5

NUMBER AND ALGEBRA

PRODUCTS AND FACTORS

In 825 ce, the Persian mathematician al-Khwarizmi used the Arabic word ‘al-jabr’ to describe
the process of adding equal quantities to both sides of an equation. When al-Khwarizmi’s
book was translated into Latin and introduced to Europe, ‘al-jabr’ became ‘algebra’ and
the word was adopted as the name for the branch of mathematics that uses formulas to
describe patterns and relationships in our world. The photo above shows a monument of
al-Khwarizmi located in Khiva, Uzbekistan.

150 New Century Maths 10 Advanced 9780170453509

 Shutterstock.com/KKulikov
Chapter outline Wordbank
Working mathematically binomial An algebraic expression that consists of 2 terms, for
example, 4a + 9, 3 – y, x2 – 4x
5.01 The index laws U F R C binomial product An algebraic expression showing 2 or more
5.02 Fractional indices* U F R C binomials multiplied together, for example, (x + 9)(3x − 4).
5.03 Adding and subtracting U F factorise To rewrite an expression with grouping symbols,
algebraic fractions by taking out the highest common factor; factorising is the
opposite of expanding, for example, 9r2 + 36r factorised is
5.04 Multiplying and dividing U F 9r(r + 4)
algebraic fractions
index laws Rules for simplifying algebraic expressions
5.05 Expanding and factorising U F involving powers of the same base, for example, am ÷ an = am–n.
expressions perfect square A square number or an algebraic expression
5.06 Expanding binomial products U F R C that represents one, for example, 64, (x + 9)2
5.07 Factorising special binomial U F quadratic expression An algebraic expression in which the
products* highest power of the variable is 2, for example, 2x2 + 5x – 3 or
x2 + 2
5.08 Factorising quadratic U F R
quadratic trinomial An algebraic expression that consists of
expressions x2 + bx + c
3 terms, for example, x2 + 2x + 6
5.09 Factorising quadratic U F R
expressions ax2 + bx + c*
5.10 Mixed factorisations* U F R
5.11 Factorising algebraic fractions* U F R
*STAGE 5.3

U = Understanding | F = Fluency | PS = Problem solving | R = Reasoning | C = Communication

9780170453509 Chapter 5 | Products and factors 151

 In this chapter you will:
• apply index laws to algebraic expressions with integer indices
• (STAGE 5.3) apply index laws to algebraic expressions with fractional indices
• add, subtract, multiply and divide simple algebraic fractions
• (STAGE 5.3) add and subtract algebraic fractions with binomial numerators
• expand and factorise algebraic expressions involving terms with indices
• expand binomial products and factorise (monic) quadratic expressions of the form
x2 + bx + c
• (STAGE 5.3) recognise and expand special binomial products that are perfect squares or the
difference of 2 squares
• (STAGE 5.3) factorise an expression with 4 terms by grouping in pairs
• (STAGE 5.3) factorise (non-monic) quadratic expressions of the form ax2 + bx + c, including
those that are perfect squares or the difference of 2 squares
• (STAGE 5.3) factorise and simplify algebraic fractions

SkillCheck ANSWERS ON P. 600

1 Simplify each expression.
a g4 × g5 b r8 ÷ r2 c (d 5)3 d (–k)2
e h × h9 f m 5 ÷ m g a1 h c0
i 3e2 × 2e5 j 18n6 ÷ 6n2 k (10w3)3 l 25q0

m (vw)5 n   o y–1
3

p p–2
m

2 Simplify each expression.

 n

a 3a 7 p 10 p 8 5t
+ ×
5 4 2 3 t 24 14 2
x y
a b − c d ÷

5.01 The index laws

WS The index laws

am × an = am + n = am−n (am)n = am × n
Homework am
Index laws
review
am ÷ an =
an

(ab)n = anbn   = n a0 = 1
n
 a an
b b
1 1
=
Presentation −1

a −1 =
Index laws −n n
 a b  a  b bn
a− n =     =  = n
an a b a b  a a

152 New Century Maths 10 Advanced 9780170453509

 Example 1
Simplify each expression.
24e 6n12  2c 2 
b   (4d4q2 r)3
Simplifying
4 with the

8en 4
index laws
a c    
 d 

d 31x0 – (31x)0 e   9x –1 f   (4q)–3

Index laws

Solution
24e 6n12 3 24 e 6n12  2c 2  2c 2
(4d4q2r)3 = 43d4×3q2×3r3
4 4 5.01
( )
8en 4 1 8 en
a = 4 b c  d  = d 4
= 64d12q6r3
= 3e n
Numbers

2 c
=
5 8
and
4 2×4 powers

d4
16c 8
=
d4

1 1
Negative

d 31x0 – (31x)0 = 31 × 1 – 1 e 9x –1 = 9 × (4q)–3 =

indices

( 4q)3
f
= 30
x
9 1
= =
x 64 q3

Example 2
Simplify:
 2
(−1 ) 2
−5 −3

3 3r
a b  

Solution
5  2 3r 
(−1 ) 2 =  − 
−5 −5 −3 3

3  3 3r  2 
a b   = 
3
=  −  =
27r 3
5

 5 8
243
=−
3125

EXERCISE 5.01 ANSWERS ON P. 600

The index laws U F R C
1 Simplify each expression.
a 8m5 × 3m9 b 30k10 ÷ 25k2 c (–y2)5
R C
EXAMPLE
1

d (3q3)2 e 32a7d4 ÷ 8ad3 f 3x6y8× 7xy4

27 w y
g 5e4g3 × (−e6g) i (4c7)3
3 8

−9w 2 y
h

45hk 5m3
j (–2p)5 l 9u3vw2 × 6uv2w8
30hk 3m2
k

Foundation Standard Complex

9780170453509 Chapter 5 | Products and factors 153

 2 Simplify each expression. R C

5
a (l3m5)6 b    m3  d  
3 5 0
 w4 
 2  2
n
c

e (–8ky5)2 f (3p2q3r4)4 g 5x0 – (7x)0 h (–5d3y5)3

 2b   3k 4 
  j –9(a2b3)0 k 12h0
4 3

3d  − 10 
i l

3 Evaluate each expression.

a 7 b 5 × 40 c (–4)3 d (23)2
R C

e 65 ÷ 63 f (–32)0 g 82 ÷ 83 h 54 × 5–3
 3
  j 75 ÷ 73 k 122 ÷ 122 l 35 × 3–5
0

5
i

m 150 ÷ 151 n (4–1)2 o (23)0 + (20)3 p (50)3 × 52

4 Find the value of (3x)0 + 3 × 20. Select the correct answer A, B, C or D.

A 6 B 1 C 7 D 4

5 Express each term as a fraction.

a 2–3 b 10–6 c 8–1 d 6–2
R C

e 3–4

6 Simplify each expression using a positive index.

a 5–4 b 8–2 w–6
R C

c d m–5
e 4y–3 f (ad)–2 g (2x)–5 h 3mp–2
i 10g–2 h–3 j (4u)–4 k 2a5q−2 l 4p2 x−3
m 10a–5c3 n (–3r6)–3 3 −2 −1 p 4(5k3)–3
4
o q p

7 Simplify 2c−3. Select A, B, C or D.

1 2 1

2c 3 8c 3
A −6c B C D
c3
8 Simplify each expression using positive indices. R C

5
a   b 5y –2 c (5m)–3
−2

d x2y3w–2
 2d 

 4 y3  6c
  g 8(3ab2)–2
−2 −1 −1

 3  7k  5k 
 a
e  f h  2 
EXAMPLE 2

9 Simplify each expression. R C

1 5  2 3
a   b    −1  d  2 
EXAMPLE
2 −2 −3 −4 −2

 4  3 3  5
c

4  3a   2  8q 2 
e     g  − 
−3 −2 −5 −2

7  5 
f h  
 m  y

 4   10 g 4   7h 6 
 d5  k  − 3 
−3 −2 −5 −4
 k2 
 3d   5 
i j l
 w 3 
−

Foundation Standard Complex

154 New Century Maths 10 Advanced 9780170453509

 10 Simplify each expression.
a (10x y) × 5x y b (20k7m2 ÷ 5k−4m)4
R C

10 3 −2 3

c (2q–3r3)4 ÷ 8(qr2)2 d 8w–5m3 ÷ (6wm2)2

 4 a5 x 6   3 g 7n6 
e  6 2 
4 −2

f  2 g 5n2 

g (4p–3h4)3 × (–2p6h9) h (8d7n5)2 ÷ (–2d3n2)3

 ax 

i (–2c3e × 5c4e5)3 5.01

Fractional indices 5.02

Fractional indices STAGE 5.3

1
1
a2 = a Any number raised to the power of is the square root of that number.
2
1
1
a3 = 3 a Any number raised to the power of is the cube root of that number.
Indices

3
squaresaw
(Advanced)

1
1
an = n a Any number raised to the power of is the nth root of that number.
n

or n am Any number raised to the power of is the nth root of that number
m

( a)
m
m

raised to the power of m.

an = n Exponential
equations
n

Note: Taking the root first often makes the calculation simpler.

