7xþ 2ð Þ 7x� 2ð Þ ¼ 7xð Þ2�22

¼ 49x2 � 4

d 4k � 5pð Þ 4k þ 5pð Þ ¼ 4kð Þ2� 5pð Þ2

¼ 16k2 � 25p2

Exercise 3-09 Difference of two squares1 Expand each expression.

a (m þ 5)(m � 5) b (c � 10)(c þ 10) c (a þ 12)(a � 12)d (6 � y)(6 þ y) e (8 � m)(8 þ m) f (p þ 1)(p � 1)g (5 þ e)(5 � e) h (v þ 11)(v � 11) i (w � 3)(w þ 3)j (x � 10)(x þ 10) k (q þ 7)(q � 7) l (9 � g)(9 þ g)m (b � 2)(b þ 2) n (15 � r)(15 þ r) o (d þ 13)(d � 13)


a (2h � 3)(2h þ 3) b (5r þ 4)(5r � 4) c (5b þ 8)(5b � 8)d (4p � 7)(4p þ 7) e (3 � 8k)(3 þ 8k) f (7x � 5)(7x þ 5)g (2 þ 9m)(2 � 9m) h (9k � 4l)(9k þ 4l) i (7n þ 8m)(7n � 8m)j (4g � 5h)(4g þ 5h) k (7u þ 3w)(7u � 3w) l (11a þ 3b)(11a � 3b)

m t þ 1t

� �t � 1

t

� �n w

3 � 2� �

w3 þ 2� �

o 1� 1r

� �1þ 1

r

� �

3 Susan’s age is p years.a What was Susan’s age last year?

b What will Susan’s age be next year?

c Write an expression for (Susan’s age last year) 3 (Susan’s age next year).

d If (Susan’s age last year) 3 (Susan’s age next year) is equal to 48, what is Susan’s age?

4 By expressing 31 3 29 as (30 þ 1)(30 � 1), use the difference of two squares to find the valueof 31 3 29.

Stage 5.3

See Example 20

Worked solutions

Difference of twosquares

MAT09NAWS10014

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5 Use the method of question 4 to evaluate each expression.

a 21 3 19 b 51 3 49 c 89 3 91 d 78 3 82

6 Use ‘sum by difference’ to evaluate each expression.

a 152 � 142 b 242 � 232 c 652 � 352

d 1012 � 992 e 232 � 172 f 502 � 482

7 A square sheet of paper of length x cm has a smaller square of length y cm cut from onecorner. It is then cut on the diagonal AB and rearranged to form a rectangle as shown.

x

y

yB

B

x

A

A

l

w

a Find an expression for:i the length (l) of the rectangle

ii the width (w) of the rectangle

iii the area of the rectangle.

b The area of the rectangle should be the same as the area of the square on the left minus thearea of the smaller square. Find an expression for this area. What does this prove?

3-10 Mixed expansions

Example 21

Expand and simplify each expression.

a (4r þ 5)(1 � 2r) b (7 þ 9x)2 c (2d � 10)(2d þ 10)d (a þ 6)(a � 6) þ (a þ 12)(a þ 3) e (m � 2)2 � (m � 2)(m þ 2)

Solutiona 4r þ 5ð Þ 1� 2rð Þ ¼ 4r 1� 2rð Þ þ 5 1� 2rð Þ

¼ 4r � 8r2 þ 5� 10r

¼ �8r2 � 6r þ 5

Stage 5.3

Worked solutions

Difference of twosquares

MAT09NAWS10014

Worksheet

Mixed expansions

MAT09NAWK00035

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b 7þ 9xð Þ2 ¼ 72 þ 2 3 7 3 9xþ 9xð Þ2

¼ 49þ 126xþ 81x2

¼ 81x2 þ 126xþ 49

c 2d � 10ð Þ 2d þ 10ð Þ ¼ 2dð Þ2�102

¼ 4d2 � 100

d aþ 6ð Þ a� 6ð Þ þ aþ 12ð Þ aþ 3ð Þ ¼ a2 � 36þ a2 þ 3aþ 12aþ 36

¼ 2a2 þ 15a

e m� 2ð Þ2� m� 2ð Þ mþ 2ð Þ ¼ m2 � 4mþ 4� m2 � 4� �

¼ m2 � 4mþ 4� m2 þ 4

¼ �4mþ 8

Exercise 3-10 Mixed expansions1 Expand and simplify each expression.

a (2m � 1)(2m þ 1) b (y þ 4) (5y � 3) c (2k � 7)2

d (d þ 9)2 e (2e � 1)(e þ 1) f (5a þ 4)(5a � 4)g (2 � p)(p � 2) h (10 � 6y)(10 þ 6y) i (h � 3z)2

j (2x � 3)(y þ 3) k (11a � 4b)(11a þ 4b) l u� 1u

� �2

2 Expand and simplify each expression.

a (m � 5)(m þ 5) þ 25 b 6y þ (y � 3)2 þ 9c (3x þ 1)(2 � x) þ 2x þ 4 d (d þ 4)2 � 8d þ 5e 16 þ (4k � 8)(4k þ 8) f (x � y)2 � (x þ y)2

g 20t � (t � 2)(t � 5) þ t2 h 2(f � 2)(f þ 2)i (2h þ 3)2 � (2h � 3)(2h þ 3) j 7xy � (2x � 3)(y þ 3)


a (8a � 1)(8a þ 1) � 4a2 þ 1 b (n þ 1)2 þ 2n þ 3c 3(4 � t)(4 þ t) þ (t � 12)(t þ 4) d (2m � n)2 þ (2m þ n)2

e (x � 2)(x þ 3) � (x � 2)(x þ 2) f 2(b � 1)2 � (2b � 1)2

g (y þ 1)2 þ (y þ 2)2 þ (y þ 3)2 h (x � 3)(x þ 3) þ (x þ 3)2 þ (x � 3)2

i (5n þ 3)(5n � 3) þ (3n � 5)(3n þ 5) j 2(a � b)(a þ b) � (a þ b)2 � (a � b)2

3-11 Factorising special binomial products

Factorising by grouping in pairsAn algebraic expression with four terms can often be factorised in pairs, that is, two terms at atime, to make a binomial product.

