Expanding brackets
and factorising
CHAPTER 8
8
Expanding brackets and factorising
CHAPTER
8.1 Expanding brackets
There are three rows. Each row has n students.
The number of students is 3 n 3n.
Two students are added to each of the three rows, so 3 2 6 students are added.
There are now n 2 students in each row.
n2
n2
n2
Total number of students 3(n 2) 3 n 3 2 3n 6
The simplest way of writing this is
3(n 2) 3n 6
Writing 3(n 2) as 3n 6 is called expanding brackets. This is also known as multiplying out
brackets.
Example 1
Expand 5(2x 3).
Solution 1
5(2x 3) 5 2x 5 3
10x 15
Multiply each term inside the bracket by the term outside the bracket.
Example 2
Expand p( p 3).
Solution 2
p( p 3) p p p 3
p2 3p
Multiply each term inside the bracket by the term outside the bracket.
Positive ( ) negative () negative ().
Example 3
Expand and simplify 5p 3( p q).
Solution 3
5p 3( p q) 5p 3 p 3 q Multiply each term inside the bracket by the 3 outside the bracket.
106
5p 3p 3q
Negative (3) negative (q) is positive (3q).
2p 3q
Collect like terms.
8.2 Factorising by taking out common factors
CHAPTER 8
Example 4
Expand 2y( y 2 4y 3).
Solution 4
2y( y 2 4y 3) 2y y 2 2y 4y 2y 3
2y 3 8y 2 6y
Multiply each term inside the bracket by
the term outside the bracket.
Exercise 8A
1 Expand
a 2(n 1)
e 5(c d )
i 5(3g 4)
b 2(2 n)
f 2(c d )
j 2(3g 4h)
c 4(p 2)
g 4(m n)
k 7(x 2y 3)
d 2(3 n)
h 2(2x 3y)
l 3(p 2q 1)
2 Expand
a n(n 1)
e y(3y y2)
i (y 1)
b b(b2 3b)
f 3(5n 4)
j 5(d 2 d )
c 2n(n 3)
g 3(2a 1)
k a(b c)
d a(4 a)
h 2(n 4)
l 2x(3x2 4x 1)
3 Expand and simplify
a 2(p 1) 5
d 2(2c 1) 3(3c 2)
g 2(2p 3q) 2(2p q)
j 3 (n 3)
b
e
h
k
4 Expand and simplify
a x(x 1) 1(x 1)
d t(t 3) 2(t 5)
b q(q 1) 3(q 1)
e a(a 3) 2(a 1)
2(t 3) t
2(m 3) 3(2m 1)
4p 2(p 1)
2q 3(q 1)
c
f
i
l
2(n 1) 3(n 2)
2(3d 4) 3(1 2d )
5y 3(2y 1)
4x 3(2x 1)
c s(s 4) 2(s 4)
f n(n 2) 4(n 2)
8.2 Factorising by taking out common factors
Expanding 3(2b 5) gives 6b 15
Factorising is the reverse process to expanding brackets.
So factorising 6b 15 gives 3(2b 5).
3 is the highest common factor of 6 and 15
The two factors of the expression 6b 15 are 3 and 2b 5
Example 5
Factorise 8c 12s.
Solution 5
8c 12s 4 2c 4 3s
4(2c 3s)
The common factors of 8 and 12 are 1, 2 and 4
The HCF of 8 and 12 is 4
Write the HCF outside the bracket and the other factor inside the bracket.
107
Expanding brackets and factorising
CHAPTER 8
Example 6
Factorise ab ac.
Solution 6
ab ac a b a c
a(b c)
The common factor of ab and ac is a.
Write the a outside the bracket.
Example 7
Factorise y2 y.
Solution 7
y2 y y y 1 y
The common factor of y2 and y is y.
yyy1
Re-writing 1 y as y 1
y(y 1)
Write the y outside the bracket.
Example 8
Factorise completely 4ax 8xb 6xc.
Solution 8
4ax 8xb 6xc 2x 2a 2x 4b 2x 3c
The common factor of 4ax, 8xb and 6xc is 2x.
2x(2a 4b 3c)
Write the 2x outside the bracket.
In questions where you are asked to ‘Factorise completely’ check that the terms of the expression
inside the bracket do not have a common factor.