On a calculator, the nth root key is or found by pressing the or key before
pressing x  or y x respectively.
√   ⁻ x
√⁻ SHIFT 2nd F

Example 3
Evaluate each expression.
1 1 1
a 9612 13313
b    c   65618
Solution
1 1
a 9612 = 961 b 13313 = 3 1331
= 31 = 11
1
6561 = 8 6561
8
On a calculator: 8 6561
because 38 = 6561
c √   ⁻

=3
=

Foundation Standard Complex

9780170453509 Chapter 5 | Products and factors 155

STAGE 5.3
Example 4
Evaluate each expression.
3 1 5
32 5 27 3 25 2
− −
Fractional
a b    c  
indices
Solution
1 1
27 3 = 25 2 =
3 1 5
32 5 = 32
3
5
( )
− −
b 1 c 5
27 25 2
a
3

= 23 1 1
=8 27
( 25 )
=3 = 5

1 1
=
3 55
=
1
=
3125

Example 5
Write each expression using a fractional index.
1
a w9 b   4 m5 c   
3
y8

Solution
1 1 1 1
a w9 = w9 ( ) 2
b
4
m5 = m5 ( ) 4 c
3
y8
= 1
8 3
5
(y )
= m4
9
= w2 1
8

= or y 3
−
8
3
y

Example 6
Simplify each expression.
b   64 k
5 3
4
(81a ) 2 5 6
( ) 3
4m 4 n2 × 3 8m2 n
−
a c  

Solution
1
(81a ) = (81a ) (64k )
5 5
4 2 5 2 4 6 −3

(64k )
a b = 5
5 5 6 3

= 81 a
2×

1
4 4

=
5
= 243a 2 5 5
64 k 3
6×
3

1
=
1024 k 10
3 3
4m 4 n2 × 3 8m2 n = 4m 4 n2 × ( 8m2 n
1
3
c )
3 1 1 1
= 4m 4 n2 × 8 3 m
2×
3
n3
3 2 1
= 4m 4 n2 × 2m 3 n 3
3 2 1
= 8m 4 3 n
2+
3
+

17 7
= 8m 12 n 3

156 New Century Maths 10 Advanced 9780170453509

 EXERCISE 5.02 ANSWERS ON P. 601 STAGE 5.3

Fractional indices U F R C
1 Evaluate each expression. R

a 64 b −27 625
EXAMPLE
1 1 1 1
d 1000
3
2 3 2 3
c
e 16 4 ( −0.00032) 5 g (0.01) 2
1 1 1 1
f h (−512) 9
( −8) 3 ( −729) 3 k 256 8
1 1 1 1
i j l 3125 5 5.02

2 Write each expression using a radical (root) sign. R C

a 7 b 15 d
1 1 1 1
3 2 4
c d y2
e ( 4 x )6 (10m )
1

(2 a ) h (12m) 3
1 1 1
3 2 7 10
f g

3 Write 4 100k using a fractional index. Select the correct answer A, B, C or D. R C

A 25k B 100k C (100k ) 4 D 5k

1 1 1 1
4 4 4

4 Write each expression using a fractional index.

80 40 20
R C

3 6 4
a b c d
b xy 25 w
c
e f 3 g 5
h 8 4y

5 Evaluate each expression. R C

a 16 4 b 25 2 ( −27) 3
3 5 5 2
d 8 3
−
c
e (−128) 7 32 5 g 1000 3
3 1 5 1
h 900 2
− − −
f
64 6 9 2 k (−216) 3
5 7 4 2
( −243) 5
− − −
i j l

6 Evaluate the expression 100000 5 . Select A, B, C or D.

7
−
R C

1 1 1 1

1000000 10000000 1000 100000
A B C D
EXAMPLE 5

7 Evaluate each expression correct to 2 decimal places.

−250 b 37 c 14
EXAMPLE
1 3
4
3 2 4 7
−
a d
4 3 200 10 g (−50) 5
−138
5 3 6
4
82
− −
e f h

8 Write each expression using a fractional index. R C

1
k x
EXAMPLE
5
4 3 3 7 3
a b c d y

1 1 1
2 a3
m

4p
e 5 2
f g 5
h 4

9 Simplify 3 ( 8d2 ) . Select A, B, C or D.

d vw
5
R C

A 32 d B 4 d C 16d
EXAMPLE
5 10 10
D 32 d
6
30 3 3 3

Foundation Standard Complex

9780170453509 Chapter 5 | Products and factors 157

 10 Simplify each expression.
STAGE 5.3
R C

(25m ) (8 y ) (16 x w )
5
8 2 3 6 2 4 2 6 6
a b c

 1 
2

(27c 6 ( 8k 9
5 4
3 3

 32 p 
) )
− −
d e f 5 15


y5
2 1

(10 000n w ) (243 x 6 p 3 n 2 × 36 p 4n5

1
4 6 −2 10 4
5
g h ) i
1
(64m ) × ( 4m ) (216a b )
1
6 −3 2 3 6 9 2

625 y d
j 3 k l
8 4
( 4
)

Adding and subtracting algebraic

5.03
fractions
Example 7
Simplify each expression.
2x 5 7
+ b    −
Adding and
­subtracting

2 3 5 3 2y
a a x
algebraic
fractions a c    +
y

Solution
a a 3×a 2×a 2x x 3 × 2x 5 × x 5 7 2 ×5 7
2 3 3×2 2 ×3 5 3 3×5 5×3 y 2y 2y 2y
Algebraic
fractions a + = + b − = − c + = +

3a 2 a 6x 5x 10 7
6 6 15 15 2y 2y
= + = − = +

5a 17
6 15 2y
Algebraic x
fractions = = =

STAGE 5.3
Example 8
Simplify each expression.
6h + 1 4 h − 7 x x +1

8 12 3 5
Algebraic a + b −
fractions

Solution
6h + 1 4 h − 7 3 ( 6h + 1) 2(4 h − 7) x + 1 5 x 3( x + 1)

8 12 24 24 3 5 15 15
x
a + = + b   − = −

18h + 3 8h − 14 5 x 3x + 3
24 24 15 15
= + = −
18h + 3 + 8h − 14
5 x − (3 x + 3)
24
15
=

26h − 11
=

5 x − 3x − 3
24
15
=
=

2x − 3
15
=

Foundation Standard Complex

158 New Century Maths 10 Advanced 9780170453509

 EXERCISE 5.03 ANSWERS ON P. 601
Adding and subtracting algebraic fractions U F
1 Simplify each expression.
3c c 4 r 5r 7y 5y
− + +
EXAMPLE
7

2 7 2 5 7 2 8 3
n n
a b c d −

4t t 5y 3y 11t 5t 3a 4 a
+ − −
3 9 16 8 12 9 10 15
e f g h +

4 3a 7d 2d 12 7 11p 3p
5.03
+ − −
3y 3y 5x 5x 12 12
i j k l +
k k
5 3 15 2 5 3 9 4
+ − +
4h h 3b 4b 2n 3p
m n o p −

2 Simplify each expression.

q r

3 x 5u 2 a 3h
+ + −
7 4 8 7 15 10 16 24
w n d r
a b c d −

4k g 5m 2 r 3h 4 5u 3a
− + +
5 9 12 5 2 5 4 7
e f g h −

5q 11w 5k 3 4 n 5n
+ +
5 9 12 18 8 11 9 6
a c
i j − k l +

5 p 3p
3 Simplify − . Select the correct answer A, B, C or D.
6 8
29p 22 p 11p 2p

24 48 24 24
A B C D
1
+ . Select A, B, C or D.
w
4 Simplify
xw xy

y+w w2 + y y + w2 y 2 +w
A B C D
4w − 3 1 + 2w
. Select A, B, C or D.
xyw xyw xy wx

5 4
5 Simplify − STAGE 5.3

6 w − 17 6w − 7 26 w − 17 26 w − 7

20 20 20 10
A B C D EXAMPLE

6 Simplify each expression.

8

m+3 m−2 3p p + 2 y + 3 2y −1

5 4 5 3 4 3
a + b + c +

x −1 x −2 5h + 6 2 h + 2 k + 3 2k − 3

7 3 2 9 10 7
d + e − f −

a −1 a 3d − 1 2 d − 9 2y +5 y −1
−
2 3 7 6 4 9
g h + i −

1 − w 3w − 2 9 − 2e e − 5 8 − 3x 2 x + 5

10 7 3 2 11 4
j − k − l −

Foundation Standard Complex

9780170453509 Chapter 5 | Products and factors 159

 Multiplying and dividing algebraic
5.04
fractions
STAGE 5.3
Example 9
Simplify each expression.
3x 2 x 4 3k 2 3 xy 3 x
     b  ×        c  ÷        d 
4 9 k 16 5 25
Algebraic a × ÷
fractions v w
puzzle

