Ch 06

Factorising

6
The area of a computer
mouse pad is 500 cm2. List
three possible dimensions
(length and width) of the
mouse pad, if it is
rectangular in shape.
Simone is designing
different sized mouse pads.
As the dimensions can vary,
she uses the algebraic
expression x2 + 5x to
represent the area, where x
can be any positive value.
If the width of the mouse
pad is x, what algebraic
expression would represent
the length of this mouse
pad?
In this chapter you will
learn how to ﬁnd the
product of factors which
make up an algebraic
expression using different
techniques of factorising.

180 Maths Quest 9 for Victoria

The highest common factor
Previously, the term expanding was defined as changing a compact form of an
expression to an expanded form. Now, we can define factorising as the reverse oper-
ation, going from an expanded form to a more compact one. This compact form is a
product of factors, or two or more factors multiplied together. First we need to know
how to find all the factors of integers.

Factors
The factors of an integer are two or more integers which, when multiplied together,
produce that integer.
3 × 2 = 6, so 3 and 2 are factors of 6.
If an integer divides into a number without a remainder, it is a factor of the number.
16 ÷ 8 = 2, so 2 and 8 are factors of 16.
The number itself and 1 are always factors of a given number.
7 and 1 are both factors of 7.

WORKED Example 1
Find all the factors of each of the following integers.
a 12 b −21 c −16
THINK WRITE
a 1 Multiplications that make 12 are a 1, 12; 2, 6; 3, 4
1 × 12, 2 × 6 and 3 × 4. All these
numbers are factors.
2 Because 12 is a positive number, it could −1, −12; −2, −6; −3, −4
be the result of multiplying 2 negative
numbers. List the negative factor pairs.
3 List the factors of 12 in ascending Factors of 12 are −12, −6, −4, −3, −2, −1, 1,
order. 2, 3, 4, 6, 12.

b 1 Multiplications that make −21 are b 1, −21; 3, −7
1 × −21 and 3 × −7.
2 Since −21 is a negative integer, the −1, 21; −3, 7
factor pairs can change signs, so list
the other pairs of factors.
3 List all the factors of −21 in Factors of −21 are −21, −7, −3, −1, 1, 3, 7,
ascending order. 21.

c 1 Multiplications that make −16 are c 1, −16; 2, −8; 4, −4
−16 × 1, −8 × 2 and −4 × 4.
2 Since −16 is a negative integer, the −1, 16; −2, 8
factor pairs can change signs, so list
the other pairs of factors. Note that
−4 and 4 is the same as 4 and −4.
3 List all the factors of −16 in Factors of −16 are −16, −8, −4, −2, −1, 1, 2,
ascending order. 4, 8, 16.

Chapter 6 Factorising 181
Finding the highest common factor
The highest common factor or HCF of two or more numbers is the largest factor that
divides into all of the given numbers without a remainder. This also applies to algebraic
terms.
The HCF of xyz and 2yz is yz because:
xyz = x × y × z
2yz = 2 × y × z.
The HCF is yz (combining the common factors of each).
For an algebraic term, the HCF is found by taking the HCF of the coefficients and
combining all common pronumerals.

WORKED Example 2
Find the highest common factor (HCF) of each of the following.
a 6 and 9 b 12, 16 and 56 c 4abc and 6bcd

THINK WRITE
a 1 Find the factors of 6 and write them in a 6: 1, 6; 2, 3
ascending order. Consider positive Factors of 6 are 1, 2, 3, 6.
factors only.
2 Find the factors of 9 and write them in 9: 1, 9; 3, 3
ascending order. Factors of 9 are 1, 3, 9.
3 Write the common factors. Common factors are 1, 3.
4 Find the highest common factor (HCF). The HCF of 6 and 9 is 3.

b 1 Find the factors of 12 and write them in b 12: 1, 12; 2, 6; 3, 4
ascending order. Factors of 12 are 1, 2, 3, 4, 6, 12.
2 Find the factors of 16 and write them in 16: 1, 16; 2, 8; 4, 4
ascending order. Factors of 16 are 1, 2, 4, 8, 16.
3 Find the factors of 56 and write them in 56: 1, 56; 2, 28; 4, 14; 7, 8
ascending order. Factors of 56 are 1, 2, 4, 7, 8, 14, 28, 56.
4 Write the common factors. Common factors are 1, 2, 4.
5 Find the highest common factor The HCF of 12, 16 and 56 is 4.
(HCF).

c 1 Find the factors of 4. c 4: 1, 2, 4
2 Find the factors of 6. 6: 1, 2, 3, 6
3 Write the common factors of 4 Common factors of 4 and 6 are 1, 2
and 6.
4 Find the HCF of the coefficients. The HCF of 4 and 6 is 2.
5 List the pronumerals that are common Common pronumerals are b and c.
to each term.
6 Find the HCF of the algebraic terms by The HCF of 4abc and 6bcd is 2bc.
multiplying the HCF of the coefficients
to all the common pronumerals.


Factorising expressions by finding the highest common
factor
An algebraic expression is made up of terms which are separated by either a + or a −
sign. For example:
4xy + 3x − 4y is an algebraic expression which has 3 terms.
4xy, 3x and 4y are all terms.
To factorise such an expression, find the HCF of the terms and ‘take it out’ by
dividing it into the terms one by one, and then place the results inside the brackets.
If both terms are positive, the HCF will be positive, so include only positive factors
of the integer.
If both terms are negative or you want to have a negative HCF, include only negative
factors of the integers.

WORKED Example 3
Factorise each of the following expressions by first finding the highest common factor (HCF).
a 5x + 15y b −14xy − 7y c 15ab − 21bc + 18bf d 6x2y + 9xy2
THINK WRITE
a 1 Find the HCF of the coefficients, using a The HCF of 5 and 15 is 5.
only positive integer factors. List the There are no common pronumerals.
pronumerals common to each term. The HCF of 5x and 15y is 5.
2 Divide each term by the HCF. 5x ÷ 5 = x so 5x = 5 × x.
15y ÷ 5 = 3y, so 15y = 5 × 3y.
3 Write the expression. 5x + 15y
4 Place the HCF outside the brackets and the = 5(x + 3y)
remaining terms inside the brackets.
b 1 Find the HCF of the coefficients, using b The HCF of −14 and −7 is −7.
negative integer factors. List the The common pronumeral is y.
pronumerals common to each term. The HCF of −14xy and −7y is −7y.
2 Divide each term by the HCF. −14xy ÷ −7y = 2x, so −14xy = −7y × 2x.
−7y ÷ −7y = 1, so −7y = −7y × 1.
3 Write the expression. −14xy − 7y
4 Place the HCF outside the brackets and the = −7y(2x + 1)
c 1 Find the HCF of the coefficients, using only c The HCF of 15, −21 and 18 is 3.
positive integer factors. List the Common pronumeral is b.
pronumerals common to each term. The HCF of 15ab, −21bc and 18bf is 3b.
2 Divide each term by the HCF. 15ab ÷ 3b = 5a, so 15ab = 3b × 5a.
−21bc ÷ 3b = −7c, so −21bc = 3b × −7c.
18bf ÷ 3b = 6f, so 18bf = 3b × 6f.
3 Write the expression. 15ab − 21bc + 18bf
4 Place the HCF outside the brackets and the = 3b(5a − 7c + 6f )
d 1 Find the HCF of the coefficients, using only d The HCF of 6 and 9 is 3.
positive integer factors. List the Common pronumerals are x and y.
pronumerals common to each term. The HCF of 6x2y and 9xy2 is 3xy.
2 Divide each term by the HCF. 6x2y ÷ 3xy = 2x, so 6x2y = 3xy × 2x.
9xy2 ÷ 3xy = 3y, so 9xy2 = 3xy × 3y.
3 Write the expression. 6x2y + 9xy2
4 Place the HCF outside the brackets and the = 3xy(2x + 3y)