Stage 5.3

See Example 21

Worked solutions

Mixed expansions

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Example 22


a 3ac þ 2bd þ 2bc þ 3ad b 4km þ 6mn � 6kp � 9np c 10xw � 6yw � 10xt þ 6yt

Solutiona 3acþ 2bdþ 2bcþ 3ad ¼ 3acþ 3adþ 2bdþ 2bc

¼ 3a cþ dð Þ þ 2b dþ cð Þ¼ cþ dð Þ 3aþ 2bð Þ

Grouping into pairs for factorising.Factorising each pair.Factorising again.

b 4kmþ 6mn� 6kp� 9np¼ 2m 2k þ 3nð Þ � 3p 2k þ 3nð Þ¼ 2k þ 3nð Þ 2m� 3pð Þ

Factorising each pair.Factorising again.

c 10xw� 6yw� 10xtþ 6yt ¼ 2 5xw� 3yw� 5xtþ 3ytð Þ¼ 2 5xw� 5xt� 3ywþ 3ytð Þ¼ 2 5x w� tð Þ � 3y w� tð Þ½ �¼ 2 w� tð Þ 5x� 3yð Þ

Factorising all terms first.Grouping into pairs for factorising.

Factorising each pair.Factorising again.

Factorising the difference of two squaresYou should recall the product (a þ b)(a � b) ¼ a2 � b2.If we use this rule in reverse, then the factors of a2 � b2 are (a � b) and (a þ b).

Summary

a2 � b2 ¼ (a þ b)(a � b)

Example 23


a x2 � 4 b 9 � 16b2 c 20d2 � 5a2 d y3 � y

Solutiona x2 � 4 ¼ x2 � 22

¼ xþ 2ð Þ x� 2ð Þb 9� 16b2 ¼ 32 � 4bð Þ2

¼ 3þ 4bð Þ 3� 4bð Þ

c 20d2 � 5a2 ¼ 5 4d2 � a2� �

¼ 5 2dð Þ2�a2h i

¼ 5 2d þ að Þ 2d � að Þ

d y3 � y ¼ y y2 � 1� �

¼ y yþ 1ð Þ y� 1ð Þ

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Exercise 3-11 Factorising special binomial products1 Factorise each expression.

a 4ab þ 5bc þ 4ad þ 5cd b 2xy � 5wy þ 2xt � 5wt

c 9ac þ 6bc þ 12ad þ 8bd d 10x2 þ 30 þ x3 þ 3x

e 3a2 þ 3ab þ 3ac þ 3bc f 6rt � 18wt þ 6rp � 18wp

g 14e � 21 þ 2de � 3d h hk � h2 � 2k þ 2h

i 3mn � 6m þ pn � 2p j 9p2 � 27 þ qp2 � 3q

k fg � fh � 10g þ 10h l 9kl � 12ml þ 9kn � 12mn

m 2p � 2c � p2 þ pc n l3 þ lm2 � 3l2 � 3m2

o a(x þ 1) þ y(x þ 1) � ka � ky p p(a � b) � 2q(a � b) þ 3qp � 6q2


a w2 � 9 b y2 � 36 c k2 � 1 d m2 � 121e p2 � 64 f t2 � 100 g 4e2 � f 2 h a2 � 9b2

i 16y2 � 1 j 4 � b2 k 25 � e2 l 1 � 16x2

m y2 � z2 n 49 � 16m2 o b2 � 121d2 p 36c2 � 25k2

q 16 � 81h2 r 25a2 � 64m2 s 100 � 49n2 t 121p2 � 144q2

u 14� 25c2 v 4t2 � 1

9 w 25h2 � 214 x 1 � m2n2


a 2a2 � 2b2 b 7k2 � 28 c 3 � 75u2 d x3 � 49x

e k � 16k3 f 50q2 � 2 g 3d2 � 12v2 h 5t5 � 125t3

i 2a2b2 � 2 j x2y2 � x2w2 k 192f 2 � 108g2 l 45d2 � 54

m 2x2 � 8a2 n 100 � 25w2 o 114� 80e2 p 9c2 � 61


a p2

16� x2

25 b x2 � 19 c v2

49�u2

81 d 2y2

9 �2m2

121

e 16a2

49 �25b2

4 f t4 � 81 g 100 � n4 h (x þ y)2 � x2

i (p � 2q)2 � (2p þ q)2 j x2

4 �y2

36k (a þ b)2 � (a � b)2

Stage 5.3

Investigation: Factorising quadratic expressions

1 a Show that (x þ 3)(x þ 5) ¼ x2 þ 8x þ 15b The quadratic expression x2 þ 8x þ 15 has three terms. The coefficient of x is 8, the

number in front of the x. How are the 3 and 5 in (x þ 3)(x þ 5) related to the 8?c The constant term in x2 þ 8x þ 15 is 15, the number with no x at the end. How are the

3 and 5 related to the 15?2 a Expand (x þ 9)(x þ 2).

b What is the coefficient of x? How are 9 and 2 related to it?c What is the constant term? How are 9 and 2 related to it?