Solution
3x 2 x 3x × 2 x
or 3x 2 x 1 3 x 12 x 4 3k 1 4 3 k
4 9 4×9 4 9 4 9 k 16 k 16 4
Algebraic a × = × = 2 × 3 b × = ×

6x2 3
fractions

x2
36 4
6
= =
=
x2
6
Upside- =
down

2 3 2 w xy 3 x x y 255
fractions

v w v 3 5 25 51 3x
c ÷ = × d ÷ = ×

2w 5y
3v 3
= =

EXERCISE 5.04 ANSWERS ON P. 601

Multiplying and dividing algebraic fractions U F
1 Simplify each product.
5 d 5 3
× ×
EXAMPLE

6 7 3 8
9 x w
a b c ×
k b
5 4 15 2 4a 5
×
3h m w 10 y 3 12c
d e × f ×

2x 9 6 10b 8h 2 h
×
3 8y 5b 18 p 3g
g h × i ×

5u 8c 9w 5p 20a 9a

12c 15u 45 pq 6 wh 3k 5 k
j × k × l ×

2 Simplify each quotient.

xy 5 x 4 5d
÷ ÷
3 7 24 6 5d 2
m g
a b c ÷

8a 5 a 9y 5y
÷ ÷
3 9 15 5 y 4 h 12
w w
d e f ÷

2c 2 8c 10 g 2 g 4 7k
÷
5e 3 7 kc 21k 3 xy 9 y
g h ÷ i ÷

5d2 d 4e 6e 2 9r 12r 2
÷
6 10 15 p2 5 p 5 35 g
j k ÷ l ÷

3 Simplify the quotient . Select the correct answer A, B, C or D.

5 xy 20
÷
3k 9ky
3y 4x 45k 3 xy 2

4x 15 y 2 60 y 4
A B C D

Foundation Standard Complex

160 New Century Maths 10 Advanced 9780170453509

 6ad 3 y 14 gm
4 Simplify the product . Select the answer A, B, C or D.
7m 4 d 27 ay
× ×

2 ak 14 gm 7m

7y 3 9 4
g
A B C D

5 Simplify each expression.

8 6 3dg 9kg 6k 12mn 8m
×
5k 4 h 5h 5 15
a ÷ b ÷ c ÷
xy xw
10a 5 8c 3n 5 9nm
÷ 4
3pf 21p 5 v 4 v 10
d ÷ e f ÷ ÷ 5.04

4 ab 3d 5bd 4 kw 8dm m 1
h 24 hw ÷
15d 8b 16 9y 3 6 12 d2
g ÷ × i ÷ ×

10 pq 6q 8ac 4 w 2c 2u 3 21 y
÷ ÷
3p 5h 5 p 9c yh 7 y 4 xy 8 x
j × k ÷ l ÷ ×
w

Mental skills 5: Maths without calculators ANSWERS ON P. 601

Multiplying by 9, 11, 99 and 101
We can use expansion when we multiply by a number near 10 or near 100.

1 Study each example.

25 × 11 = 25 × (10 + 1) 14 × 9 = 14 × (10 – 1)
= 25 × 10 + 25 × 1 = 14 × 10 – 14 × 1
a b

= 250 + 25 = 140 – 14
= 275 = 126
32 × 12 = 32 × (10 + 2) 7 × 99 = 7 × (100 – 1)
= 32 × 10 + 32 × 2 = 7 × 100 – 7 × 1
c d

= 320 + 64 = 700 – 7
= 384 = 693
27 × 101 = 27 × (100 + 1) 18 × 8 = 18 × (10 – 2)
= 27 × 100 + 27 × 1 = 18 × 10 – 18 × 2
e f

= 2700 + 27 = 180 – 36
= 2727 = 144

2 Now evaluate each product.

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

Foundation Standard Complex

9780170453509 Chapter 5 | Products and factors 161

 5.05 Expanding and factorising expressions

Example 10
Expand and simplify by collecting like terms.
a −3r2(4r + 2) − 5r3 b 9(m − 3) + m(m − 10)
Algebra
using
diagrams

Solution

−3r2(4r + 2) − 5r3 = −3r2 × 4r + (−3r2) × 2 – 5r3 Expanding

HCF by
factor trees

a
= −12r + (−6r ) – 5r
3 2 3

Factorising
= −17r3 − 6r2 Collecting like terms to simplify.
using
diagrams

b 9(m − 3) + m(m − 10) = 9 × m – 9 × 3 + m × m – m × 10

= 9m – 27 + m2 – 10m
WS

= m2 + 9m – 10m – 27 It’s neater to place m2 at the front.

Expanding
Homework
and

= m2 – m – 27 Collecting like terms to simplify.

factorising

WS

Example 11
Algebra
Homework4

Factorise each expression.

The distributive a 18xy2 – 24xy    b 3m(5 + 2d) + 7(5 + 2d)   c −5k2 + 15k
law

Solution
a The HCF of 18xy2 and 24xy is 6xy.
∴ 18xy2 – 24xy = 6xy × 3y – 6xy × 4 Rewrite the expression using the HCF 6xy.
Factorising
puzzle

= 6xy(3y – 4) Write the HCF at the front of the brackets.

b The HCF of 3m(5 + 2d) + 7(5 + 2d) is (5 + 2d).
∴ 3m(5 + 2d) + 7(5 + 2d ) = (5 + 2d) × 3m + (5 + 2d) × 7
Factorising
expressions

= (5 + 2d)(3m + 7)
c When factorising expressions that begin with a negative term, we use the ‘negative’ HCF.
The highest ‘negative’ common factor of –5k2 and 15k is –5k.
∴ −5k2 + 15k = (−5k) × k + (–5k) × (–3) (–5k) × (–3) = + 15k
= (−5k)[k + (–3)]
= −5k(k – 3)

162 New Century Maths 10 Advanced 9780170453509

 Example 12
Factorise each expression.
a 8a3 + 4a2 b 20h3k + 25h4k − 10h2k

Solution
a The HCF of 8a3 and 4a2 is 4a2.
∴ 8a3 + 4a2 = 4a2 × 2a + 4a2 × 1 Rewrite the expression using the HCF 4a2.
= 4a2(2a + 1)
5.05

b The HCF is 5h2k.

∴ 20h3k + 25h4k – 10h2k = 5h2k × 4h + 5h2k × 5h2 – 5h2k × 2.
= 5h2k(4h + 5h2 – 2)

EXERCISE 5.05 ANSWERS ON P. 601

Expanding and factorising expressions U F
1 Expand each expression.
a 5(d + 11) b −3(r + 10) c 7(x − 9y) d −4(a – 5w)
e −(2 − p )
2
f −10e(2e + 3)
2
g 6y(1 + 7y) h 4xy(3xy − 1)
i 8rq(2q − r) j 3ab(4b − 7a) k −6h (1 − 3h)
2
l −5x(5x2 + 4y)
m –(3 + 8a) n –2m2(3m – 4n) o 5g(3 + 7g2) p –1(5e – 12)

2 Expand −2y(5 + 7y). Select the correct answer A, B, C or D.

A 3y – 5y2 B −10y + 14y2 C −10y − 14y2 D −10y – 5y2

3 Use the substitution x = 2 to test whether each equation is correct or incorrect.

a 4(x + 10) = 4x + 40 b 5(x − 1) = 5x − 6 c x(3 − x) = 3x − x2

4 Expand and simplify by collecting like terms.

a 4k(3k – 5) – 9k2 b 5h − 7h(4 − h)
c 9w − 3w(5 + 2w )
3 2
d 24x3 − 5x2(2x2 − 5x)
e 8 – 3(2 – 7d) f 4n(3 – 5n) – 6n2
g 4y(y − 4) + 5(2y + 1) h 5(1 − 2a) − a(3 + 4a)
EXAMPLE
10

i 7(3 + 6w) – (5 – 8w) j 4y2(5y + 5) + 4(2 − 7y2)

k −v(2v + 7) + 6(v − 1) l 2(4 – 3a) − a(3 − a)
m 2c(5c − 1) − 4(7c − 5) n 3m(m + 5m2) − m2(1 − 3m)
o 4x(2y + 5) − 6y(10 − 3x)

Foundation Standard Complex

9780170453509 Chapter 5 | Products and factors 163

 5 Factorise each expression.
a 15y − 20 b 21 + 35w c 2p + p2
EXAMPLE
11

d 30y − 20y2 e 36d2 + 24d f 28k2 – 21k

g 8(c − 5) − c(c − 5) h m(3 + 2m) + 7(3 + 2m) i −q2 − 36q
j −8x + 12x2 k b(3b + 5) − 2(3b + 5) l −12cd2 + 8cd
m −hn2 + h2 n n −15g2 – 18g o 48q2 – 54q

6 Factorise −16pw + 10xw. Select A, B, C or D.

A −2w(8p – 10x) B −2w(8p – 5x) C −8w(2p − x) D −4w(5p + 2x)

7 Factorise each expression.

a 8m2y2 – 12my b 36ab2c + 27bc
c 24m2n − 108mn2 d 20dg2 – 35ag
EXAMPLE

e 40wy3 + 24w2y2 f 75g3h2 − 125gh

12

g −4p3 − 8p2 + p h 6mn2 + 3mn + 48m2 n

i 32p3 g + 8pg2 − 8pg j 18a5 − 12a2 + 15a4
k 28m3h2 – 21mh2 l 15kwp – 24wp2 – 9kw

Did you know?