The key to HCF factoring is to find and divide each term by the HCF and place that
result inside the brackets, leaving the HCF outside the brackets.
Each factorisation can be checked by expanding your factored answer. For example,
in the previous worked example:
a 5(x + 3y) = 5x + 15y b −7y(2x + 1) = –14xy − 7y
c 3b(5a − 7c + 6f ) = 15ab − 21bc + 18bf d 3xy(2x + 3y) = 6x2y + 9xy2

remember
remember
1. Factorising is the ‘opposite’ of expanding, going from an expanded form to a
more compact form.
2. Factor pairs of a term are numbers and pronumerals which, when multiplied
together, produce the original term.
3. The number itself and 1 are factors of every integer.
4. The highest common factor (HCF) of given terms is the largest factor that
divides into all terms without a remainder.
5. An expression is factorised by finding the HCF of each term, dividing it into
each term and placing the result inside the brackets, with the HCF outside the
brackets.

6A The highest common factor

WORKED 1 Find all the factors of each of the following integers.
Example
1 a 36 b 17 c 51 d −14
e −8 f 100 g −42 h 32
Math
i −32 j −9 k −64 l −81

cad
m 29 n −92 o 48 p −12 HCF
WORKED 2 Find the highest common factor (HCF) of each of the following.
Example
2 a 4 and 12 b 6 and 15 c 10 and 25 d 24 and 32 L Spread
XCE
e 12, 15 and 21 f 25, 50 and 200 g 17 and 23 h

sheet
E
6a and 12ab
i 14xy and 21xz j 60pq and 30q k 50cde and 70fgh l 6x2 and 15x HCF
m 6a and 9c n 5ab and 25 o 3x2y and 4x2z p 4k and 6
3 multiple choice GC pro
gram

What is 5m the highest common factor of? HCF
A 2m and 5m B 5m and m C 25mn and 15lm
D 20m and 40m E 15m2n and 5n2
WORKED 4 Factorise each of the following expressions, by first finding the highest common factor Math
Example
cad

3
(HCF).
a 4x + 12y b 5m + 15n c 7a + 14b Factorising
using the
d 7m − 21n e −8a − 24b f 8x − 4y HCF
g −12p − 2q h 6p + 12pq + 18q i 32x + 8y + 16z 6.1
j 16m − 4n + 24p k 72x − 8y + 64pq l 15x2 − 3y HEET
SkillS

m 5p2 − 20q n 5x + 5 o 56q + 8p2
p 7p − 42x y2
q 16p + 20q + 4
2
r 12 + 36a2b − 24b2


5 Factorise each expression by taking out the highest common factor.
a 9a + 21b b 4c + 18d2 c 12p2 + 20q2
d 35 − 14m n 2
e 25y − 15x
2
f 16a2 + 20b
g 42m + 12n
2
h 63p + 81 − 27y
2
i 121a2 − 55b + 110c
j 10 − 22x2y3 + 14xy k 18a2bc − 27ab − 90c l 144p + 36q2 − 84pq
m 63a b − 49 + 56ab
2 2 2
n 22 + 99p q − 44p r
3 2 2
o 36 − 24ab2 + 18b2c
6 Factorise by taking out the highest negative common factor.
a −x + 5 b −a + 7 c −b + 9
d −2m − 6 e −6p − 12 f −4a − 8
g −3n2 + 15m h −7x2y2 + 21 i −7y2 − 49z
j −12p2 − 18q k −63m + 56 l −12m3 − 50x3
m −9a2b + 30 n −15p − 12q o −18x2 + 4y2
p −3ab + 18m − 21 q −10 − 25p2 − 45q r −90m2 + 27n + 54p3
reads
L Sp he
7 Factorise by taking out the highest common factor.
a a2 + 5a b m2 + 3m c x2 − 6x
et
EXCE

Factorising d 14q − q 2
e 18m + 5m2 f 6p + 7p2
ax2 + bx g 7n − 2n
2
h a2 − ab + 5a i 7p − p2q + pq
j xy + 9y − 3y 2
k 5c + 3c2d − cd l 3ab + a2b + 4ab2
m 2x y + xy + 5xy
2 2
n 5p2q2 − 4pq + 3p2q o 6x2y2 − 5xy + x2y
8 Factorise each of the following expressions.
a 5x2 + 15x b 10y2 + 2y c 12p2 + 4p
d 24m − 6m
2
e 32a2 − 4a f −2m2 + 8m
g −5x + 25x
2
h −7y2 + 14y i −3a2 + 9a
j −12p − 2p
2
k −15b2 − 5b l −26y2 − 13y
m 4m − 18m2 n −6t + 36t2 o −8p − 24p2

QUEST
S
M AT H

GE

1 This Noughts and Crosses game is to be ﬁnished. The squares labelled p,
EN

q, r, s and t are empty. What is the best play if:
CH L a O is to play next?
AL b X is to play next?

O O X

X p q

r s t

2 The following is a famous problem studied by an Egyptian mathemat-
ician called Hypatia.
‘Find a number that is the sum of two squares such that its square is
also the sum of two squares.’
There are a number of solutions to this problem. Can you ﬁnd 3 poss-
ible solutions?

I just want you to remember
you remember
one thing!
Factorise the expressions
to find the puzzle’s code.

7y (2 – x) b 2(2 + a) a (a + b) b 2(a – b) 4a (1 – b) 3x (y – 4) –a (3 – b) –(x + 1) 4x (2 – x) –(xy + 6) x 2(1 –x) –2(y – 3)

’
7x (2 – x) –x (3 + x) 4(5 – 2x) 2y (3 + 7x) –3b (a + b) 2xy (2 + x) xy (x – 1) 2(2x + 1) x (x + 7) y (1 – x) b (1 – b)

’
a (a – b) 2b (b – a) 11(2x – 3) y (x + y) 2(x – 4) 8(2 –x) 5y (5 + y) 6(3 – 2x) 3a (2 + a) 5(x – 3)

= a 2 + ab = 4x + 2 = x2y – xy
= = =

= 5x – 15 = 14x – 7x 2 = b – b2
= = =

= 2x – 8 = 6a + 3a 2 = y – xy
= = =

= 7x + x 2 = 4a – 4ab = 16 – 8x
= = =

= 2b 2 + ab 2 = 3xy – 12x = 25y – 5y 2
= = =

= 14y – 7xy = 20 – 8x = x2 – x3
= = =

= 2b 2 – 2ab = –3x – x 2 = 18 – 12x
= = =

= –3a + ab = xy + y 2 = –3ab – 3b 2
= = =

= –x – 1 = 8x – 4x 2 = ab 2 – b 3
= = =

= 6y + 14xy = –2y + 6 = 22x – 33
= = =

= 4xy + 2x 2y = –xy – 6 = a 2 – ab
= = =


More factorising using the highest
common factor
Remember, whenever factorising, look for a common factor first.