Example 9
Factorise completely 9a2b 12ab2c.
Solution 9
9a2b 12ab2c 3 3 a a b 3 4 a b b c
108
The common factor is 3 a b.
3ab(3 a 4 b c)
Write the 3ab outside the bracket.
3ab(3a 4bc)
Note that 3a and 4bc do not have a
common factor.
Exercise 8B
1 Factorise these expressions.
a 3y 6
b 2x 2
e 6x 2y
f 3a 12b
i 10c 6d
j 8x 12y
m 9d 6e 12f
n pq pr
q db b
r pq py p
2
u 4y 3y
v y3 2y
y ay3 by2 cy
c
g
k
o
s
w
5p 10
12p 6q
10m 15n 5
xy zy
y2 yz
y3 5y
d
h
l
p
t
x
3d 6e
4d 6e
8p 6q 4
ab 7b
y2 y
p2 pq pr
8.3 Expanding the product of two brackets
2 Factorise completely
a 2xy 4x
d 5ac abc
g 14y4 7y2
j 10x2y 15y2
m 8a2b 16a3 12ab2
CHAPTER 8
3pq 9qr 6pqr
4pq 6pqr
9d3 6d2
3xy2 8x2y5
4a4b2 6a3b3 12a2b2
b
e
h
k
n
c
f
i
l
o
4xy 8y2
rt rst
ab3 a2b
12pq2r 16qp3
(p2q)2 p3q
8.3 Expanding the product of two brackets
Consider the areas of these two rectangles.
4
x
x
x
2
2
Area x(x 2)
Area 4(x 2)
Combining these two rectangles gives a single rectangle with length (x 4) and width (x 2)
x
4
x
2
Area (x 4)(x 2)
The area of the large rectangle is equal to the sum of the areas of the two smaller rectangles so
(x 4)(x 2) x(x 2) 4(x 2)
To expand two brackets, multiply each term in the first bracket by the second bracket so
(x p)(x q) x(x q) p(x q)
x2 qx px pq
Expanding (x p)(x q) gives x2 qx px pq.
Example 10
Expand and simplify (x 2)(x 3).
Solution 10
(x 2)(x 3) x(x 3) 2(x 3)
Multiply each term, x and 2 in the first
bracket by the second bracket, (x 3).
x2 3x 2x 6
x2 x 6
Collect like terms.
109
Expanding brackets and factorising
CHAPTER 8
Example 11
Expand and simplify (n 2)2.
Solution 11
(n 2)2 (n 2)(n 2)
(n 2)2 (n 2) (n 2).
n(n 2) 2(n 2)
Multiply each term in the first bracket by the second bracket.
n2 2n 2n 4
n2 4n 4
Collect like terms.
Example 12
Expand and simplify (4p 5q)(3p 2q).
Solution 12
(4p 5q)(3p 2q) 4p(3p 2q) 5q(3p 2q)
Multiply each term in the first bracket by
the second bracket.
12p2 8pq 15qp 10q2
Collecting like terms; note 15qp is the
same as 15pq.
12p2 7pq 10q2
Exercise 8C
1 Expand and simplify
a (x 1)(x 2)
e (x 2)(x 1)
i (x 2)(x 1)
b (q 3)(q 1)
f (x 3)(x 3)
j (x 3)(x 4)
2 Expand and simplify
a (x 1)2
d (x 5)2
b (x 3)2
e (x 8)2
3 Expand and simplify
a (x y)(x 2y)
d (3p q)(p q)
g (p q)(p 3q)
j (4x 5y)(2x 3y)
m (3p 5q)2
b
e
h
k
n
4 Expand
a (a b)(c d)
b (e f)(g h)
c (r 1)(r 3)
g (r 1)(r 3)
(2x y)(x y)
(x 3y)(2x 5y)
(2x 3y)(3x 4y)
(x y)2
(6a 5b)(6a 5b)
d (x 4)(x 3)
h (x 2)(x 4)
c (x 7)2
f (x 10)2
c
f
i
l
(p q)(3p q)
(a b)(a b)
(3x 2y)(x y)
(2x y)2
8.4 Factorising by grouping
In Section 8.2, an expression was factorised by taking out the common factor of the terms in the
expression. The common factor was a single term, for example 3, 4a, ab. For some expressions the
common factor can involve the sum or difference of terms, for example x 3 or 2a 4b.