CAPTCHA
For security purposes on a website, have you ever been asked to enter letters, words or
numbers that have been displayed in a wavy, difficult-to-read format like this?

This process is called CAPTCHA, which stands for Completely Automated Public Turing
Test to Tell Computers and Humans Apart, invented in 1999. It is designed to ensure that
the person accessing the website is a human and not another computer that could be hacking
into the site, sending spam or viruses. CAPTCHA uses optical character recognition (OCR),
which is a technology that can convert images of text into editable text.
How does CAPTCHA prove that you are a human and not a computer?
What is a Turing test, named after English computer scientist and mathematician
Alan Turing?

Foundation Standard Complex

164 New Century Maths 10 Advanced 9780170453509

 Expanding binomial products 5.06
Binomial means ‘2 terms’. (m + 8) and (m − 3) are binomial expressions because they each
have exactly 2 terms. (m + 8)(m − 3) is called a binomial product because it is a product of 2
STAGE 5.3

binomial expressions. WS

5.06
Example 13 Binomial
Homework
products

Expand each binomial product.

a (m + 8)(m − 3) b (4y − 5)(3y − 2)
WS

Area
Homework
diagrams

Solution
a (m + 8)(m − 3) = m(m – 3) + 8(m – 3) Each term in (m + 8) is multiplied by (m – 3)
= m2 – 3m + 8m – 24 Expanding
Expanding

= m2 + 5m – 24 Simplifying
binomial
products 1

One way of remembering which pairs of

terms to multiply together in a binomial
O
(m + 8)(m − 3) = m2 – 3m + 8m – 24
F
product is called the FOIL method
Expanding
binomial

I L = m2 + 5m – 24
(First-Outer-Inner-Last), as shown.
products 2

WS

(4y – 5)(3y − 2) = 12y2 – 8y – 15y + 10 Using FOIL

Algebra
Homework2

b
= 12y2 – 23y + 10 Simplifying
WS

Algebra
Homework6

Perfect squares
The formulas for expanding the perfect square of a binomial are:
Expanding

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

(a – b)2 = a2 – 2ab + b2 Technology

Expanding
binomials

Example 14
Expand each perfect square.
a (n − 5) b   (k + 4) c   (3y − 8)
Expanding

2 2 2
binomials

Solution
(n − 5)2 = n2 − 2 × n × 5 + 52 1st term squared – double product +
Special

= n2 − 10n + 25 2nd term squared

a binomial
products

(k + 4)2 = k2 + 2 × k × 4 + 42 (3y − 8)2 = (3y)2 − 2 × 3y × 8 + 82

= k2 + 8k + 16 = 9y2 − 48y + 64
b c

9780170453509 Chapter 5 | Products and factors 165

STAGE 5.3
Difference of 2 squares
(a + b)(a – b) = a2 − b2

The answer is called the difference of two squares.

Difference

When the sum of 2 terms is multiplied by their difference, the answer is the square of the first
of
2 squares

term minus the square of the second term (the difference of 2 squares).

Special
Example 15
Expand each expression.
binomial
products

a (m + 6)(m − 6) b (10 − e)(10 + e)

Products
c (8g − 3)(8g + 3) d (2x + 7w)(2x − 7w)
and factors
squaresaw
Solution
a (m + 6)(m − 6) = m2 − 62 b (10 − e)(10 + e) = 102 − e2
= m2 − 36 = 100 − e2
(8g − 3)(8g + 3) = (8g) − 3 (2x + 7w)(2x − 7w) = (2x)2 − (7w)2
Factoromi-

2 2
noes
c d
= 64g2 − 9 = 4x2 − 49w2
Trinominoes

Example 16
Expand and simplify each expression.
Enough a x(5y − x)2 b (2a − 3)(2a + 3) − (a + 3)2
time

Solution
x(5y − x)2 = x(25y2 – 10xy + x2)
WS
a
= 25xy2 – 10x2y + x3
Special
Homework
products

b (2a − 3)(2a + 3) − (a + 3)2 = 4a2 − 9 − (a2 + 6a + 9)

= 4a2 − 9 − a2 − 6a − 9
= 3a2 − 6a − 18

EXERCISE 5.06 ANSWERS ON P. 602

Expanding binomial products U F R C
1 Expand (5 + 3b)(2b – 7). Select the correct answer A, B, C or D.
A 6b2 + 11b – 35 B 6b2 – 11b – 35 C 6b2 – 31b – 35 D 11b – 6b2 – 35
EXAMPLE
13

2 Expand each binomial product.

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

Foundation Standard Complex

166 New Century Maths 10 Advanced 9780170453509

 3 Copy and complete the expansion of each perfect square.
STAGE 5.3

a (y + 7)2 = y2 + _____ + 49 b (h − 5)2 = _____ − 10h + 25

c (g − m) = g – 2gm + _____
2 2
d (u + 11)2 = u2 _____ + 121 EXAMPLE

e (3k + 5)2 = 9k2 _____ + 25 f (8x – 3w)2 = 64x2_____ + _____

14

4 Expand each perfect square.

a (h + 7)2 b (p − 5)2 c (w − 6)2 d (4 − c)2
(11 + y)2 f (d − w)2 g (10 − x)2 h (d + q)2
5.06
e
i (5u + 1)2 j (3b + 7)2 k (7e − 11)2 (2 – 5h)2
2
l
m (8x – 3y)2 n (6m + 5c)2 o (g + 12w)2
2

 2 + 

p
6 3
q  − m
a
2 2

 + 3 x 

r
m  x

5 Expand (4a – 5w)2. Select A, B, C or D.

A 16a2 + 25w2 B 16a2 − 9aw − 25w2
C 16a2 + 18aw − 25w2 D 16a2 − 40aw + 25w2

6 Expand each expression.

a (k + 7)(k − 7) b (x + 8)(x − 8) c (g − 1)(g + 1)
EXAMPLE
15

d (12 + w)(12 − w) e (4 + d)(4 – d ) f (7b − 5)(7b + 5)

g (3h + 7)(3h − 7) h (10e − 1)(10e + 1) i (11 + 4y)(11 – 4y)
j (1 − 5p)(1 + 5p) k (3a – 7c)(3a + 7c) l (9h – 5m)(9h + 5m)
4 4 1  1 4a 4a
m  − m  + m n  3 g +   3 g −  o  + 2   − 2 
 3  3


7 Expand (8g – 3y)(8g + 3y). Select A, B, C or D.

m m   g  g 

A 16g2 – 9y2 B 64g2 – 9y2

C 16g2 − 6w2 D 64g2 – 48gy + 25w2

8 Expand each expression.

a (3t − d)2 b (7k + 3)(3 + 7k) c (9 − 4g)(4g + 9)
d (5 + 2a)(2a − 5) e (ab – 8)(ab + 8) f (4n + 5p)(5n − 4p)
1 1
g (10x – 7y)2 h (2 − 11hm)(2 + 11hm)

i  xy − y   xy + y 

9 Expand and simplify each expression.

a 8y – (4y – 1)(1 – 2y) b 5(3w + 7) – (2w − 7)(w + 5)
EXAMPLE
16

c (a – b)2 – (b – 2a)2 d 4(p − 5)2 – 5(p – 3)(p + 8)

e (3 − y)2 – (3 + y)2 f 8x2 – (2x − 5)(2x + 5)
g (m + 2)2 – (m + 2)(m – 2) − (m − 2)2 h 2(6 − d)2 − (d + 5)2 − (6 − d) (d + 5)

Foundation Standard Complex

9780170453509 Chapter 5 | Products and factors 167

 10 Expand and simplify (4d − m)2 – (4d − m)(4d + m). Select A, B, C or D.
STAGE 5.3

A 8dm B 32d2 – 8dm C 2m2 – 8dm D 32d2 − 2m2

11 This diagram shows a house (3x – 10) metres long 30 m

and (2x +15) metres wide on a block of land with (3x – 10) m
dimensions 30 m × 20 m. R C

(2x + 15) m
Write down a binomial expression for the area of 20 m
the house in square metres.
a

b Expand and simplify this expression.