3 a Expand (x þ 8)(x � 3).b What is the coefficient of x? How are 8 and �3 related to it?c What is the negative constant term? How are 8 and �3 related to it?

See Example 22

See Example 23

Worked solutions

Factorising specialbinomial products

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3-12 Factorising quadratic expressionsA quadratic expression such as x2 � 5x þ 7 is called a trinomial because it has three terms. Otherexamples of quadratic trinomials are x2 þ x � 15, 2x2 � 3x þ 9 and �4x2 þ 9x þ 20.The expansion of (x þ 2)(x þ 4) is x2 þ 6x þ 8.[ The factorisation of x2 þ 6x þ 8 is (x þ 2)(x þ 4).

Summary

In the factorisation of a quadratic trinomial such as x2 þ 6x þ 8:

• each factor must have an x term to give x2x2 + 6x + 8 = (x + 2)(x + 4)

• 2 þ 4 ¼ 6, which is the coefficient of x, the number infront of the x

x2 + 6x + 8 = (x + 2)(x + 4)

• 2 3 4 ¼ 8, which is the constant term with no x. x2 + 6x + 8 = (x + 2)(x + 4)

Example 24

Factorise x2 þ 11x þ 24.

SolutionFind the two numbers that have a sumof 11 and a product of 24.It is best to test numbers that have aproduct of 24 and then check if theirsums equal 11:

Pair of numbers Product Sum6, 4 6 3 4 ¼ 24 6 þ 4 ¼ 10

2, 12 2 3 12 ¼ 24 2 þ 12 ¼ 14

8, 3 8 3 3 ¼ 24 8 þ 3 ¼ 11The correct numbers are 8 and 3.[ x2 þ 11x þ 24 ¼ (x þ 8)(x þ 3)

Stage 5.3 4 a Expand (x � 4)(x � 1).b What is the negative coefficient of x? How are �4 and �1 related to it?c What is the positive constant term? How are �4 and �1 related to it?

5 In the expansion of any binomial product, how are the coefficient of x and the constantterm related to the numbers in the binomials?

6 Copy and complete:a (x þ ___)(x þ ___) ¼ x2 þ 5x þ 4 b (x þ ___)(x þ ___) ¼ x2 þ 8x þ 15c (x þ ___)(x þ ___) ¼ x2 þ 7x þ 12 d (x þ ___)(x � ___) ¼ x2 � 4x � 32e (x þ ___)(x � ___) ¼ x2 þ 2x � 3 f (x � ___)(x � ___) ¼ x2 � 9x þ 20

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Stage 5.3Example 25

Factorise x2 � 5x � 6.

SolutionFind two numbers with a sum of�5 and a product of �6.Since their product is �6, one ofthe numbers must be negative.Test numbers that have a productof �6 and check their sums.

Pair of numbers Product Sumþ3, �2 3 3 (�2) ¼ �6 3 þ (�2) ¼ 1

�3, þ2 �3 3 2 ¼ �6 �3 þ 2 ¼ �1

þ6, �1 6 3 (�1) ¼ �6 6 þ (�1) ¼ 5

�6, þ1 �6 3 1 ¼ �6 �6 þ 1 ¼ �5The correct numbers are �6 and 1.[ x2 � 5x � 6 ¼ (x � 6)(x þ 1)

Summary

To factorise quadratic trinomials of the form x2 þ bx þ c:

• find two numbers that have a sum of b and a product of c

• use these two numbers to write a binomial product of the form (x ___ )(x ___ )

Example 26

Factorise each quadratic expression.

a x2 þ 6x � 16 b m2� 7m � 18 c y2 � 2y þ 1

Solutiona x2 þ 6x � 16

Find two numbers that have a product of �16 and a sum of 6.Since the product is negative, one of the numbers must be negative.They are þ8 and �2.[ x2 þ 6x � 16 ¼ (x þ 8)(x � 2)

b m2� 7m � 18Product ¼ �18, sum ¼ �7.Since the product is negative, one of the numbers must be negative.They are �9 and þ2.[ m2 � 7m � 18 ¼ (m � 9)(m þ 2)

c y2 � 2y þ 1Product ¼ 1, sum ¼ �2.Since the sum is negative, one of the numbers must be negative.Since the product is positive, both of the numbers must be negative.They are �1 and �1.

) y2 � 2yþ 1 ¼ y� 1ð Þ y� 1ð Þ¼ y� 1ð Þ2 Note: (y � 1)2 is a perfect square.

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Example 27


a 3g2 þ 12g � 36 b 48 � 8p � p2

Solutiona 3g2 þ 12g � 36 ¼ 3 g2 þ 4g � 12

� �

¼ 3 g � 2ð Þ g þ 6ð ÞTaking out the HCF of 3 first.

Product ¼ �12, sum ¼ 4

b 48� 8p� p2 ¼ �p2 � 8pþ 48

¼ �1 p2 þ 8p� 48� �

¼ � pþ 12ð Þ p� 4ð Þ

Rearranging the terms to make the p2 term first.

Taking out a common factor of �1.