110
8.4 Factorising by grouping
CHAPTER 8
Example 13
Factorise completely 12(x 2)2 9(x 2).
Solution 13
12(x 2)2 9(x 2)
3 4 (x 2) (x 2) 3 3 (x 2)
3 is the HCF of 12 and 9
3 (x 2) 4 (x 2) 3 (x 2) 3
3 and (x 2) are both common factors.
3(x 2)[4(x 2) 3]
So write 3(x 2) outside the square bracket.
3(x 2)[4x 8 3]
Simplify the terms inside the square bracket.
3(x 2)(4x 5)
Expanding (a b)(c d) gives ac ad bc bd.
Since factorising is the reverse process to expanding brackets,
factorising ac ad bc bd gives (a b)(c d).
In the expression ac ad bc bd there is no common factor of all four terms.
Pairs of terms with a common factor can be grouped together and factorised to give
a(c d) b(c d). These two terms have a common bracketed factor, (c d).
Using this common factor gives (c d)(a b) which can also be written as (a b)(c d).
This is called factorising by grouping.
Example 14
Factorise pr qs ps qr.
Solution 14
pr qs ps qr pr ps qs qr
Group the terms in pairs so that each pair has a common factor –
in this case p and q.
p(r s) q(r s)
Factorise each pair. The bracketed term must be the same.
(r s)(p q)
(r s) is a common factor.
This answer could be written as (p q)(r s).
Exercise 8D
1 Factorise completely
a y(x 3) 2(x 3)
c p(x 2y) q(x 2y)
e (x 1)2 4(x 1)
g (x 5) 3(x 5)2
i 4(x 2)2 2(x 2)
k 6(x 4) 4(x 4)2
b
d
f
h
j
l
a(x y) b(x y)
2p(x 4) 3q(x 4)
(x y)2 b(x y)
(x 3y) 2(x 3y)2
6(x y)2 3(x y)
6y(x 3y) 9(x 3y)2
2 Factorise completely
a ab ac db dc
c x2 ax 2x 2a
e x2 3x 2x 6
g 2x2 8x 3x 12
b
d
f
h
pq 2r pr 2q
ps qr pr qs
x2 3x x 3
2x2 3x 2x 3
111
Expanding brackets and factorising
CHAPTER 8
8.5 Factorising expressions of the form x2 bx c
Expanding (x 4)(x 2) gives
(x 4)(x 2) x(x 2) 4(x 2)
x2 2x 4x 8
x2 6x 8
Since factorising is the reverse process to expanding brackets, factorising x2 6x 8 gives (x 4)(x 2).
In the expression x2 6x 8 there is no
common factor of all three terms.
Taking out a common factor of just two of the three terms is not
factorising the expression. An answer x(x 6) 8 is incorrect.
From Section 8.3 (x p)(x q) x2 qx px pq.
So
(x p)(x q) x2 (p q)x pq.
(p q)x is the same as x(p q).
The diagrams show the expansions of
(x p)(x q)
x
x
q
(x 4)(x 2)
and
p
x
x
px
x
qx
pq
2
2
4
x
4x
2x
8
2
For these expansions to be the same, obviously x2 x2 but also
pq 8
and
px qx 4x 2x 6x
and the sum of p and q is 6,
since px qx (p q)x.
that is, the product of p
and q is 8
To factorise x2 6 x 8
find two numbers whose product is 8 and whose sum is 6.
The two numbers are 2 and 4.
So
x2 6x 8 x2 2x 4x 8
x(x 2) 4(x 2)
(x 2)(x 4)
Example 15
Factorise x2 7x 12
Solution 15
x2 7x 12
4 3 12
4 3 7
x2 7x 12 x2 4x 3x 12
Find two numbers whose product is 12 and whose sum is 7
Write 7x as 4x 3x.
x(x 4) 3(x 4)
Factorise by grouping. The bracketed term must be the same.
(x 4)(x 3)
(x 4) is a common factor.
This answer could also be written as (x 3)(x 4).