The binomial common factor
Sometimes the common factor may itself be in brackets.
Consider the following expression:
5(x + y) + 6b(x + y).
While it may appear at first that no factorising can be done, consider the fact that
both terms contain the bracketed expression (x + y) as a factor. This is called a binomial
factor because it contains two terms.

WORKED Example 4
Factorise each of the following expressions by taking out the binomial common factor.
a 5(x + y) + 6b(x + y) b 2b(a − 3b) − (a − 3b)

THINK WRITE
a 1 Identify the common factor. a Common factor is (x + y).
2 Divide each term by the common 5(x + y) ÷ (x + y) = 5
factor. 6b(x + y) ÷ (x + y) = 6b
3 Write the expression. 5(x + y) + 6b(x + y)
4 Factorise, by taking out the = (x + y)(5 + 6b)
binomial common factor and
placing the remaining terms
inside brackets.

b 1 Identify the common factor. b Common factor is (a − 3b).
2 Divide each term by the common 2b(a − 3b) ÷ (a − 3b) = 2b
factor. (a − 3b) ÷ (a − 3b) = 1
3 Write the expression. 2b(a − 3b) − (a − 3b)
4 Factorise, by taking out the = (a − 3b)(2b − 1)
binomial common factor and
placing the remaining terms
inside brackets.

These answers can also be checked by performing an expansion.
When factorising expressions that already have brackets in them, look to this method
of using the bracketed part as the common factor or, in some cases, as part of the
common factor.

Factorising by grouping terms
If an algebraic expression has 4 terms and no common factor in all the terms, it may be
possible to group the terms in pairs and find a common factor in each pair.

WORKED Example 5
Factorise each of the following expressions by grouping the terms in pairs.
a 5a + 10b + ac + 2bc b x − 3y + ax − 3ay c 5p + 6q + 15pq + 2

THINK WRITE

a 1 Write the expression. a 5a + 10b + ac + 2bc
2 Look for a common factor of all 4 terms.
(There isn’t one.) If necessary, rewrite
the expression so that the terms with
common factors are next to each other.
3 Take out a common factor from each = 5(a + 2b) + c(a + 2b)
group.
4 Factorise by taking out a binomial = (a + 2b)(5 + c)
common factor.

b 1 Write the expression. b x − 3y + ax − 3ay
2 Look for a common factor of all 4 terms.
the expression so that the terms
with common factors are next to
each other.
3 Take out a common factor from each = 1(x − 3y) + a(x − 3y)
pair of terms.
4 Factorise by taking out a binomial = (x − 3y)(1 + a)
common factor.

c 1 Write the expression. c 5p + 6q + 15pq + 2
2 Look for a common factor of all 4 terms. = 5p + 15pq + 6q + 2
the expression so that the terms
with common factors are next to
each other.
3 Take out a common factor from each = 5p(1 + 3q) + 2(3q + 1)
pair of terms. = 5p(1 + 3q) + 2(1 + 3q)
4 Factorise by taking out a binomial = (1 + 3q)(5p + 2)
common factor.

The answers found in worked example 5 can each be checked by expanding.

There are only 3 possible pair groupings to consider with this technique:
1st and 2nd terms + 3rd and 4th terms; 1st and 4th terms + 2nd and 3rd terms; 1st and
3rd terms + 2nd and 4th terms.
So in the worst case you would have to try all 3.


remember
remember
1. When factorising any number of terms, look for a factor which is common to
all terms.
2. A binomial factor has 2 terms.
3. The HCF of an algebraic expression may be a binomial factor, which is in brackets.
4. When factorising expressions with 4 terms which have no factor common to all 4:
(a) group the terms in pairs with a common factor
(b) factorise each pair
(c) factorise the expression by taking out a binomial common factor.
5. (x + y) = 1(x + y)

More factorising using the
6B highest common factor
WORKED 1 Factorise each of the following expressions by taking out the binomial common factor.
Example
4
a 2(a + b) + 3c(a + b) b 4(m + n) + p(m + n)
c 7x(2m + 1) − y(2m + 1) d 4a(3b + 2) − b(3b + 2)
e z(x + 2y) − 3(x + 2y) f 12p(6 − q) − 5(6 − q)
g 3p2(x − y) + 2q(x − y) h 4a2(b − 3) + 3b(b − 3)
i p (q + 2p) − 5(q + 2p)
2
j 6(5m + 1) + n2(5m + 1)
d WORKED 2 Factorise each of the following expressions by grouping the terms in pairs.
hca Example
a xy + 2x + 2y + 4 b ab + 3a + 3b + 9
Mat

5
Factorising c xy − 4y + 3x − 12 d 2xy + x + 6y + 3
using e 3ab + a + 12b + 4 f ab − 2a + 5b − 10
grouping g m − 2n + am − 2an h 5 + 3p + 15a + 9ap
i 15mn − 5n − 6m + 2 j 10pq − q − 20p + 2
SHE
ET 6.1 k 6x − 2 − 3xy + y l 16p − 4 − 12pq + 3q
m 10xy + 5x − 4y − 2 n 6ab + 9b − 4a − 6
Work

o 5ab − 10ac − 3b + 6c p 4x + 12y − xz − 3yz
q 5pr + 10qr − 3p − 6q r ac − 5bc − 2a + 10b

1
1 List all the factors of −36.
2 Find the highest common factor of 45 and 21.
3 Find the highest common factor of 8ab and 12ad.
4 Factorise 27xy − 45y.
5 Factorise 25y2 − 85xy + 35x2y2.
6 Factorise −27abc + 36ac − 18a2bc.
7 Factorise 9(x − 2y) + b(x − 2y).
8 Factorise 7x + 14 − 2xy − 4y.
9 Factorise 3abc + 6ac + b + 2.
10 Factorise 12xy − 8 + 3x2y − 2x.

Factorising using the difference of two
squares rule
In chapter 5, we saw expansions of the form (x + 2)(x − 2) = x2 − 4. The result of this
expansion is called a difference of two squares because one square is subtracted from
another square.
To factorise a difference of two squares, we use the rule or formula in reverse.
a2 − b2 = (a + b)(a − b)
We recall that a and b are any terms, and that the expression x2 − 4 can be written as
x − 22 , so that it can be factorised as (x + 2)(x − 2) or (x − 2)(x + 2) as the order of the
2

two factors is unimportant.