Product ¼ �48, sum ¼ 8

Exercise 3-12 Factorising quadratic expressions1 Find two numbers whose:

a product is 6 and their sum is �7 b product is �12 and their sum is 1c product is �15 and their sum is �2 d product is 12 and their sum is 7e product is 20 and their sum is �9 f product is �14 and their sum is 5g product is �10 and their sum is 3 h product is �25 and their sum is 0i product is �2 and their sum is �1 j product is �18 and their sum is �7

2 Factorise each quadratic expression.

a x2 þ 7x þ 12 b x2 þ 12x þ 35 c x2 þ 5x þ 4d x2 þ 7x þ 10 e x2 þ 9x þ 20 f t2 þ 6t þ 5g e2 þ 5e þ 6 h h2 þ 4h þ 4 i n2 þ 11n þ 10j a2 þ 11a þ 30 k d2 þ 10d þ 24 l y2 þ 15y þ 44


a y2 � 2y � 3 b r2 � 5r � 14 c h2 � 3h � 4d w2 � 7w � 18 e e2 � 6e � 27 f a2 � 4a � 12


a x2 þ 3x � 4 b t2 þ 5t � 24 c m2 þ 2m � 15d a2 þ a � 2 e k2 þ 5k � 14 f w2 þ 4w � 12g m2 � 5m þ 4 h w2 � 6w þ 8 i x2 � 12x þ 35j p2 � 10p þ 24 k n2 � 3n þ 2 l a2 � 6a þ 9

5 Factorise each expression given it is a perfect square.

a m2 þ 4m þ 4 b p2 þ 20p þ 100 c a2 � 10a þ 25

6 Factorise each quadratic expression. Look for the highest common factor first.

a 3m2 þ 9m þ 6 b 2y2 þ 2y � 4 c 5t2 � 10t � 400d 5e4 þ 25e3 � 120e2 e x3 � x2 � 110x f 4b2 � 4b � 168g 4w2 þ 4w � 48 h 3a3 � 9a2 � 12a i 2e2 þ 18e þ 40j 24 � 5t � t2 k 42 þ u � u2 l 28 þ 3x � x2

m 12 � b � b2 n 7k � 12 � k2 o 12x � 35 � x2

Stage 5.3

See Example 24

See Example 25

See Example 26

See Example 27

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a h2 � 3h � 18 b 18 � 7t � t2 c w2 þ 8w þ 7d k2 � 12k � 45 e v2 þ 8v � 20 f 3c2 þ 30c þ 75g q2 � 6q þ 5 h a2 � 4a þ 3 i 6x2 þ 36x � 96j x2 � 16x þ 64 k 8u2 � 24u � 32 l b2 þ 11b þ 30m y2 � 12y þ 36 n 5r2 � 5r � 10 o 4l2 � 8l � 32p g2 � 24g þ 80 q 108 � 6d � 2d2 r 26 þ 11n � n2

3-13Factorising quadratic expressionsof the form ax2 þ bxþ c

We will now factorise quadratic trinomials such as 6x2 þ 19x þ 15, where x2 has a coefficient thatis not a common factor. We do this by splitting the middle term into two terms and then factoriseby grouping in pairs.

Example 28

Factorise 3x2 þ 8x þ 4.

SolutionWe need to split up the middle term 8x.

Find two numbers that have a product of 12 and a sum of 8.

3x2 + 8x + 4 = 3x2 + 8x + 4

product of 12 (3 × 4)

sum of 8

Investigation: Factorising quadratic trinomials bygrouping in pairs

1 We can factorise the quadratic trinomial 6y2 þ 19y þ 15 by rewriting it as 6y2 þ 10y þ9y þ 15, splitting up the middle term 19y.a Factorise 6y2 þ 10y þ 9y þ 15 by grouping in pairs.b Factorise 6y2 þ 9y þ 10y þ 15 by grouping in pairs. Is your answer the same as that

for part a?c In 6y2 þ 19y þ 15, 6 is the coefficient of y2 while 15 is the constant term. Find the

product of 6 and 15 and the product of 10 and 9. What do you notice?2 a Factorise 2c2 þ 13c þ 21 by rewriting it as 2c2 þ 6c þ 7c þ 21.

b Factorise 2c2 þ 13c þ 21 by rewriting it as 2c2 þ 7c þ 6c þ 21.c Find the product of 2 and 21 and the product of 6 and 7. What do you notice?

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The two numbers are þ6 and þ2, so we will split 8x

into 6x and 2x.) 3x2 þ 8xþ 4 ¼ 3x2 þ 6xþ 2xþ 4

¼ 3x xþ 2ð Þ þ 2 xþ 2ð Þ¼ xþ 2ð Þ 3xþ 2ð Þ

Factorising by grouping in pairs.Factorising again.

Summary

To factorise quadratic trinomials of the form ax2 þ bx þ c:

• find two numbers that have a sum of b and a product of ac

• use these two numbers to split the middle term bx into two terms• factorise by grouping in pairs

Example 29


a 5k2 � 12k þ 4 b 9m2 � 9m � 4 c 6t2 þ t � 12

Solutiona 5k2 � 12k þ 4

5 3 4 ¼ 20. Find two numbers that have a product of 20 and a sum of �12.Since the sum is negative, one of the numbers must be negative.Since the product is positive, both of the numbers must be negative.They are �10 and �2. Split �12k into �10k and �2k.5k2 � 12k þ 4 ¼ 5k2 � 10k � 2k þ 4

¼ 5k k � 2ð Þ � 2 k � 2ð Þ¼ k � 2ð Þ 5k � 2ð Þ

Factorising by grouping in pairs.