The green area of the block of land not covered by
the house is to be turfed. Write a simplified expression for this area in square metres.
c

12 A photograph frame is (k + 5) cm wide and (k + 5) cm

3k cm long. The gap between the photograph
and frame is 4 cm at the top and bottom and 4 cm
2 cm on each side. R C
a What is the area of the photograph frame?
b Write down expressions for the length and
width of the photograph. 2 cm 2 cm
Write down a binomial expression for the
3k cm
area of the photograph.
c

d Expand and simplify your expression for the

area of the photograph.
Find an expression for the area of the frame
not taken up by the photograph.
e

4 cm
13 Find an expression for the area of each shape in
simplest form. R C
4k + 3
2d + 3
a b c

3k + 2 5a + 4
5d – 7
7k – 9

Investigation
Squaring a number ending in 5
Study this mental shortcut for squaring a number ending in 5:
•• To evaluate 352, calculate 3 × 4 = 12, add ‘25’ to the end: 352 = 1225.
•• To evaluate 752, calculate 7 × 8 = 56, add ‘25’ to the end: 752 = 5625.
•• To evaluate 1052, calculate 10 × 11 = 110, add ‘25’ to the end: 1052 = 11 025.
Let n stand for the tens digit of the number ending in 5 being squared.
Expand (10n + 5)2 and investigate why the above method works.

Foundation Standard Complex

168 New Century Maths 10 Advanced 9780170453509

 Factorising special binomial products 5.07

Factorising by grouping in pairs STAGE 5.3

An algebraic expression with 4 terms can often be factorised in pairs, that is, 2 terms at a
time, to make a binomial product.
Grouping 5.07

Example 17
Factorise each expression.
a 6ae + 4cd + 8ad + 3ce b 3xy – 4px – 6y2 + 8py
c 12aw + 20cx – 8cw – 30ax

Solution
6ae + 4cd + 8ad + 3ce = 6ae + 8ad + 4cd + 3ce Grouping into pairs
= 2a(3e + 4d) + c(4d + 3e) Factorising each pair
a

= (3e + 4d)(2a + c) Factorising again

b 3xy – 4px – 6y2 + 8py = x(3y – 4p) – 2y(3y – 4p) Factorising each pair
= (3y – 4p)(x – 2y) Factorising again
12aw + 20cx – 8cw – 30ax = 2(6aw + 10cx – 4cw – 15ax) Factorising all terms first
= 2(6aw – 15ax – 4cw + 10cx) Grouping into pairs
c

= 2[3a(2w – 5x) – 2c(2w – 5x)] Factorising each pair

= 2(2w – 5x)(3a – 2c) Factorising again

Factorising the difference of 2 squares

The difference of 2 squares
Difference
of 2 perfect

a2 – b2 = (a + b)(a – b)
squares

Example 18 Difference of

Factorise each expression.

2 squares

a m2 – 25 b   36 – 49y2 c  48k2 – 3x2 d  w4 – 4w2

Solution
m2 – 25 = m2 – 52 36 – 49y2 = 62 – (7y)2
= (m + 5)(m – 5) = (6 + 7y)(6 – 7y)
a b

48k2 – 3x2 = 3(16k2 – x2) w4 – 4w2 = w2(w2 − 4)

= 3[(4k) – x ] = w2(w2 − 22)
c d
2 2

= 3(4k + x)(4k – x) = w2(w + 2)(w – 2)

9780170453509 Chapter 5 | Products and factors 169

STAGE 5.3 EXERCISE 5.07 ANSWERS ON P. ###
Factorising special binomial products U F
1 Factorise each expression.
a 3px + 2qx + 3py + 2qy b 2wh + 2wk – 3uh – 3uk
EXAMPLE
17

c 15mk + 20mg + 6nk + 8ng d 4yx – 8ay + 7ax – 14a2

e 14ak – 35af + 8k – 20f f cd + cy − hd – hy
g 4am + 4at + 4em + 4et h 3yk – 6by + 12k – 24b
i 12ac − 21mc – 35mw + 20aw j 21hk – 28he + 6pk − 8pe
k 30y – 6yq – 15y + 3q
2
l 32gh + 24gk – 4mh − 3mk
m 6ng – 10ch − 4cg + 15nh n p2w – p3 − w + p
o m(d − 3) – mx + a(d − 3) − ax p ek + k(x + y) + d(x + y) + ed

2 Factorise each expression.

d2 – 16 b x2 – 25 y2 – 169
EXAMPLE
18
a c
d p2 − 81 e 25 − h2 f 121 – a2
g 4r2 − 9d2 h 25g2 – 4e2 i 144 −49m2
j 81y2 – 16k2 k 1 – 4d 2 l m2 – 25n2
m 25q2 – 9b2 n 64b2 – u2 o 144x2 − 1
p 36k – w2 2
q g – 16p
2 2
r 4e2 – 121d2
1
m2p2 – 25 a2 − q2w2 x2 −
16
s t u
25 1 9
− h2 w 6 − 9c 2 16 v2 −
9 4 4
v x

3 Factorise each expression.

a 4m2 – 16p2 b 3y2 – 27 c w3 – 36w
d q – 64q3 e 12 – 48u2 f rh2 – r3
g 125a2b2 − 20v2 h 3x4 – 27p2 i 54g3p – 6gp3
1
j 28c – 63cw2 k 144e5 – 36e3 9d 2 − 2
4
l
1 7
m 100k 2 − 6 n 98a2 − 2w2 o c 2 − 1
4 9
p 108m3n2 −27m

4 Factorise each expression.

1 48
− x2 3p2 −
d2 w 2
25 9 49
a b c
p2
−

64b2 4 y 2
u4 − 16 f (m + 3)2 – 25
25 81
d − e
g (x – y)2 – y2 h 2a5 – 162a i (p + x)2 – (p – x)2
4c 2 e 2
− k 4g2 – (g – 2k)2 l 5w5 – 80w
9 36
j

Foundation Standard Complex

170 New Century Maths 10 Advanced 9780170453509

 Factorising quadratic expressions
5.08
x2 + bx + c
A quadratic expression is an algebraic expression in which the highest power of the variable
is 2, such as y2 + 12y − 4, −m2 – 8m, 5d2 – 7d + 1 and 3x2 + 10x.
A quadratic expression such as y2 + 12y – 4 is called a trinomial because it has 3 terms.
The expansion of (x + 2)(x + 9) is x + 11x + 18, a quadratic trinomial.
Factorominoes

2 5.08

∴ The factorisation of x2 + 11x + 18 is (x + 2)(x + 9).

Factorising quadratic trinomials

In the factorisation of a quadratic trinomial such as x2 + 11x + 18: Trinominoes

• each factor must have an x term to give x2     x2 + 11x + 18 = (x + 2)(x + 9)

• 2 + 9 = 11, which is the coefficient of x,

the number in front of the x                
x2 + 11x + 18 = (x + 2)(x + 9)

• 2 × 9 = 18, which is the constant term with no x  x2 + 11x + 18 = (x + 2)(x + 9)

Example 19
Factorise each quadratic trinomial.
a a2 + 7a + 12 b x + 9x + 8
Factorising

2
quadratic
expressions1

Solution
Find the 2 numbers that have a sum
of 7 and a product of 12.
a Pair of numbers Product Sum

It is best to test numbers that have

6, 2 6 × 2 = 12 6+2=8

a product of 12 and then check if

3, 4 3 × 4 = 12 3+4=7

their sums equal 7.

The correct numbers are 3 and 4.
∴ a2 + 7a + 12 = (a + 3)(a + 4)
b Find 2 numbers with a sum of 9 and a product of 8.
Test numbers that have a product of 8 and check if their sums equal 9.
The correct numbers are 8 and 1.
∴ x2 + 9x + 8 = (x + 8)(x + 1)

Factorising quadratic expressions of the form x2 + bx + c

• Find 2 numbers that have a sum of b and a product of c
• Use these 2 numbers to write a binomial product of the form (x ___ )(x ___)
Factorising
quadratic
expressions 1

9780170453509 Chapter 5 | Products and factors 171

 Example 20
Factorise each quadratic expression.
a x2 + 2x – 24 b c2 – c – 42 c k2 − 11k + 30

Solution
x2 + 2x − 24
Find 2 numbers that have a product of –24 and a sum of 2.
a

Since the product is negative, one of the numbers must be negative.