WORKED Example 6
Factorise each of the following expressions using the difference of two squares rule.
a x2 − 9 b 25x2 − 49 c 64m2 − 25n2 d (c + 7)2 − 16 e 3x2 − 48
THINK WRITE
a 1 Write the given expression. a x2 − 9
2 Rewrite, showing the two squares. = x2 − 32
3 Factorise using the formula = (x + 3)(x − 3)
a2 − b2 = (a + b)(a − b) where a = x
and b = 3.
b 1 Write the given expression. b 25x2 − 49
2 Rewrite, showing the two squares. = (5x)2 − 72
3 Factorise, using the formula = (5x + 7)(5x − 7)
a2 − b2 = (a + b)(a − b) where a = 5x
and b = 7.
c 1 Write the given expression. c 64m2 − 25n2
2 Rewrite, showing the two squares. = (8m)2 − (5n)2
3 Factorise, using the formula = (8m + 5n)(8m − 5n)
a2 − b2 = (a + b)(a − b) where
a = 8m and b = 5n.
d 1 Write the expression. d (c + 7)2 − 16
2 Rewrite, showing the two squares. = (c + 7)2 − 42
3 Factorise, using the formula = (c + 7 + 4)(c + 7 − 4)
a2 − b2 = (a + b)(a − b) where
a = c + 7 and b = 4.
4 Simplify. = (c + 11)(c + 3)
e 1 Write the expression. e 3x2 − 48
2 Look for a common factor and factorise, = 3(x2 − 16)
taking out the common factor.
3 Rewrite the two terms in the bracket = 3(x2 − 42)
showing the two squares.
4 Factorise, using the formula = 3(x + 4)(x − 4)
a2 − b2 = (a + b)(a − b) where
a = x and b = 4.


remember
remember
When factorising an algebraic expression with 2 terms follow these steps.
1. Look for a common factor ﬁrst. If there is one, factorise by taking it out.
2. Rewrite the expression showing the two squares and identifying the a and b
parts of the expression.
3. Factorise, using the rule a2 − b2 = (a + b)(a − b).

Factorising using the
6C difference of two
squares rule
6.2 WORKED 1 Factorise each of the following expressions using the difference of two squares rule.
HEET Example
a x2 − 25 b x2 − 81 c a2 − 16 d c2 − 64
SkillS

6a, b
e y − 144
2
f 16 − x 2
g 25 − p 2
h 121 − a2
i 36 − y 2
j 4b − 25
2
k 9a − 16
2
l 25d 2 − 1
reads WORKED 2 Factorise each of the following expressions.
L Sp he Example
a x2 − y2 b a 2 − b2 c p 2 − q2 d c2 − d2
et
EXCE

6c
Difference e 9x − y
2 2
f 16p2 − q2 g 25m2 − n2 h 81x2 − y2
of two i p2 − 36q2 j m2 − 4n2 k a2 − 49b2 l y2 − 100z2
squares rule
m 36m − 25n
2 2
n 16q2 − 9p2 o 4m2 − 49n2

d WORKED 3 Factorise each of the following expressions.
hca Example
a (x + 9)2 − 16 b (p + 8)2 − 25 c (p − 2)2 − q2 d (c − 6)2 − d2
Mat

6d
Difference e (x + 7) − y
2 2
f (p + 5) − q
2 2
g (r − 9)2 − s2 h (a − 1)2 − b2
of two i (x + 1) − y
2 2
j (a + 3) − b
2 2
k (a − 3)2 − 1 l (b − 1)2 − 36
squares rule
WORKED 4 Factorise each of the following expressions, using the difference of two squares rule,
Example
6e
after removing any common factors.
a 2m2 − 32 b 5y2 − 45 c 6p2 − 24 d 4m2 − 100
e 288 − 2x 2
f 80 − 5a 2
g 36x − 9y
2 2
h 9y2 − 81z2
i 4m2 − 36n2 j 100x2 − 25y2 k 144p2 − 4q2 l 32y2 − 2x2
m 3m2 − 27n2 n 4(x + 2)2 − 36 o 3(m + 1)2 − 12 p 7(y − 3)2 − 28
q 2(y − 4)2 − 50 r 3(b + 5)2 − 48
5 multiple choice
a The expression x2 − 121 factorises to:
A (x − 11)2 B (x − 11)(x + 12) C (x + 11)(x − 11)
D (x − 11)(x − 11) E (x + 11)(x − 12)
b What does the expression 36m2 − 1 factorise to?
A (6m + 1)(6m − 1) B 36(m + 1)(m − 1) C (6m − 1)2
D 6(6m − 1)(6m − 1) E (6m + 1)(6m + 1)
c What does the expression 16a2 − 25b2 factorise to?
A 16(a + 5b)(a − 5b) B 4(4a − 5b)2 C (16a + 25b)(16a − 25b)
D (4a + 5b)(4a − 5b) E (4a − 5b)(4a − 5b)
d The expression 5c2 − 20 factorises to:
A 5(c + 2)(c − 2) B 5(c + 4)(c − 4) C (5c + 4)(5c − 4)
D (5c + 20)(5c − 20) E (5c + 10)(5c − 10)

e The expression (x + 4)2 − 9 factorises to:
A 9(x + 4)(x − 4) B (x + 13)(x − 5) C (x + 7)(x + 1)
D (x − 1)(x + 5) E (x − 7)(x + 11)
6 A circular pool with a radius of r metres is surrounded by a circular path 1 m wide.

a Find the surface area of the pool.
b Find, in terms of r, the distance from the centre of the pool to the outer edge of the
path.
c Find the area of the circle that includes the path and the pool. (Don’t expand the
expression.)
d Write an expression for the area of the path.
e Simplify this expression.
f If the pool had a radius of 5 m, what would be the area of the path, to the nearest
square metre?
g If the pool had a radius of 7 m, what would be the area of the path, to the nearest
square metre?

QUEST
S
M AT H

GE

1 Work out the values of the following without using a calculator or long
EN

multiplication.
CH L a 92 – 7 2
AL b 852 – 152
c 7652 – 2352
2 The difference between the squares of two consecutive odd numbers is
48. What are the numbers?
3 The difference between the squares of two consecutive even numbers is
92. What are the numbers?


What has area got to do with factorising?

Copy the diagram shown at right on to paper a
and use scissors to cut around the outline of
the large square.
1 What is the area of the large square in
terms of the pronumeral a?
2 What is the area of the small square in
a
terms of the pronumeral b? b

b

Cut out the small square from the large square. a
3 Write an expression for the area of the
shape left over.
4 Find the lengths, in terms of a and b, of the
two sides not yet labelled on your shape.
Include these on your shape.
a
b

b

Now cut your shape into two pieces as shown a
in the diagram at right. (Cut along the dotted
line.) Rearrange the two pieces into a rectangle.
5 Write an expression for the length of the
rectangle.
6 Write an expression for the width of the
a
rectangle. b
7 Use your answers from parts 5 and 6 to
write an expression for the area of the
rectangle. b
8 Relate this expression for the area of this
shape to the one found earlier in part 3.
9 Where have you seen this formula before?

Quadratic trinomials
An expression with 3 terms is called a trinomial, tri meaning three.
A quadratic trinomial is an algebraic expression that has 3 terms and the highest
power is a squared term.
The expressions y2 + 3y − 4 and 5x2 − 2x + 4 are both quadratic trinomials.
The expression x3 + 3x2 − 4 is not a quadratic trinomial, because x2 is not the highest
power.
Therefore, we could write a quadratic trinomial as:
constant × (pronumeral)2 + constant × pronumeral + constant.
Note: The + signs could be replaced by − signs or the constants could be positive or
negative.