112
8.6 Factorising the difference of two squares
CHAPTER 8
Example 16
Factorise x2 x 6
Solution 16
x2 1x 6
2 3 6
2 3 1
Find two numbers whose product is 6 and whose sum is 1
x2 x 6 x2 2x 3x 6
Write x as 2x 3x.
x(x 2) 3(x 2)
Factorise by grouping. The bracketed term must be the same.
(x 2)(x 3)
(x 2) is a common factor.
This answer could also be written as (x 3)(x 2).
Exercise 8E
1 Write down the two numbers
a whose product is 10 and whose sum is 7
b whose product is 7 and whose sum is 8
c whose product is 24 and whose sum is 11
d whose product is 6 and whose sum is 5
e whose product is 8 and whose sum is 2
f whose product is 12 and whose sum is 1
g whose product is 12 and whose sum is 4
h whose product is 9 and whose sum is 6
i whose product is 20 and whose sum is 9
j whose product is 16 and whose sum is 0
2 Factorise these expressions.
a x2 3x 2
d x2 5x 4
g x2 11x 24
j x2 7x 8
m x2 2x 1
p x2 5x 36
b
e
h
k
n
q
x2 4x 3
x2 7x 10
x2 5x 6
x2 17x 18
x2 6x 9
x2 16x 36
c
f
i
l
o
x2 6x 5
x2 8x 7
x2 2x 8
x2 3x 18
x2 10x 25
8.6 Factorising the difference of two squares
x2, 4, 4x2, 9, 1, p2, A2, B2,
(2t 1)2 and 9(t 6)2
are all squares.
x2 4
A B2
4x2 9
1 p2
All these expressions show
the difference of two squares.
2
(2t 1)2 9(t 6)2
An expression that is the difference of two squares can be factorised using the method in
Section 8.5
113
Expanding brackets and factorising
CHAPTER 8
Example 17
Factorise x2 4
Solution 17
x2 4 x2 0x 4
x2 0x 4
2 2 4
2 2 0
x2 4 x2 2x 2x 4
Write x2 4 in the form x2 bx c.
Find two numbers whose product is 4 and whose sum is 0
Write 0x as 2x 2x.
x(x 2) 2(x 2)
Factorise by grouping. The bracketed term must be the same.
(x 2)(x 2)
(x 2) is a common factor.
This answer could also be written as (x 2) (x 2).
Example 18
Factorise x2 n2.
Solution 18
x2 n2 x2 0x n2
x2 0x n2
n n n2
n n 0
x2 n2 x2 nx nx n2
Find two numbers whose product is n2 and whose sum is 0
Write 0x as nx nx.
x(x n) n(x n)
Factorise by grouping. The bracketed term must be the same.
(x n)(x n)
(x n) is a common factor.
This answer could also be written as (x n)(x n).
To factorise the difference of the squares of two terms, multiply the sum of the two terms by the
difference of the two terms.
So, for any two terms A and B,
A2 B2 (A B)(A B)
Example 19
Factorise 4x2 9
Solution 19
4x2 9 (2x)2 32
(2x 3)(2x 3)
114
4x2 and 9 are squares since 4x2 (2x)2 and 9 32, so this expression is
the difference of two squares. Write 4x2 9 in the form A2 B2.
Use A2 B2 (A B)(A B) with A 2x and B 3
8.6 Factorising the difference of two squares
CHAPTER 8
Example 20
Factorise 18x2 50y2
Solution 20
18x2 50y2 2(9x2 25y2)
Take out the common factor, 2, and 9x2 and 25y2 are squares since
9x2 (3x)2 and 25y2 (5y)2
2[(3x)2 (5y)2]
Write 9x2 25y2 in the form A2 B2.
2[(3x 5y)(3x 5y)]
Use A2 B2 (A B)(A B) with A 3x and B 5y.
2(3x 5y)(3x 5y)
Example 21
Factorise (2t 1)2 9(t 6)2
Solution 21
(2t 1)2 9(t 6)2
(2t 1)2 [3(t 6)]2
Write (2t 1)2 9(t 6)2 in the form A2 B2.
[(2t 1) 3(t 6)][(2t 1) 3(t 6)]
Use A2 B2 (A B)(A B) with A 2t 1 and B 3(t 6).
[2t 1 3t 18][2t 1 3t 18]
[5t 17][t 19]
Expand and simplify the terms in the squared brackets.