They are +6 and −4.
∴ x2 + 2x − 24 = (x + 6)(x − 4)
c2 − c − 42
Product = −42, sum = −1.
b

Since the product is negative, one of the numbers must be negative.

They are −7 and +6.
∴ c2 − c − 42 = (c − 7)(c + 6)
k2 − 11k + 30
Product = 30, sum = −11.
c

Since the sum is negative, at least one of the numbers must be negative.
Since the product is positive, both of the numbers must be negative.
They are −6 and −5.
∴ k2 − 11y + 30 = (k − 6)(k − 5)

EXERCISE 5.08 ANSWERS ON P. 603

Factorising quadratic expressions x2 + bx + c UFR

1 Find 2 numbers whose:

product is 24 and their sum is −11 b product is −40 and their sum is 6
R

a
c product is 15 and their sum is 8 d product is −48 and their sum is −8

2 Factorise each quadratic expression.

x2 + 12x + 32 b w2 + 9w + 14 y2 + 12y + 27
R
EXAMPLES
19,20
a c
d k2 + 13k + 36 e a2 + 19a + 90 f g2 + 10g + 21
g m2 – 11m + 28 h p2 – 8p + 15 i c2 – 15c + 54
j u2 – 5u + 4 k h2 – 14h + 48 l e2 − 8e + 16
m d2 + 3d − 28 n q2 + 2q – 80 o n2 + 4n − 45
p x2 – 6x − 7 q g2 – 11g − 80 r w2 – 4w − 96
s r2 – 12r − 28 t b2 – 3b − 54 u m2 – 4m − 12
v y2 – 14y + 49 w d2 – d − 30 x e2 + 9e − 90

Foundation Standard Complex

172 New Century Maths 10 Advanced 9780170453509

 Factorising quadratic expressions
5.09
of the form ax2 + bx + c
We have factorised quadratic trinomials of the type x2 + bx + c. For example, the factorisation of
x2 + 6x + 8 is (x + 2)(x + 4). These are called monic quadratic expressions because x2 does not
STAGE 5.3

have a coefficient (or its coefficient is 1).

We will now factorise quadratic trinomials of the type ax2 + bx + c, such as 8x2 + 14x − 15,
where x2 has a coefficient. These are called non-monic quadratic expressions.
5.09
Product
and factors
squaresaw

Example 21
Factorise each quadratic expression.
a 4y2 + 8y – 140 60 + 7d − d2
Perfect
squares
b
Solution
4y2 + 8y – 140 = 4(y2 + 2y − 35) Taking out the HCF of 4 first.
Technology
Factorising
a
= 4(y + 7)(y − 5) Product = −35, sum = 2
trinomials

b 60 + 7d − d = −d + 7d + 60
2 2
Rearranging the terms to make the d2 term first.
= −1(d2 – 7d − 60) Taking out a common factor of –1.
= −(d – 12)(d + 5) Product = −60, sum = −7
Factorising
trinomials

Example 22
Factorise 5k2 – 12k + 4. Factorising
quadratic

Solution expressions 2

There is no HCF, so we need to split up the middle term –12k.

Find 2 numbers that have a product of 20 and a sum of –12.
sum of –12
5k2 – 12k + 4 = 5k2 – 12k + 4

product of 5 × 4 = 20
Since the sum is negative, at least one of the numbers must be negative.
Since the product is positive, both of the numbers must be negative.
The 2 numbers are –10 and –2, so we will split –12k into –10k and –2k.
∴ 5k2 – 12k + 4 = 5k2 – 10k – 2k + 4
= 5k(k – 2) – 2(k – 2) Factorising by grouping in pairs
= (k – 2)(5k – 2) Factorising again

Factorising quadratic trinomials of the form ax2 + bx + c

• Find 2 numbers that have a sum of b and a product of ac
• Use these 2 numbers to split the middle term bx into 2 terms
• Factorise by grouping in pairs

9780170453509 Chapter 5 | Products and factors 173

STAGE 5.3
Example 23
Factorise each quadratic expression.
a 3x2 + 8x + 4 b   3x2 – 11x + 10 c  4x2 – 3x – 7
Factorising
Solution
3x2 + 8x + 4
quadratic
expressions
(Advanced)

3 × 4 = 12.
a

Find 2 numbers that have a product of 12 and a sum of 8. They are 6 and 2.
Split 8x into 6x and 2x.
Advanced
algebra

3x2 + 8x + 4 = 3x2 + 6x + 2x + 4
= 3x(x + 2) + 2(x + 2) Factorising by grouping in pairs
= (x + 2)(3x + 2)
3x2 – 11x + 10
3 × 10 = 30.
b

Find 2 numbers that have a product of 30 and a sum of –11.

Since the sum is negative and the product is positive, both numbers must be negative.
They are −6 and −5.
Split −11x into −6x and −5x.
3x2 – 11x + 10 = 3x2 – 6x – 5x + 10
= 3x(x − 2) − 5(x − 2)
= (x − 2)(3x − 5)
4x2 – 3x – 7
4 × (−7) = −28.
c

Find 2 numbers with a product of −28 and a sum of −3.

Since the product is negative, one of the numbers must be negative. They are −7 and 4.
4x2 – 3x – 7 = 4x2 – 7x + 4x – 7
= x(4x – 7) + 1(4x – 7)
= (4x – 7) (x + 1)

Example 24
Factorise each quadratic expression.
a 24k2 – 54k – 15 b 14 + 29a – 15a2
Solution
24k2 – 54k − 15 = 3(8k2 – 18k − 5) Taking out the HCF of 3 first
= 3(8k2 – 20k + 2k − 5) Product = −40, sum = −18
a

= 3[4k(2k − 5) + 1(2k − 5)]

= 3(2k − 5)(4k + 1)
14 + 29a – 15a = −15a2 +29a + 14
2
Rearranging to make the a2 term first
= −(15a2 − 29a − 14) Taking out a common factor of –1
b

= −(15a2 + 6a – 35a − 14) Product = −210, sum = −29

= −[3a (5a + 2) − 7(5a + 2)]
= −(5a + 2)(3a − 7)

174 New Century Maths 10 Advanced 9780170453509

 EXERCISE 5.09 ANSWERS ON P. 603 STAGE 5.3

Factorising quadratic expressions ax + bx + c U F R 2

1 Factorise each quadratic expression. Look for the highest common factor first.
a 2k2 + 16k + 30 b 3q2 – 3q – 90 c 5x2 – 55x + 140
R
EXAMPLE
21

d 6y2 + 12y − 18 e n3 – 16n2 + 63n f 24 + 2a − a2

g 4p + 32 − p2 h –x4 – 4x3 + 21x2 i –20 + 24h – 4h2

2 Factorise each quadratic expression.

5.09

a 6d2 + 19d + 15 b 8m2 + 10m + 3 c 2y2 + 7y + 5

R
EXAMPLE
22

d 2w2 + 31w + 15 e 6k2 + 23k + 20 f 3u2 + 22u + 35

g 4x2 + 21x + 27 h 5e2 + 39e + 28 i 12h2 + 13h + 3

3 Factorise each quadratic expression.

a 4k − 11k + 6 b 6w – 17w + 5 c 5p2 – 23p +12
R
EXAMPLE
2 2 23

d 6g2 – 35g + 49 e 3x2 – 32x + 20 f 21m2 – 26m + 8

g 20r2 – 23r + 6 h 10a2 – 7a – 6 i 20u2 – 23u − 21
j 2p2 – 17p – 9 k 6y2 − y – 40 l 12b2 – 17b − 5
m 12q2 – 8q – 15 n 4h2 + 13h – 12 o 10n2 + 31n − 14

4 Factorise each expression given it is a perfect square.

a 16w − 56w + 49 b 121h + 176h + 64 c 9p2 – 90p + 225
R

2 2

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

a 6m – 9m – 60 b 8x − 18x + 4x 6 −5e − 29e
R
EXAMPLE
2 3 2 2 24
c
d −18w2 – 51w + 42 e 4 – 4k − 48k2 f 8 + 2n − 15n2
g 48d2 + 156d + 90 h 30h − 24h2 + 9 i 3 + 5p – 28p2

6 Factorise the quadratic expression 8a2 + 14ax – 15x2. Select A, B, C or D.

A (4a − 3x)(2a + 5x) B (2a − 5x)(4a + 3x)
R

C (5a + 4x)(3a – 2x) D (4a − 5x)(2a + 3x)

7 Factorise each quadratic expression.

a 28d − 11d + 1 b 6 – 2y – 20y2 c 30w2 + 41w − 15
R
EXAMPLE
2 21

d 4p2 + 23p + 15 e 23a – 12 −10a2 f 9r2 + 29r + 20

g 28k2 + 19ky + 3y2 h 72p – 6p2 – 3p3 i 4 − 21m – 18m2
j 10g – 54g2 – 36g3 k 44u2 − u – 3 l 32h2 + 28hm – 15m2