The general case
Using the above definition, a general quadratic trinomial can be written, replacing the
words ‘constant’ with a, b and c respectively and perhaps x as the pronumeral.
ax2 + bx + c
Note: We can allow for minus signs in the trinomial by giving a, b and c negative
values.
There are only two cases, each of which will require a different technique:
a = 1 The coefficient of the squared term is 1.
a ≠ 1 The coefficient of the squared term is not equal to 1.

Factorising quadratic trinomials
Factorising is the reverse of expanding. Consider the expansion below:
(x + 3)(x + 5) = x(x + 5) + 3(x + 5)
= x2 + 5x + 3x + 15
= x2 + 8x + 15
Now, if we were to factorise x + 8x + 15, the answer must be (x + 3)(x + 5), since
2

factorising is the reverse of expanding. Looking for a relationship between the quad-
ratic and its factors, we can see that 3 + 5 = 8 and 3 × 5 = 15.
Now consider the more general expansion:
(x + p)(x + q) = x(x + q) + p(x + q)
= x2 + qx + px + pq
= x2 + (p + q)x + pq
If p and q are the constant terms in the factors, we can see that the coefficient of the
x term is the sum of p and q, while the constant term is the product of p and q. In other
words, we are looking for two numbers p and q which multiply to give c and add to
give b.

Factorising ax2 + bx + c when a = 1
The following method works for every possible trinomial (with a = 1) that can be
factorised.
Step 1 Put the trinomial into the correct order or standard form x2 + bx + c.
Step 2 Find all the factor pairs of c (the constant term).
Step 3 Identify the factor pair whose sum equals b; that is, p + q = b.
Step 4 The factorisation for x2 + bx + c = (x + p) (x + q).


Let us perform these 4 steps on the trinomial y2 + 5y + 4.
Step 1 The quadratic is already in standard form, with b = 5 and c = 4.
Step 2 Find the factor pairs of 4 (the c).
1 and 4
2 and 2
−1 and −4
−2 and −2
Step 3 Find the sum of each factor pair.
1 + 4 = 5 (equals b)
2+2=4
−1 + −4 = −5
−2 + −2 = −4
Step 4 The ﬁrst pair’s sum equals 5 (the b), so p = 1, q = 4 and the factorisation is
(y + 1)(y + 4).

WORKED Example 7
Factorise each of the following quadratic trinomials.
a y2 + 6y + 8 b y2 − 5y + 4 c y2 − 3y − 10 d y2 − 4 + 3y
THINK WRITE
a 1 Make sure that the expression is in a y2 + 6y + 8
the correct order.
2 Find the factor pairs of c (+8). 8: 1 and 8, 2 and 4, −1 and −8, −2 and −4
3 Find the sum of each factor pair and 1+8=9
identify the pair with a sum of b (+6). 2+4=6 (equals b)
−1 + −8 = −9
−2 + −4 = −6
4 Write the expression and its y2 + 6y + 8 = (y + 2)(y + 4)
factorised form.
b 1 Make sure that the expression is in b y2 − 5y + 4
the correct order.
2 Find the factor pairs of c (+4). 4: 1 and 4, 2 and 2, −1 and −4, −2 and −2
3 Identify the pair with a sum of b (−5). −1 + −4 = −5
4 Write the expression and its y2 − 5y + 4 = (y − 1)(y − 4)
factorised form.
c 1 Make sure that the expression is in c y2 − 3y − 10
the correct order.
2 Find the factor pairs of c (−10). 10: 1 and −10, 2 and −5, −1 and 10, −2 and 5
3 Identify the pair with a sum of b (−3). 2 − 5 = −3
4 Write the expression and its y2 − 3y − 10 = (y + 2)(y − 5)
factorised form.
d 1 Make sure that the expression is in d y2 + 3y − 4
the correct order.
2 Find the factor pairs of c (−4). −4: 1 and −4, 2 and −2, −1 and 4
3 Identify the pair with a sum of b (+3). −1 + 4 = 3
4 Write the expression and its y2 + 3y − 4 = (y − 1)(y + 4)
factorised form.

Again, each of the factorisations can be checked by expanding the pair of brackets.
Note: In the four parts of the previous worked example, we have covered all cases,
namely b either positive or negative and c either positive or negative, without resorting
to any different method.

remember
remember
1. A quadratic trinomial consists of 3 terms, the highest power being a squared
term.
2. The general form of a quadratic trinomial is ax2 + bx + c.
3. There are 2 main cases: either a = 1 or a ≠ 1.
4. Always look for a common factor ﬁrst.
5. To factorise a quadratic trinomial when a = 1, (x2 + bx + c) use the following
steps:
(a) put the trinomial into standard form x2 + bx + c
(b) ﬁnd all the factor pairs of c
(c) identify the factor pair; call one factor p and the other q, whose sum equals
b; that is, p + q = b
(d) the factorisation for x2 + bx + c = (x + p) (x + q).

6D Quadratic trinomials

WORKED Factorise each of the following quadratic trinomials. 6.3
Example
1 x2 + 4x + 3 2 x2 + 12x + 11 3 a2 + 6a + 5 HEET
7

SkillS
4 b2 − 8b + 7 5 x2 − 5x + 6 6 y2 − 7y + 12
7 c2 + c − 20 8 x2 + 4x − 21 9 x2 + 9x − 10
10 p2 − 3p − 28 11 q2 − 6q − 27 12 x2 − 6x − 16
13 y2 + 10y + 9 14 x2 − 12x + 32 15 c2 − 5c + 36 L Spread
XCE

sheet
E
16 m2 − 5m − 14 17 x2 − 6x − 55 18 m2 − 14m + 24
Factorising
19 x2 + 12x + 35 20 x2 − 17x + 72 21 x2 + 10x + 25 x2 + bx + c
22 p2 + 4p + 4 23 a2 + 8a + 16 24 y2 − 6y + 9
25 x2 − 16x + 64 26 x2 − 10x + 25 27 m2 − 7m + 6
28 x2 − 12x + 27 29 k2 − 9k − 22 30 x2 + 11x − 12 Math
31 x − 13x + 42
2
32 a + a − 6
2
33 a − 3a − 4
2
cad

Factorising
34 x − 3x − 10
2
35 x + 2x − 8
2
36 x2 − 5x − 6 x2 + bx + c
37 x2 + 4x − 5 38 b2 − 2b − 35 39 c2 − 15c − 16
40 x2 − 16x − 36 41 x2 + x − 30 42 d2 + 5d − 14
43 f 2 − 3f − 40 44 m2 + 5m − 6 45 q2 + 7q − 18 SHE
ET 6.2
Work

46 x + 5x − 24
2
47 x − x − 2
2
48 y + 3y − 28
2

49 a + 7a − 18
2
50 x + x − 12
2


Mouse pad dimensions
At the start of the chapter, we considered
the dimensions (length and width) of a
rectangular mouse pad.
1 If the area of a rectangular mouse pad
is 500 cm2 and the width is 20 cm,
what is the length? Explain how you
are able to work this out.
2 If the area of the mouse pad is
(x2 + 5x) cm2 and the width is x cm,
what is the length? Explain how
factorising the expression helps you
find the length.
3 For a particular mouse pad, x is 22 cm.
a Find the width and length of this
mouse pad.
b Multiply the length and width
together to find the area.
c Substitute x = 22 into the area
expression x2 + 5x to calculate the area.
Verify that this gives the same result as that obtained in part b.
4 Simone also designs rectangular floor mats for under desk chairs. She uses the
algebraic expression x2 + 2x − 15.
a If the length of the mat is (x + 5) cm, find an expression for the width.
b If the length of the mat is 70 cm, what is the width?
c If the width of the mat is 1 m, what is the length?