b 9m2 � 9m � 4

9 3 (�4) ¼ �36. Find two numbers with a product of �36 and a sum of �9.Since the product is negative, one of the numbers must be negative.They are �12 and 3.9m2 � 9m� 4 ¼ 9m2 � 12mþ 3m� 4

¼ 3m 3m� 4ð Þ þ 1 3m� 4ð Þ¼ 3m� 4ð Þ 3mþ 1ð Þ

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c 6t2 þ t � 126 3 (�12) ¼ �72. Find two numbers with a product of �72 and a sum of 1.They are �8 and þ9.6t2 þ t � 12 ¼ 6t2 � 8t þ 9t � 12

¼ 2t 3t � 4ð Þ þ 3 3t � 4ð Þ¼ 3t � 4ð Þ 2t þ 3ð Þ

Example 30


a 18a2 � 18a � 8 b 10 � 7x � 12x2

Solutiona 18a2 � 18a� 8 ¼ 2 9a2 � 9a� 4

� �

¼ 2 9a2 � 12aþ 3a� 4� �

¼ 2 3a 3a� 4ð Þ þ 1 3a� 4ð Þ½ �¼ 2 3a� 4ð Þ 3aþ 1ð Þ

Taking out the HCF of 2 first.

Product ¼ �36, sum ¼ �9

b 10� 7x� 12x2 ¼ �12x2 � 7xþ 10

¼ � 12x2 þ 7x� 10� �

¼ � 12x2 þ 15x� 8x� 10� �

¼ � 3x 4xþ 5ð Þ � 2 4xþ 5ð Þ½ �¼ � 4xþ 5ð Þ 3x� 2ð Þ

Rearranging the terms to make thex2 term first.Taking out a common factor of �1.

Product ¼ �120, sum ¼ �7

Exercise 3-13 Factorising quadratic expressionsof the form ax

2 þ bx þ c


a 2x2 þ 11x þ 5 b 4x2 þ 13x þ 3 c 5x2 þ 17x þ 6d 6x2 þ 19x þ 10 e 2w2 þ 31w þ 15 f 4e2 þ 15e þ 9g 8f 2 þ 14f þ 3 h 3d2 þ 5d þ 2 i 2b2 þ 9b þ 7j 5y2 þ 16y þ 11 k 8g2 þ 26g þ 15 l 6a2 þ 23a þ 21


a 2y2 � 11y þ 12 b 10k2 � 19k þ 6 c 6e2 � 13e þ 6d 4b2 � 13b þ 3 e 6w2 � 23w þ 15 f 8t2 þ 26t þ 15g 9x2 � 12x þ 4 h 12f 2 � 25f þ 12 i 4h2 � 36h þ 81j 5y2 � 6y � 11 k 4d2 � d � 5 l 2m2 � 3m � 9m 3t2 � t � 30 n 6h2 � h � 7 o 2y2 � 5y � 12p 8a2 � 2a � 3 q 15u2 � 7u � 4 r 9c2 � 12c � 5

Stage 5.3

See Example 28

See Example 29

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a 5m2 þ 2m � 7 b 6g2 þ g � 12 c 3p2 þ 4p � 4d 7w2 þ 6w � 1 e 5y2 þ 14y � 3 f 3n2 þ 10n � 8g 4b2 þ 9b � 9 h 8m2 þ 10m � 3 i 3x2 þ 2x � 16

4 Factorise each expression given it is a perfect square.

a 81w2 � 180w þ 100 b 4y2 þ 8y þ 4 c 25h2 � 40h þ 16

5 Factorise each quadratic expression by first taking out a common factor.

a 6t2 þ 10t � 4 b 6g2 þ 15g � 36 c 24e2 � 28e � 12d 8a2 � 10a � 12 e 12t2 þ 20t � 8 f �25q2 � 5q þ 6g �12m2 þ 14m � 4 h 20 � h � 12h2 i 18 þ 48c þ 24c2

j 15 þ 9z � 6z2 k 12d2 þ 2d � 30 l 22x � 12 � 6x2


a 2a2 þ 5a þ 3 b 12m2 � 32m þ 5 c 4x2 þ 11x � 3d 7w2 � 8w þ 1 e 4h2 � 7h � 15 f 8x2 � 2x � 3g 5r2 þ 26r þ 5 h 2d2 � 15d þ 7 i 6n2 � 7n � 3j 8 � 6m � 9m2 k 3 � 2c � 5c2 l 15g2 þ 19g þ 6m 15 þ 14q � 8q2 n 3x2 � 13x þ 14 o 16 � 8d � 3d2

3-14 Mixed factorisations

Summary

Factorisation strategies• Look for any common factors and factorise first• If there are two terms, try factorising using the difference of two squares• If there are three terms, try factorising as a quadratic trinomial• If there are four terms, try factorising by grouping in pairs

Algebraic expression

Take out any common factors

Three terms

If quadratic trinomial,try to factorise

Four terms

Try to factorise bygrouping in pairs

Two terms

Factorise if differenceof two squares

Stage 5.3

See Example 30

Worked solutions

Factorising quadraticexpressions of theform ax 2 þ bx þ c

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Worksheet

Mixed factorisations

MAT09NAWK00036

Puzzle sheet

Factorominoes

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Puzzle sheet

Products and factorssquaresaw

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Example 31


a 3a2 � 27 b 5a2 þ 100 c 20b2 � 52b þ 24 d d3 � d2 � d þ 1

Solutiona 3a2 � 27 ¼ 3 a2 � 9

� �

¼ 3 aþ 3ð Þ a� 3ð ÞTaking out the HCF of 3 first.