Foundation Standard Complex

9780170453509 Chapter 5 | Products and factors 175

5.10 Mixed factorisations

STAGE 5.3
Factorisation strategies
• Look for any common factors and factorise first
• If there are 2 terms, try factorising using the difference of two squares
• If there are 3 terms, try factorising as a quadratic trinomial
Factor­
ominoes

• If there are 4 terms, try factorising by grouping in pairs

Algebraic expression
Mixed
­factorisations

Take out any common factors

2 terms 3 terms 4 terms

Factorise if difference If quadratic trinomial, Try to factorise by

of 2 squares try to factorise grouping in pairs

Example 25
Factorise each quadratic expression.
a 5k2 – 80 b 4y2 + 60d2
c 24u2 – 68u + 20 d 9m3 – 18m2 – 4m + 8
Solution
5k2 − 80 = 5(k2 − 16) Taking out the HCF of 5 first.
= 5(k + 4)(k − 4) Difference of 2 squares
a

4y2 + 60d2 = 4(y2 + 15d2) 2 terms but not a difference of 2

squares.
b

24u2 – 68u + 20 = 4(6u2 – 17u + 5) Taking out the HCF of 4 first.

= 4(6u2 – 15u – 2u + 5) Product = 30, sum = −17
c

= 4[3u(2u − 5) – 1(2u − 5)] Factorising each pair.

= 4(2u − 5)(3u − 1) Factorising again.

9m3 – 18m2 – 4m + 8 = 9m2(m − 2) − 4(m − 2) Factorising by grouping in pairs

= (m − 2)(9m2 − 4)
d

= (m − 2)(3m + 2)(3m − 2) Difference of 2 squares

176 New Century Maths 10 Advanced 9780170453509

 EXERCISE 5.10 ANSWERS ON P. 603 STAGE 5.3

Mixed factorisations U F R
1 Factorise each expression.
a 60y2 − 15 b 15m2 – 41m + 14 k2 – 14k + 49
R
EXAMPLE

c 25

d 11a – 12 – 2a2 e 6gy + 12cp + 8cy + 9gp f 15 – 17x – 4x2

g 64q2 – 16q + 1 h 81r2 – 36w2 i 3bn2 – 3np + 2bnd −2dp
j 6k2 + 23k + 20 k 42a2 + 73a + 28 4 – 100h2
5.10
l
m 36c2 + 53c − 63 n 16 – 46d – 35d2 o 25u2 + 10u + 4
p 3xy – 8w + 4x – 6wy q 1 – b2 + b3 − b r 28c2 – 9ch – 4h2
s 15n2 – 19n − 56 t a3p – ap u 16r2 – 146rk + 18k2

2 Factorise each expression.

a 48c − 75 b 8q2 + 22q − 40 c 15x2 + x3 – 2x4
R

d 27p2 – 24pr – 16r2 e 9w2 + 24w − 105 f n4 – 4n2

g y – y2 – y3 + y4 h 36u2 + 105u + 49 i 121k2 – 198k + 81
j 4(c – d)2 – (c + d)2 k 20y2 + 18d + 12y + 30dy l ah2 – 6ah – ah3
m 18b2 – 54b + 27b − 81 n 32g – 8g3y2 o 8 – 42e – 36e2
p 27q2p – 3p q 30h – 15h2 – 15h3 r w4 – w3 – w2 + 1
s a5 – 256a3 t m2 – k2 + 9m – 9k u 30a2 + 65a – 25

3 Factorise the expression w6 – w4 – w2 + 1. Select the correct answer A, B, C or D.

A w (w – 1)(w + 1) B (w – 1)(w + 1)
R

4 2 2 2

C (w – 1) 4 2
D (w2 – 1)2(w2 + 1)

Factorising algebraic fractions 5.11

Example 26 STAGE 5.3

Simplify each expression.

12 x + 8 y 30m − 18 2 h 2 + 8h 5 k 2 + 9k − 2

WS

4 25m2 − 9 −3h − 12 k 2 + 7 k + 10
a b c d
Homework
Simplifying

Solution
algebraic
fractions

12 x + 8 y 4(3 x + 2 y ) 30m − 18 6(5m − 3)

4 4 25m2 − 9 (5m + 3)(5m − 3)
a b
= =

6
= 3x + 2y
5m + 3
=

c 2 h2 + 8h 2 h( h + 4) d 5k 2 + 9k − 2 (5k − 1)( k + 2)
−3h − 12 −3( h + 4) k 2 + 7 k + 10 ( k + 2)( k + 5)
= =

2h 5k − 1

3 k +5
=− =

Foundation Standard Complex

9780170453509 Chapter 5 | Products and factors 177

STAGE 5.3
Example 27
Simplify each expression.
5 2 6m m2 − 16 h2 + 5 h + 6 h2 + 2 h

x + x − 2 x2 − x m + 4m 9 h + 4h + 3 h − 2h − 3
a 2
− b  2
× c  2
÷ 2

Solution
a 5
−
2
=
5
−
2 Factorising denominators

Using common denominators

x 2 + x − 2 x 2 − x ( x + 2)( x − 1) x( x − 1)
5x 2( x + 2)
x ( x + 2)( x − 1) x ( x + 2)( x − 1)
= −

5 x − 2( x + 2)
x ( x + 2)( x − 1)
=

5x − 2x − 4
x ( x + 2)( x − 1)
=

3x − 4
x ( x + 2)( x − 1)
=

6m m2 − 16 6m (m − 4)(m + 4)
m + 4m 9 m(m + 4) 9
b 2
× = ×

6m
2
(m + 4)(m − 4)
m(m + 4) 3
9
= ×

2(m − 4)
3
=

h2 + 5 h + 6 h2 + 2 h h2 + 5 h + 6 h2 − 2 h − 3
h + 4h + 3 h − 2h − 3 h + 4h + 3 h + 2h
c 2
÷ 2 = 2 × 2

( h + 2)( h + 3) ( h − 3)( h + 1)
( h + 1)( h + 3) h( h + 2)
= ×

h−3
=
h

EXERCISE 5.11 ANSWERS ON P. 603

Factorising algebraic fractions U F R
1 Simplify each expression. R

15 x + 9 y 6 4 k − 8d

EXAMPLE
26

3 12r − 18 4
a b c

a2 − 4 3 g + 3h 4 p2 − 16

a+2 12 p − 24
d e f
g 2 − h2
w 2 − 49 7− y c 2 + 11c + 30

4( w + 7)2 y−7 6c + 30
g h i

k 2 + 5k − 24 9 y 2 − 4 q2

2 k 2 − 18 3 y 2 − 4 yq − 4 q2
he − hd + pe − pd
j k l
e 2 − d2
r − 2r − 24 40 + 11b − 2b2 6m − 30 − pm + 5 p