2
1 Find the HCF of 48a and 54.
2 Factorise 12x2 − 32x3y.
3 Factorise 8(x − 2y) − 3q(x − 2y).
4 Factorise x2 − 81.
5 Factorise (x + 6)2 − y2.
6 Factorise 49x2 − 100y2.
7 Factorise y2 − 7y − 8.
8 Factorise y2 + y − 12.
9 Factorise c2 + 8c + 15.
10 Factorise x2 − 10x + 16.

More quadratic trinomials
Removing a common factor
Sometimes a quadratic trinomial will have a common factor in each term. After
removing the common factor, the remaining factor is a case where a = 1.

WORKED Example 8
Factorise each of the following quadratic trinomials by first taking out a common factor.
a 2x2 − 16x − 18 b −x2 − 5x + 6
THINK WRITE
a 1 Write the expression and take out the a 2x2 − 16x − 18 = 2(x2 − 8x − 9)
common factor.
2 List the factor pairs of the last term. −9: 1 and −9, 3 and −3, −1 and 9
3 Find the pair with a sum equal to the 1 + −9 = −8
coefficient of the middle term.
4 Write the expression and its factorised 2x2 − 16x − 18 = 2(x + 1)(x − 9)
form with the common factor outside
the brackets.
b 1 Write the expression and take out the b −x2 − 5x + 6 = −(x2 + 5x − 6)
common factor, which, in this case, is −1.
2 List the factor pairs of the last term. −6: 1 and −6, 2 and −3, −1 and 6, −2 and 3
3 Find the pair with a sum of the −1 + 6 = 5
coefficient of the middle term.
4 Write the expression and its factorised −x2 − 5x + 6 = −(x − 1)(x + 6)
form with the common factor outside
the brackets.

Factorising ax2 + bx + c when a ≠ 1
When no common factor can be removed, as in the previous worked example, a modi-
fied method for factorisation can be used.
The factors of ac must be considered instead of just the factors of c. We need to find
2 numbers which multiply to give ac and add to give b. These 2 numbers will be used
to rewrite the middle term of the expression, before using the method of factorising by
grouping terms.
Remember to check that the terms are in the correct order before you start.
WORKED Example 9
Factorise each of the following quadratic trinomials.
a 3x2 + 5x + 2 b 2x2 − 11x − 6 c 4x2 − 19x + 12
THINK WRITE
a 1 Write the expression. a 3x2 + 5x + 2
2 List the factor pairs of ac (6). 6: 1 and 6, 2 and 3, −1 and −6, −2 and −3
3 Identify which pair has a sum of b (5). 2+3=5
4 Rewrite the expression by breaking 3x2 + 5x + 2
the middle term into 2 terms using = 3x2 + 3x + 2x + 2
the factor pair from step 3 .
5 Factorise by grouping terms. = 3x(x + 1) + 2(x + 1)
= (x + 1)(3x + 2)
Continued over page


THINK WRITE
b 1 Write the expression. b 2x2 − 11x − 6
2 List the factor pairs of ac (−12). −12: 1 and −12, −1 and 12, 2 and −6, −2 and
6, 3 and −4, −3 and 4
3 Identify which pair has a sum of b 1 + −12 = −11
(−11).
4 Rewrite the expression by breaking 2x2 − 11x − 6
the middle term into 2 terms using = 2x2 − 12x + x − 6
the factor pair from step 3 .
5 Factorise by grouping terms. = 2x(x − 6) + 1(x − 6)
= (x − 6)(2x + 1)
c 1 Write the expression. c 4x2 − 19x + 12
2 List the factor pairs of ac (48). 48: −4 and −12, −6 and −8, −2 and −24, −3
Consider only the negative factors as and −16
we are looking for a pair with a
negative sum.
3 Identify which pair has a sum of b −3 + −16 = −19
(−19).
4 Rewrite the expression by breaking
the middle term into 2 terms using 4x2 − 19x + 12
the factor pair from step 3 . = 4x2 − 16x − 3x + 12

5 Factorise by grouping terms. = 4x(x − 4) − 3(x − 4)
= (x − 4)(4x − 3)

With practice you will be able to reduce the number of factor pairs to consider, instead
of looking at all of them. For example, in part c of worked example 10, we considered
only negative factor pairs as we were looking for a negative sum. Similarly, in part a
we need only have considered the positive factor pairs.
Note: This method can also be used for expressions where a = 1.

remember
remember
1. Always look for a common factor before factorising.

2. To factorise any quadratic trinomial of the form ax2 + bx + c, follow these
steps.

Step 1 List the factor pairs of ac.
Step 2 Identify which pair has a sum of b.
Step 3 Rewrite the expression by breaking the middle term into two terms
using the factor pair from Step 2.
Step 4 Factorise by grouping the terms.

3. You can always check your answer by expanding.


6E More quadratic trinomials

WORKED 1 Factorise each of the following quadratic trinomials by first taking out a common L Spread
Example XCE
factor.

sheet
E
8
a 2x2 + 10x + 12 b 3x2 + 15x + 12 c 7x2 + 42x + 56 Factorising
d 6x + 54x + 120
2
e 2x − 12x + 16
2
f 3x2 − 12x + 9 ax2 + bx + c
g 5x2 − 45x + 70 h 6x2 − 42x + 72 i 4x2 − 4x − 24
j 3x + 9x − 30
2
k 2x + 8x − 42
2
l 7x2 + 35x − 252
m 3x − 9x − 54
2
n 8x + 40x − 48
2
o 5x2 + 20x − 60
WORKED 2 Factorise each of the following quadratic trinomials. Math
Example
a 2x2 + 7x + 3 b 2x2 + 7x + 6 c 3x2 + 7x + 2

cad
9
d 3x + 10x + 3
2
e 5x2 + 11x + 2 f 6x2 + 13x + 6 Factorising
g 2x − 7x + 3
2
h 3x2 − x − 2 i 5x2 + 3x − 2 ax2 + bx + c
j 7x − 17x + 6
2
k 10x2 − 11x − 6 l 2x2 + 5x − 12
m 7x − 26x + 15
2
n 11x2 − 28x + 12 o 2x2 + x − 36
p 4x + 8x − 5
2
q 2x2 + 5x − 18 r 7x2 − 3x − 4
3 multiple choice
a The expression 3x2 + 21x + 36 factorises to:
A 3(x + 6)(x + 2) B 3(x + 4)(x + 3) C (3x + 9)(x + 4)
D (3x + 2)(x + 18) E (3x + 2)(x + 9)
b What does the expression 5x2 + 20x − 60 factorise to?
A (5x − 12)(x + 5) B (5x + 15)(x − 4) C 5(x + 6)(x − 2)
D 5(x − 6)(x + 2) E 5(x − 6)(x − 2)
c The expression 2x2 − 16x + 14 factorises to:
A 2(x − 1)(x − 7) B 2(x + 1)(x + 7) C (2x − 7)(x − 2)
D (2x − 2)(x − 7) E (2x − 2)(2x − 7)
d What does the expression 3x2 − 13x + 4 factorise to?
A (3x − 2)2 B (3x − 2)(x + 2) C (3x − 1)(x − 4)
GAME
D (3x − 4)(x − 1) E (3x − 4)(x + 1)

time
e What does the expression 12x2 + 58x − 10 factorise to? Factorising
A (12x − 2)(x − 5) B 2(6x + 1)(x − 5) C 2(6x + 5)(x − 1) — 001
D 2(6x − 1)(x + 5) E 2(2x − 5)(3x + 1)