Difference of two squares.

b 5a2 þ 100 ¼ 5(a2 þ 20) Two terms but not a difference oftwo squares.

c 20b2 � 52bþ 24 ¼ 4 5b2 � 13bþ 6� �

¼ 4 5b2 � 10b� 3bþ 6� �

¼ 4 5b b� 2ð Þ � 3 b� 2ð Þ½ �¼ 4 b� 2ð Þ 5b� 3ð Þ

d d3 � d2 � d þ 1 ¼ d2 d � 1ð Þ � 1 d � 1ð Þ¼ d � 1ð Þ d2 � 1

� �

¼ d � 1ð Þ d þ 1ð Þ d � 1ð Þ¼ d � 1ð Þ2 d þ 1ð Þ

Factorising by grouping in pairs.

Difference of two squares.

Exercise 3-14 Mixed factorisations1 Factorise each expression.

a m2 � 16m þ 64 b 3d2 � 3d c 3d2 � 4d � 15d 3k � 15 � 5h þ hk e 25y2 � 64 f 100f 2 � 64g q2 þ 3q � 3pq h 3 þ 2g � g2 i 24b2 þ 44b � 40j 25r2 � 1 k b3 þ b2 þ b þ 1 l 4x2 � 20x þ 25m 4 � d � 5d2 n b3 � b2 � b þ 1 o 8 � 2v2

p mn2 þ mnp þ 3mn þ 3mp q 2w2 � 24w þ 72 r 36h2 þ 12h þ 1


a 15r2 � 31rt � 24t2 b 4d2 þ 4d þ 1 c 9g2 � 36k2

d e3 � 3e2 � 10e e 5(p þ q)2 � 125(p � q)2 f 28x2 � 7g a2 � b2 þ 4a � 4b h c3 � 2c2 � 4c þ 8 i 6a2 þ 13a � 5j t2 � 3t þ 5t � 35 k 18p2 þ 24p þ 8 l 1 � 2a � 24a2

m 9x2 � 27x þ 18x � 54 n 2a2b � 6ab � 3a þ 9 o 2a2 þ 12a þ 18p 25u2 � 10u þ 1 q 4k2 � 5k � 21 r 48 � 3w2

s 3 � 27s2 t k3 þ 4k2 � 16k � 64 u 5y3 � 10y2 þ 15y

v m3n � 4mn w 8 � 2a2 x 32c2 � 40c � 12

Stage 5.3

See Example 31

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3-15 Factorising algebraic fractions

Example 32


a 10aþ 25b5 b 9y2 � 16

6yþ 8c x2 þ x�4x� 4 d t2 � 3t þ 2

3t2 � 5t � 2

Solution

a 10aþ 25b5 ¼ 5 aþ 5bð Þ

5¼ aþ 5b

b 9y2 � 166yþ 8

¼ 3yþ 4ð Þ 3y� 4ð Þ2 3yþ 4ð Þ

¼ 3yþ 4ð Þ 3y� 4ð Þ2 3yþ 4ð Þ

¼ 3y� 42

c x2 þ x�4x� 4 ¼

x xþ 1ð Þ�4 xþ 1ð Þ

¼ x xþ 1ð Þ�4 xþ 1ð Þ

¼ � x4

d t2 � 3t þ 23t2 � 5t � 2

¼ t � 2ð Þ t � 1ð Þt � 2ð Þ 3t þ 1ð Þ

¼ t � 2ð Þ t � 1ð Þt � 2ð Þ 3t þ 1ð Þ

¼ t � 13t þ 1

Example 33


a 4x2 þ x

� 2x2 � 1

b 3m� 64 3

8mm2 � 2m

c d2 þ 3d þ 2d2 � 9

4d2 þ d3d þ 9

Solutiona 4

x2 þ x� 2

x2 � 1¼ 4

x xþ 1ð Þ �2

xþ 1ð Þ x� 1ð Þ

¼ 4 x� 1ð Þx xþ 1ð Þ x� 1ð Þ �

2xx xþ 1ð Þ x� 1ð Þ

¼ 4x� 4� 2xx xþ 1ð Þ x� 1ð Þ

¼ 2x� 4x xþ 1ð Þ x� 1ð Þ

¼ 2 x� 2ð Þx xþ 1ð Þ x� 1ð Þ

Factorising denominators.

Using common denominators.

Stage 5.3

NSW

9780170193085

Chapter 1 2 3 4 5 6 7 8 9 10 11 12 13


106

b 3m� 64 3

8mm2 � 2m

¼ 3 m� 2ð Þ4 3

8mm m� 2ð Þ

¼ 3 m� 2ð Þ6 41

32 6 86m

6m m� 2ð Þ¼ 6

c d2 þ 3d þ 2d2 � 9

4d2 þ d3d þ 9 ¼

d2 þ 3d þ 2d2 � 9

33d þ 9d2 þ d

¼ d þ 2ð Þ d þ 1ð Þd þ 3ð Þ d � 3ð Þ 3

3 d þ 3ð Þd d þ 1ð Þ

¼ d þ 2ð Þ d þ 1ð Þd þ 3ð Þ d � 3ð Þ 3

3 d þ 3ð Þd d þ 1ð Þ

¼ 3 d þ 2ð Þd d � 3ð Þ

Exercise 3-15 Factorising algebraic fractions1 Simplify each expression.