2

r + 12r + 32 8b2 + 34b + 35 m2 + 3m − 40

m 2 n o

x 2 y2 − 4

xy − 3mxy − 2 y + 6m
25 − 4a2 5n2 − 4n − n3
p q r 2
6a − 11a − 10
2
n − 2n + 1
2

Foundation Standard Complex

178 New Century Maths 10 Advanced 9780170453509

 2 Simplify each expression. R

7 4 8 3

EXAMPLE
27
k ( k − 1) k ( k + 1) ( x + 4)( x − 7) x ( x + 4)
a − b +

1 4 5 1

(m + 3)(2m + 1) (2m + 1)(m − 2) 3h + 24 h2 − 64
c + d −

10 3 4

p2 + 4 p p2 − 4 p a2 − 25 ( a + 5)2
a
e + f −

2 3 7

w 2 − 9w + 20 w 2 + 2 w − 24 d 2 − 3d − 4 2 d − 8
d 5.11
g + h −

9m 5 3u 5

16 − 25m2 32m − 40m2 u2 + 5u − 14 2u2 − 98
i − j −

c +1 n +1 2

10c 2 + 7c − 12 5c 2 − 39c + 28 4n3 − 36n 5n2 + 15n
c
k + l +

3 Simplify each expression. R

3r r +5 4 y − 12 y 2 + 3 y

5r + 25 9 8 y − 24
a × b ×
y
w 2 + 6 w w 2 − 36 8a − 6 12 a − 9

2 10 w 3a 2 + 3a a 3 − a
c ÷ d ÷

5n 2n2 + 4n x 2 + 4 x − 5 x 2 − 16

n + 10n + 21 n − 2n − 15 x2 + 4 x 3 x + 15
e 2
÷ 2 f ×

15 y − 12 k 8 y + 4k m2 − 4 m + 4 5m − 10

2 y + ky − 2 y − k 25 y 2 − 16k 2 7m 21m2 + 42m
g 2
× h ÷

1 − 9m2 2m + 10 c 3 + 8c 2 c 2 − 6c + 9

m + 9m + 20 3m2 + 5m − 2 6c − 18 c − 64
i 2
× j × 2

3e 5e 2p − 6 p2 + 3p − 18

e 2 + 5e − 6 4 − 4 e 2 2 p2 − 7 p − 4 p2 + 3p − 28
k ÷ l ÷

d2 + 10d + 21 d2 + 4 d − 12 h2 + 8h + 12 h2 + 10h + 24

5d2 − 20 d + 6d − 7 4 h + 12 h + 4h + 3
m × 2 n ÷ 2

Power plus ANSWERS ON P. 604

1 Expand and simplify each expression.
+

a (x + 5)3 b (y – 2)3 c (a + b)3 d (3d + 10)3

2 Use the given expansion to evaluate each square number without using a calculator.
a 212 = (20 + 1)2 b 452 = (40 + 5)2
c 292 = (30 − 1)2 d 592 = (60 − 1)2
e 1022 = (100 + 2)2 f 982 = 100 − 2)2
3 By expressing 31 × 29 as (30 + 1)(30 − 1), evaluate 31 × 29 without using a calculator.
4 Use the method of question 3 to evaluate each expression.
a 21 × 19 b 51 × 49 c 89 × 91 d 78 × 82

Foundation Standard Complex

9780170453509 Chapter 5 | Products and factors 179

 5 Expand and simplify each expression.
y + 1)2 + (y + 2)2 + (y + 3)2
a ( 
b (x − 3)(x + 3) + (x + 3)2 + (x − 3)2
c (5n + 3)(5n − 3) + (3n − 5)(3n + 5)
d 2(a − b)(a + b) − (a + b)2 − (a − b)2
6 Factorise each expression.
a p4 – 16q4 b w8 − 256 c (m – 2n)2 – (2n + m)2
7 a  Expand (a – b)(a2 + ab + b2)
b Hence, factorise :
i x3 – y3 ii 8m3 – 27p3
8 a  Expand (a + b)(a2 − ab + b2)
b Hence, factorise:
i d3 + w3 ii u3 + 64h3

180 New Century Maths 10 Advanced 9780170453509

 CHAPTER 5 REVIEW
Language of maths
algebraic fraction base binomial binomial product
Qz

coefficient constant term denominator difference of 2 squares

Numbers
and powers

expand factorise fractional grouping in pairs

CHAPTER 5 REVIEW
highest common index law indices perfect square
Qz

factor (HCF)
Advanced
algebra and
functions

power product quadratic expression reciprocal

root sum term trinomial
Qz

1 Explain the meaning of quadratic and trinomial. Give an example of a quadratic

Algebra

WS
trinomial.
2 Any number raised to the power of 0 is equal to what?
3 What is the difference between expand and factorise?
Algebra 6
Homework

4 What power is associated with the reciprocal of a term or number?

5 In the quadratic expression 2x2 – 3x + 6, what is:
a the constant term? b the coefficient of x?
6 Copy and complete:
To factorise quadratic expressions of the form x2 + bx + c, first find 2 numbers
that have a ________ of b and a ________ of c.

Topic summary
• What was this topic about? What was the main theme?
• What content was new and what was revision for you?
• Write 3 index laws in both words and symbols
• Write 10 questions (with solutions) that could be used in a test for this chapter. Include some
questions that you have found difficult to answer.
• List the sections of work in this chapter that you did not understand. Follow up this work.
Print (or copy) and complete this mind map of the topic, adding detail to its branches and using
WS

pictures, symbols and colour where needed. Ask your teacher to check your work.
Mind map:
Homework
Products and
factors

Expanding and
The index Algebraic
factorising
laws fractions
expressions

Factorising algebraic PRODUCTS AND Expanding

fractions FACTORS binomial products

Factorising quadratic Factorising special

expressions binomial products

9780170453509 Chapter 5 | Products and factors 181

 TEST YOURSELF 5 ANSWERS ON P. 604

1 Simplify each expression.

24r 8h 8
a 3v4w3 × 2v5w8 c (5xy2)2
5.01
3rh2
b
2p  2p 
d   e (4k)–1
0 3

 3  3 
f

 5b 8 y 6 
g (4m)–2 h 4m–2
TEST YOURSELF 5

i  b2 y 3 

 45ab 4 
j (4t4u5)3 × 8t2u k 45c6d8 ÷ (–3cd2)2
−1

l  54 a2b 3 

2 Simplify each expression.

1 3  4g  9
a   b    
5.01
−5 −2 −3 −2

 10   5 7  2d 
c d  

STAGE 5.3 3 Evaluate each expression.

400 2 b ( −27 )3 64 2
1 1
d ( −32 )5
5 − 2

5.02
a c

4 Evaluate each expression.

1
(16a ) (81m ) (125 x y )
3 2
4 2 12 5 3 6 −3
5.02
4

8p
a b c 6
d
( 3
)
5 Simplify each expression.
r 3r 5 g 3g 11 x 2 x 3 2m
− + −
5.03

4 5 3 2 24 3 4 7
a b c d −

STAGE 5.3
6 Simplify each expression.
3k − 4 2 k w +1 w − 3 2p + 5 7p 5 +2y 1− y
+ −
5 3 4 9 3 12 11 8
5.03 a b − c d +

7 Simplify each expression.

4m 3m 6 4 de 5 10 vw 3 3v 4
÷
5.04

5 8 d 21 5 20 3w
a × b × c d ÷ ×

8 Expand and simplify each expression.

a b

a 8pq(7p − 5q) b −5h2(3h + 7)

5.05

c 4eg(g – 6e) – 6e2g d 12(9 – n) – 5(2n + 3)

e x (6x + x ) +2x(5x + x )
2 2 3 2
f 3(7 – 2y) – 5y(3 +4y)

9 Factorise each expression.

a 16ar2 + 24ar b −24p + 18q
5.05

c 2(5x – 1) – 3x(5x – 1) d 15xy2 – 30x3y3

e 6pt2 + 12p2t – 48p3 f 32r2m4 + 12r4m3
g 50x4y3 – 75x3y4 h −8p3q3 + 48p3q6
i n(n2 + 6) – (n2 + 6)

Foundation Standard Complex

182 New Century Maths 10 Advanced 9780170453509

10 Expand each binomial product.
a (b + 3)(b + 10) b (d + 8)(d − 10) c (w − 6)(9 – w)
STAGE 5.3

d (5x + 7)(4x − 3) e (7y − 3)(7y + 3) f (5p − 8)(5p − 8) 5.06

11 Expand each binomial product.

a (n + 9)(n − 9) b (3y + 2d)2 c (4n − 11)(4n + 11)
5.07

12 Factorise each expression.

TEST YOURSELF 5
a 3pu + 2qt + 2qu + 3pt b 4ab + 6bc – 6ad – 9dc b2 − 100
5.07
c
d 25 – 16y2 e 20x2 − 125 f 3r3 – 27r

13 Factorise each expression.

y2 + 10y + 25 b q2 – 21q + 20 n2 + 8n − 33
5.08
a c
d a2 – 11a + 28 e m2 – 5m – 84 f p2 + 3p – 54

14 Factorise each expression.

a 3w2 + 5w + 2 b 2y2 – 3y – 9 60 – 5b – 5b2
5.09
c
d 3p2 + 10p – 8 e 12x2 – 46x + 14 f 6n2 – 13n + 6
g 40y2 + 49y – 24 h 48c2 − 64c –35 i 10e2 − 7e + 1

15 Factorise each expression.

a 5q2 – 45 b 20x2 – 52x + 24 c3 − c2 − c + 1
5.10
c
d 4 – 47k – 12k2 e 6m2 + m – 40 f a4 – a2 + a3 − a

16 Simplify each expression.

20d − 35 4r + 12 y

5.11

10 7r 2 + 21ry
a b

u2 − 4u − 32 1 3

2u2 − 128 25 − h2 h2 + 13h + 40
c d −

4 2 6m2 − 24

p − 9 p2 + 6 p + 9 7m + 14
e 2
+ f

3 3w − 15 20 w

y 2 + 4 y y 2 − 16 5 3w − 15 w
y
g − h × 2

b2 + 3b + 2 b2 + b 6a 2

b2 − 9 3b + 9 a2 + 5a + 6 a2 − 5a − 14
i ÷ j −

3 − 12 q2 q2 + q − 2 r 3 − 25r r 2 + 12r + 35

q + 4 q + 4 6q − 12 q2 r2 − 9 12r − 36
k 2
× l ÷

Foundation Standard Complex

9780170453509 Chapter 5 | Products and factors 183