QUEST
S
M AT H

GE

1 In numbering the pages of a book, it was determined that 348 digits are
EN

needed to completely number the book. If the numbering begins at the
first page (that is, the first page is numbered 1), how many pages are in
CH L the book?
AL
2 Volumes 15 and 16 of a series of books stand side by side, in order, on a
bookshelf. Volume 15, without its cover, is 3.2 cm thick and volume 16,
without its cover, is 2.8 cm thick. The cover of each book is 3 mm thick. If
an insect begins at the first page of volume 15 and eats its way in a straight
line to the last page of volume 16, how far will the insect have travelled?


Isaac Newton demonstrated this in 1665.
Newton 1665.
Factorise the trinomials on
the left of the page. Join
the dot next to each with
= 3x 2 + 14x + 8
the dots next to its factors,
=
using straight lines. The
= 2x 2 + 3x – 9 two lines will pass through
=
a letter and number
= 2x 2 – x – 1 giving the puzzle’s code.
=

= 2x 2 + 5x – 3
= G (x + 1)

= 2x 2 – 3x – 2 T
4
= C (x – 1)

= 3x 2 + 2x – 1
11 (x – 2)
= D 6
= 2x 2 – 9x + 4 7
= E (x + 3)
= 3x 2 + x – 2 12
= H (x – 4)
U 10
= 3x 2 – 13x – 10 9
= (x + 4)
S 13
= 4x 2 – 4x – 15
= (x – 5)
15
= 3x 2 – 11x – 4 X
= (2x + 1)
N
= 2x2 – x – 3 F
= L (2x – 1)
= 3x 2 – 8x + 4 19
= 14 5 (2x + 3)
M
= 3x 2 + 4x + 1 1 17
= (2x – 3)
= 3x 2 – 10x – 8
O R 18
= (2x – 5)
= 2x 2 + 5x – 12 16
= (3x + 1)
= 2x 2 – 7x + 6 P W 2 8
= I
(3x – 1)
= 2x 2 – 11x + 12
= A (3x + 2)
3
= 3x 2 – 7x + 2
(3x – 2)
=

1 2 3 4 5 6 3 7 2 4 3 8 9 10 11 5

12 13 14 15 10 9 3 16 4 12 17 5 14 15

11 3 15 15 5 17 5 18 4 19 14 6 14 12 17 8

Mixed factorising practice
You may like to review the techniques demonstrated in the previous worked examples,
or go over the solutions in your workbook before attempting this set of mixed
problems.

remember
remember
1. Always look for a common factor to take out if possible.
2. For an expression with 2 terms: try the difference of two squares formula
a2 − b2 = (a + b)(a − b).
3. For an expression with 3 terms: if it is a quadratic trinomial of the form
ax2 + bx + c, rewrite the expression by breaking the middle term into 2 terms
using a factor pair which multiplies to give ac and adds to give b. Factorise by
grouping terms.
4. For an expression with 4 terms: try factorising by grouping terms.

6F Mixed factorising practice

Factorise each of the following. Math

cad
1 2m + 8n 2 −m + 6 3 ab − 3a + 2b − 6
Mixed
4 p2 − 16 5 m2 + 3m + 2 6 2m2 − 18 factorising

7 3a − 9b 8 x2 + 6x + 5 9 −m2 − 3m
10 8m2 + 4 − 16q 11 x2 + 4x − 21 12 2ab − 6b − ac + 3c
13 2m2 − 5m − 3 14 18a2 − 3a 15 x2 − 9x + 18
16 (x + 2)2 − 1 17 5m2 + 33m − 14 18 m2 − 9m + 20
19 x2 − x − 30 20 1 − x2 21 2m2 + 3m − 2
22 4a2 + 4a + 1 23 3x2 − x − 2 24 7 − 14a2
25 8p + 9p2 26 2mn + 2m + 3n + 3 27 2x2 + 2x − 12
28 p2 − 36q2 29 −5x2y − x 30 m2 + 8m + 16
31 10p2 − 25q 32 5x2 − 30x + 40 33 (x + 3)2 − 25
GAME
34 2x2 + 6x − 20 35 3a2 − 48b2 36 −18m − 12n2
time

37 2ac + bc − 6a − 3b 38 (a − 7)2 − 36 39 2m2 − 9m + 10 Factorising
— 002
40 110a b + 121b − 10a − 11
2 2


Simplifying algebraic fractions
In this section we look at using factorisation techniques to simplify more complex
algebraic fractions. After factorising a numerator or denominator, terms may be
cancelled, resulting in a simpliﬁed fraction, or even no fraction at all.

Simplifying after taking out a common factor
Look for common factors in the numerators and denominators before cancelling.

WORKED Example 10
Simplify each of the following fractions by factorising the numerator and denominator
and cancelling as approriate.
3x + 9 – 25 10 x + 15 x2 + 4 x
a --------------- b --------------------- c --------------------- d -----------------
-
15 10 x + 20 6x + 9 x2 – 5 x
THINK WRITE
3x + 9
a 1 Write the fraction. a ---------------
15
3( x + 3)
2 Factorise both numerator and denominator. = -------------------
-
15
x+3
3 Cancel any common factors (3). = ----------- -
5
– 25
b 1 Write the fraction. b --------------------
-
10x + 20
– 25
2 Factorise both numerator and denominator. = ----------------------
-
10 ( x + 2 )
–5
3 Cancel any common factors (5). = -------------------
-
2( x + 2)
10x + 15
c 1 Write the fraction. c --------------------
-
6x + 9
5 ( 2x + 3 )
2 Factorise both numerator and denominator. = ----------------------
-
3 ( 2x + 3 )
5
3 Cancel any common factors [the binominal factor (2x + 3)]. = -- -
3
x 2 + 4x
d 1 Write the fraction. d -----------------
x 2 – 5x
x( x + 4)
2 Factorise both numerator and denominator. = -------------------
-
x( x – 5)
x+4
3 Cancel any common factors (x). = ------------
x–5

Simplifying after factorising quadratic trinomials
If there is a quadratic trinomial in either the numerator, the denominator or both, then
factorise before cancelling.