a 3xþ 3y3 b 5

10t � 10rc ab� ac

a2

d y� 11� y

e w2 � 16wþ 4 f 5d � 5t

d2 � t2

g k þ 5ð Þ2k2 � 25

h 6c2 � 62cþ 2 i am� anþ m� n

m2 � n2

j y2 þ 9yþ 202yþ 10 k k2 � 3k � 4

k2 � 16l 16a2 � 25c2

4a2 � 9acþ 5c2

m s2 þ 4sþ 4s2 � s� 6

n 1� c� 2c2

3c2 þ 2c� 1o apþ 4p� 2a� 8

2p2 � 82 Simplify each expression.

a 5m mþ 1ð Þ þ

2mþ 1ð Þ mþ 2ð Þ b 6

wþ 5ð Þ wþ 3ð Þ �4

w wþ 3ð Þ

c 3bþ 2ð Þ b� 1ð Þ þ

1b� 1ð Þ b� 3ð Þ d 2

k2 þ k� 3

k2 � 1

e 54hþ 4þ

3h2 þ h

f 3d2 þ 3d þ 2

� 4d þ 2

g 3r2 � 36

� 54r þ 24 h 3

d2 þ 2dþ d

d2 � 4

i 5k2 � 3k � 4

� kk2 � 1

j 2q2 � 1

þ 3qþ 1

Stage 5.3

See Example 32

See Example 33

9780170193085


107


a 3mþ 92 3

4mmþ 3 b 5d � 10

3d � 9 35d � 158d � 16

c 4eþ 2 3

e2 þ 2e8e

d 3k þ 65 3

10kk þ 2 e 5h

3hþ 9 36hþ 18h2 þ h

f 4a2 � b2 3

3aþ 3b8

g r þ tt2 � r2 3

r2 � rt5r þ 5t

h 20mþ 167m� 7 3

7m5mþ 4 i p2 þ 2pþ 1

p2 � 13

4p� 4p2 þ p

j yþ 25y

47yþ 14

15yk 5

x2 � 44

152xþ 4 l 4nþ 8

nþ 5 46nþ 125nþ 25

m d2 þ dd þ 3 4

6dd2 � 9

n 1f 2 � 6f þ 9

44

f 2 � 9o 3f þ 6

f 2 þ f � 64

f 2 � 2f � 8f 2 � f � 12

Power plus

1 For this composite shape, write an expression for:

a the length of CD

b the length of BC

c its perimeterd its area

Bu + t

3u + 2t

2u + 5t

u – t

C D

EF

A

2 Find the average of 6r, 2r þ 8, r � 5, 2r, r þ 7 and 3r þ 8.

3 A rectangular garden has its longer sides each 5 m longer than its shorter sides. If itslonger sides each have length y m, then write a simplified expression for:

a the area of the garden b the perimeter of the garden

4 How many hours does it take a car to travel a distance of 20t2 kilometres at an averagespeed of 4t km/h?


a (a þ b)(a þ b þ c) b (x � 2)(x þ y � 3) c (p þ 3)(p � 4)(p � 5)d (x � 1)(y � 1)(z � 1) e (wt � 3r) 2 f (x3 þ 2)2


a n2 þ 4mn þ 4m2 b x2 � 2xy þ y2 c 25x2 � 40xy þ 16y2

d 5a2 � 30ab þ 45b2 e c4 þ 2c2 þ 1 f 8 � 7t2 � t4

g x4 � 1 h (a þ b)2 � c2 i (p þ q)2 � (p � q)2

7 If x2 þ bx þ c ¼ (x þ p)(x þ q), what are the signs of p and q if:

a b and c are both positive? b b and c are both negative?c b is positive and c is negative? d b is negative and c is positive?

Stage 5.3

9780170193085

Chapter 1 2 3 4 5 6 7 8 9 10 11 12 13


108

Chapter 3 review

n Language of maths

area diagram binomial binomial product coefficient

common factor constant term difference of two squares distributive law

expand expression factor factorise

formula grouping in pairs highest common factor (HCF) like terms

perfect square pronumeral product quadratic trinomial

simplify sum variable

1 Why is it possible to add or subtract terms like 6n and 4n but not 7x and 2y?

2 In the expression 2x2 � 3x þ 6, what is:

a the constant term? b the coefficient of x2?

3 In what type of algebraic expression is the highest power of the variable equal to 2?

4 Explain what product means.

5 In algebra, what is the opposite of expand?

6 Complete: A binomial is an algebraic expression with ____ _________.

n Topic overview• Write in your own words the new things you have learnt.• What parts of this topic did you like?• What parts of the topic did you find difficult or you did not understand?• Make a list of the skills you have learned in this topic, such as substituting and expanding.

Copy (or print) and complete this mind map of the topic, adding detail to its branches and usingpictures, symbols and colour where needed. Ask your teacher to check your work.