(5t 17)( t 19)
Exercise 8F
1 Factorise these expressions.
a p2 1
c x2 36
e 144 b2
g 4p2 1
i 25y2 4
k 49 (1 x)2
b
d
f
h
j
y2 9
a2 100
1 p2
9x2 1
(x 1)2 25
2 a Factorise x2 p2.
b Hence find the value of
ii 7.642 2.362
i 512 492
3 Find the value of 10062 9942
4 Factorise these expressions completely.
b
a 2p2 32
2
2
d
c 3y 75x
2
2
e 12p 27q
f
2
2
g 8(x 2) 2(x 1)
h
27a2 48
4a2 64b2
9(p 1)2 4p2
50(2x 1)2 18(1 x)2
115
Expanding brackets and factorising
CHAPTER 8
Chapter summary
You should now be able to:
expand (multiply out) brackets by multiplying each term inside the bracket by the term
outside the bracket, for example 3x(x 2) 3x2 6x
factorise by taking out a common factor, for example 4c 6cs 2c(2 3s); factorising is
the opposite process to expanding brackets
expand two brackets by multiplying each term in the first bracket by the second bracket,
for example (x p)(x q) x(x q) p(x q) x2 qx px pq and then simplify
if possible
factorise by grouping
factorise an expression of the form x2 bx c
factorise the difference of two squares using the general rule A2 B2 (A B)(A B).
Chapter 8 review questions
1 a Multiply out 4(3x 2)
2 Simplify
a 3(a 2) 5
b Simplify 2(3x 1) 3(x 2)
b 4(b 3) 3b
3 Simplify
a 5(2n 3m) 3(n 4m)
b 4(2n 2m 1) 3(2n 4p 3)
4 Expand and simplify 3(5x 2) 2 (2x 5)
5 a Simplify i p2 p7
b Expand t (3t2 4)
c 2(3c d) 3(c d )
ii x8 x3
(1388 March 2002)
y4 y3
iii y5
6 a Expand the brackets p(q p2)
b Expand and simplify 5(3p 2) 2(5p 3)
(1387 November 2003)
(1387 November 2004)
7 Expand y(3y2 5y)
8 Factorise
a 2r 6
b 4s 10t
9 Factorise
a 6a 12b 30
b 8x 12y 16z
10 a Expand and simplify 3(2x 1) 2(2x 3)
b Factorise y2 y
11 Factorise k2 k
12 Factorise
a ab 2bc
116
b 2x 3ax3
(1387 November 2003)
Chapter 8 review questions
13 a Factorise x2 3x
CHAPTER 8
b Simplify k5 k2
(1387 June 2004)
14 Expand and simplify (y 5)(y 3)
(1388 January 2005)
15 Expand and simplify (x 9)(x 4)
16 a Simplify 5p 4q 3p q
c Factorise 4x 6
e Simplify 2x3 x5
(1388 March 2005)
7
x
b Simplify 2
x
d Multiply out and simplify (x 3)(x 2)
(1386 November 2002)
17 a Expand and simplify ( x 7)(x 4)
c Factorise p2 6p
b Expand y( y3 2y)
d Factorise completely 6x2 9xy
(1387 June 2005)
18 a Simplify k5 k2
b Expand and simplify
ii (x 3y)(x 2y)
i 4(x 5) 3(x 7)
2
c Factorise (p q) 5(p q)
d Simplify (m4)2
e Simplify 2t 2 3r 3t 4
(1387 June 2004)
19 Expand and simplify
a 3b 1 4(b 2)
c (x 2)(x 7)
b y(2y y3)
d 4 (m 1)2
20 a Expand (x 5)(x 8)
b Factorise x 2 5x 14
(1388 March 2004)
21 a Expand and simplify (x 1)(x 7)
b Factorise y2 3y 10
(1388 November 2005)
22 a Factorise x2 4x 21
b Factorise 4x2 25
c Factorise ab 2ay bx 2xy
23 a Factorise m2 n2
b Hence find the value of 1982 22
24 Factorise completely 6p2 4p 3pq 2q
25 Factorise completely
a 3a2 12b2
(1385 June 1998)
b 8(n 1)2 2(n 3)2
26 a Factorise x2 4x 4
b Hence, factorise completely (3x 4)2 (x2 4x 4)
117