WORKED Example 11
Simplify each of the following algebraic fractions by ﬁrst factorising the numerator and
denominator.
x2 + 3 x – 4 x2 – 7 x – 8 x2 – 6 x + 5
a --------------------------
- b ---------------------------
2 + 3x + 2
c ------------------------------------
-
x–1 x 2 x 2 – 16 x + 30

THINK WRITE

a 1 Write the numerator, making sure a x2 + 3x − 4
that the expression is in the correct
order.
2 Find the factor pair of c (−4) which Factor sum of −4: −1 + 4 = 3
adds to b (3).
3 Write the expression and its factorised x2 + 3x − 4 = (x − 1)(x + 4)
form.
x 2 + 3x – 4
4 Write the original fraction. --------------------------
x–1
( x – 1)( x + 4)
5 Replace the numerator with the = ---------------------------------
-
x–1
factorised form.
6 Cancel common factors and simplify if =x+4
appropriate.
b 1 Write the numerator, making sure b x2 − 7x − 8
that the expression is in the correct
order.
2 Find the factor pair of c (−8) which Factor sum of −8: 1 − 8 = −7
adds to b (−7).
3 Write the expression in its factorised x2 − 7x − 8 = (x + 1)(x − 8)
form, using the appropriate factor pair.
4 Write the denominator, making sure that x2 + 3x + 2
the expression is in the correct order.
5 Find the factor pair of c (2) that adds to Factor sum of 2: 1 + 2 = 3
b (3).
6 Factorise using the appropriate factor x2 + 3x + 2 = (x + 1)(x + 2)
pair.
2
x – 7x – 8
7 Write the original fraction. --------------------------
2
-
x + 3x + 2
8 Replace the numerator and denominator ( x + 1)( x – 8)
= ---------------------------------
-
with the factorised form. ( x + 1)( x + 2)
9 Cancel any common factors and x–8
= -----------
-
simplify if appropriate. x+2

Continued over page


THINK WRITE
c 1 Write the numerator, making sure that the c x2 − 6x + 5
expression is in the correct order.
2 Find the factor pair of c (5) which adds to Factor sum of 5: −1 + −5 = −6
b (−6).
3 Factorise using the appropriate factor pair. x2 − 6x + 5 = (x − 1)(x − 5)
4 Write the denominator, making sure that the 2x2 − 16x + 30 = 2(x2 − 8x + 15)
expression is in the correct order and take out
the common factor of 2.
5 Find the factor pair of c (15) that adds to b Factor sum of 15: −3 + −5 = −8
(−8).
6 Factorise using the appropriate factor pair. 2x2 − 16x + 30 = 2(x − 3)(x − 5)
x 2 – 6x + 5
7 Write the original fraction. -----------------------------------
-
2x 2 – 16x + 30
( x – 1)( x – 5)
8 Replace the numerator and denominator with = ------------------------------------
-
2( x – 3)( x – 5)
the factorised form.
x–1
9 Cancel any common factors and simplify if = ------------------- -
2( x – 3)
appropriate.

x–8
Note: We cannot simplify a fraction like ----------- any further. It may be tempting to cancel
-
x+2
x in the numerator or denominator, but x is not a factor of x − 8 or x − 2 and hence we
cannot cancel further. (The same applies to trying to cancel 8 and 2 by dividing by 2.)
We can cancel only terms or expressions that are factors; that is, they are multiplied to
another term or expression.

Factorising will also help when we have to multiply or divide algebraic fractions.

WORKED Example 12
2x + 4 3x – 3
Simplify --------------- × -------------- , by factorising the numerator and denominator and cancelling as
-
3 x+2
appropriate.
THINK WRITE
2x + 4 3x – 3
1 Write the expression. -------------- × --------------
- -
3 x+2
2( x + 2) 3( x – 1)
2 Factorise each numerator and denominator as appropriate and = ------------------- × -------------------
- -
3 x+2
cancel common factors; one in any numerator and one in any
denominator. The factors that can be cancelled are 3 and (x + 2).
3 Multiply numerators and multiply denominators and simplify. = 2(x − 1)

Note: If 2 fractions are divided, you must change the division sign to a multiplication
sign and tip the second fraction (times and tip).

remember
remember
When simplifying algebraic fractions:
1. factorise the numerator and the denominator
2. cancel factors where appropriate
3. if 2 fractions are multiplied, factorise where possible then cancel any factors,
one from the numerator and one from the denominator
4. if 2 fractions are divided, remember to times and tip before factorising and
cancelling.

Simplifying algebraic
6G fractions
WORKED 1 Simplify each of the following by factorising the numerator and denominator and 6.4
Example
cancelling as appropriate. HEET
10a, b, c

SkillS
2a + 2 3a + 6 4a – 4 7 p – 21
a --------------- b --------------- c --------------
- d ------------------
2 9 4 7
4x + 8 3x – 12 9x + 54 12x – 96
e --------------- f -----------------
- g -----------------
- h --------------------
- Math
8 6 3 4

cad
12x – 24 9x – 45 2 –6 Simplifying
i --------------------
- j -----------------
- k -----------------
- l -----------------
-
18 15 2x + 10 6x – 24 algebraic
fractions
2x + 2 3x – 6 5x – 15 9x + 18
m --------------- n --------------
- o -----------------
- p ------------------
3x + 3 4x – 8 6x – 18 3x + 6
WORKED 2 Simplify each of the following fractions by factorising numerator and denominator and
Example
10d
cancelling as appropriate.
x 2 + 3x p2 – 5 p y 2 – 4y a 2 + 10a
a ----------------- b -----------------
- c ----------------
- d --------------------
-
x p y a
7x 5b 9y 4m
e ----------------- f ----------------- g ----------------
- h -------------------
-
5x – x 2 5b – b 2 y 2 – 9y m 2 – 4m
7x 3m a – 2a 2 2b 2 + b
i ----------------
- j -------------------
- k -----------------
- l --------------------
-
x – 2x 2 2m 2 – m a 2 + 3a 5b – 8b 2
3x 2 + x 7b – 2b 2 9m 2 + 18m 32a – 16a 2
m ----------------
- n --------------------
- o --------------------------
- p --------------------------
-
9x – x 2 b 2 + 5b 12m 2 – 3m 24a + 16a 2
WORKED 3 Simplify each of the following algebraic fractions by ﬁrst factorising the numerator and
Example
11
the denominator.
x 2 + 5x + 6 x 2 + 7x + 12 x 2 – 9x + 20 b 2 – 8b + 7
a --------------------------
- b -----------------------------
- c ----------------------------- d --------------------------
-
x+3 x+4 x–5 b–1
a 2 – 18a + 81 a 2 – 22a + 121 x 2 – 49 m 2 – 64
e --------------------------------
- f -----------------------------------
- g ----------------
- h ------------------
a–9 a – 11 x–7 m–8
a 2 – 7a + 12 p2 – 4 p – 5 x 2 + 6x + 9 m 2 – 2m + 1
i -----------------------------
- j --------------------------
- k --------------------------
- l ----------------------------
-
a 2 – 16 p 2 – 25 x 2 + 2x – 3 m 2 + 5m – 6
y 2 – 4y – 12 x 2 – 4x + 4 x 2 + 3x – 40 x 2 + 3x – 18
m ----------------------------
- n -------------------------- o ----------------------------- p -----------------------------
y 2 – 36 x2 – 4 x 2 + 6x – 16 x 2 – 6x + 9

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