Expandingexpressions

Products andfactors

Factorisingexpressions

Multiplying anddividing terms

Algebraic fractionsAdding and subtractingterms

Factorising algebraicfractions

Puzzle sheet

Algebra crossword

MAT09NAPS10035

Quiz

Algebra

MAT09NAQZ00004

Worksheet

Mind map: Productsand factors(Advanced)

MAT09NAWK10037

9780170193085 109


a 9y þ 7y b 4ab � 7ab c 3x2 þ 5x2

d 7p þ 2q � 5p e 4w � 5 þ 2w f 8x2 � 13x � 5x2

g 2m þ 3n þ 5m þ 7n h 5u þ 7 � 2u � 3 i 12d2 þ 5d � 7d þ 4


a 9 3 2n b g 3 3h c a 3 d

d (�3) 3 2r e 12y2 4 3y f 2xy 4 y

g 36a 4 9a h 15t2 4 5t2 i 20pq 4 (�5q)


a h6þ 4h

6b 11k

8 �7k8 c 7

m�5m

d w4 �

2w5 e 4k

3 �k2 f 5

2d� 4

d

g d5k� 3

kh 8

mþ 1�3

mþ 14 Simplify each expression.

a 5m 3

4i

b hk

33k

c 143y

39y2d

d r4 4

r5 e 5

v 42v f a

b4

ab

g m5 4

5m10d

h 4x 4

12xp i ad

63

5a2 4

ad8

j 3c4d

49c

16d4

68 k 15a

4b4

43a2b

3abc

l 5y8 4

y4 3

15a2y


a 5ðr � 2Þ b aða� 10Þ c 3yð2x� 5yÞd 3pð4p� 7qÞ e �6nðn� 9Þ f �6t(t þ 7)g bðbþ 1Þ þ 5ðbþ 1Þ h 3xð2x� 7Þ � 5ð2x� 7Þ i 12� 6ð2x� 3Þ


a 4m � 12 b xy þ xz c �3g � 30d 16w2 þ 24w e �8qþ 16 f 15k � 10hk2

g 3ðy� 6Þ þ yðy� 6Þ h 3xð2x� 1Þ � 2ð2x� 1Þ i pðpþ 6Þ � 2ðpþ 6Þ

7 Expand each binomial product.

a (a þ 2)(a þ 3) b (y þ 3)(y � 7) c (t � 3)(t þ 8)d (h � 5)(h � 4) e (4 � g)(3 þ g) f (3p þ 1)(2p þ 3)g (5x þ 1)(4x � 3) h (4r � 5)(3r þ 4) i (5 � 3q)(2q � 1)


a (n þ 5)2 b (u � 3)2 c (9 þ x)2 d (6 � k)2

e (3y þ 1)2 f (2x � 5)2 g (6 � 7b)2 h (1 � 4h)2


a (n þ 6)(n � 6) b (b � 1)(b þ 1) c (10 þ m)(10 � m)d (3r þ 1)(3r � 1) e (4w � 5)(4w þ 5) f (1 � 8w)(1 þ 8w)

See Exercise 3-01

See Exercise 3-02

See Exercise 3-03

See Exercise 3-04

See Exercise 3-05

See Exercise 3-06

See Exercise 3-07

Stage 5.3

See Exercise 3-08

See Exercise 3-09

9780170193085110

Chapter 3 revision


a (2x þ 5)(x � 1) þ (x þ 3)2 b (y þ 4)2 � (y � 1)(y þ 2)c (3a þ 1)(3a � 1) � (3a þ 1)2 d (2p � 3q)2 þ (2p þ 3q)2


a 5mn � 3n þ 10m � 6 b ab � a � b þ 1 c x2y þ xy3 � 5x � 5y2

d 10p2 þ 30 þ p3þ 3p e w2 � 9 f m2 � n2

g 36 � r2 h 16a2 � 9 i 81x2 � y2

j 49a2 � 25b2 k 3q2 � 27 l 50m2 � 32m 24 � 150d2


a m2 � 11m � 12 b y2 � 4y þ 3 c t2 � 8t � 65d p2 þ 6p � 27 e 3n2 þ 9n þ 6 f 4y2 þ 4y � 8g 18 þ 3x � x2 h 60 � 5b � 5b2 i 11k � 4 � 6k2


a 6x2 þ x � 2 b 5s2 � 7s � 6 c 8w2 � 14w þ 3d 3e2 þ 12e þ 12 e 6y2 þ 15y � 36 f 24a2 � 28a � 12g 36x � 27x2 � 12 h 52t � 30 � 16t2 i 12g2 � 46g þ 14


a 64y2 � 25 b x2 � 16x þ 64 c 16r � 48d 3g � 15 � 5h þ hg e a3 � a2 � a þ 1 f 3b3 � 3b

g 4y2 � 5y � 21 h 18q2 þ 24q þ 8 i 1 � 2p � 24p2


a 9y2 � 2512y� 20 b 2x2 þ 3x� 5

6x2 þ 19xþ 10c 3w2 � 75

3w2 � 30wþ 7516 Simplify each expression.

a 4x2 þ x

� 3x2 � 1

b 4x2 � 9

þ 23xþ 9

c k2k þ 6

þ 5k2 � 9

d 1x� 3ð Þ xþ 2ð Þ þ

4xþ 2ð Þ xþ 3ð Þ

e 4dd2 � 3d þ 2

� 3d2 � 2d

f 3m2 � 2m� 3

� 2m2 þ 5mþ 4

g 3d � 6d þ 3 3

3d þ 95d � 10 h y2 � 5yþ 6

y3 � 4y3

2y2 � 8y2 � 2y

i 4d2 � 4

48

d þ 2 j mm2 þ m

44

5mþ 5

Stage 5.3

See Exercise 3-10

See Exercise 3-11

See Exercise 3-12

See Exercise 3-13

See Exercise 3-14

See Exercise 3-15

See Exercise 3-15

9780170193085

Chapter 3 revision

111

Documents

Number and algebraProducts and factorsweb2.hunterspt-h.schools.nsw.edu.au/studentshared/MATHEMATICS/Year... · a 2a 3 3b 3 5c b 4p 3 3 3 2q c 40n2 4 5n d 8b2 4 40b2 e 6h 3 2k 3 (