Download Topic 2: Factorisation 3 3 3( ) Factorising a b a b + = +

Topic 2: Factorisation
Factorising is the opposite of expanding. Factorising (sometimes called factoring) is based on the
word Factors, so factorising is about producing factors. Just like the factor pairs of 36 will multiply
to give 36, the factor of a polynomial must multiply to give the polynomial.
Product
Sum
3a + 3b = 3(a + b)
Factorising
Factorising is about starting with a sum (or difference) of terms and finishing with a product of
factors. In the question above, the start is a sum(+) of two terms(3a, 3b) and finishing with a
product (x) of two factors [3 and (a+b)].
Factorising by common factor
Factorise 2a + 6
The first step is to look at the two terms and identify the highest common factor (HCF). In this case
the HCF is 2. Write each term with 2 as a factor. With experience, some of the steps may be
omitted.
2a + 6
= 2× a + 2×3
= 2(a + 3)
Examples
Factorise
H.C.F.
3 f − 21
3
3 f − 21
= 3× f − 3× 7
= 3( f − 7)
2ab − a
a
2ab − a
= a × 2b − a × 1
= a (2b − 1)
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Solution
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3 xy − 6 xz
3x
3 xy − 6 xz
= 3x × y − 3x × 2 z
= 3x × ( y − 2 z )
= 3x( y − 2 z )
ab 2c + 3ab
ab
ab 2c + 3ab
= ab × bc + ab × 3
= ab(bc + 3)
−
25 x 2 − 10 x
Hint: take out a
negative common
factor if the first term
is negative
−
5x
−
25 x 2 − 10 x
= − 5x × 5x + − 5x × 2
= − 5 x(5 x + 2)
Remember, you can check your answer by expanding the answer you obtained to get the question
back. For example:
ab(bc + 3)
= ab × bc + ab × 3
= ab 2c + 3ab
= the original expression
Factorising by common factor (grouping factor)
Consider the question:
Factorise
x(a + b) + 4(a + b)
You notice that this expression is basically two terms. Looking at both terms you will notice that
(a+b) is common to both terms. This common factor is called a grouping factor. It is a factor just
like the questions above.
x(a + b) + 4(a + b)
=
(a + b)( x + 4)
Factorise
p (r − s ) − 4(r − s )
p (r − s ) − 4(r − s )
=
(r − s )( p − 4)
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Factorise
2e(a − b) − 5(b − a )
Notice in this example that the grouping factors are not the same, but they are similar. Factorising
by grouping factor cannot happen until they are both the same.
The binomial (b - a) can be rewritten as –(a - b). Check by expanding if you are unsure.
2e(a − b) − 5(b − a )
= 2e(a − b) − 5 × −(a − b)
= 2e(a − b) + 5(a − b)
(a b)(2e + 5)
=−
Factorise
y ( a − b) + a − b
This question becomes clearer if some extra detail is added, a-b can also be expressed as 1(a-b).
Then the expression becomes:
y ( a − b) + a − b
= y (a − b) + 1(a − b)
=
(a − b)( y + 1)
Factorise
ab(a − b) + b − a
ab(a − b) + b − a
= ab(a − b) + (b − a )
= ab(a − b) − 1(a − b)
=−
(a b)(ab − 1)
Often the questions require some factorising before a grouping factor becomes clear.
For example: Factorise ab + 4b − 4a − 16
By looking at the first two terms you notice a common factor of b and another factor (a+4). When
looking at the last two terms you notice a common factor of -4 and another factor (a+4).
ab + 4b − 4a − 16
= b(a + 4) − 4(a + 4)
=
(a + 4)(b − 4)
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Factorise
a 2 − 3ab − 3a + 9b
a 2 − 3ab − 3a + 9b
= a (a − 3b) − 3(a − 3b)
=−
(a 3b)(a − 3)
Factorise 15 pr
− 3rq + 5 ps − qs
15 pr − 3rq + 5 ps − qs
= 3r (5 p − q ) + s (5 p − q )
=(5 p − q )(3r + s )
Factorising by common factor (exponents)
Sometimes a good knowledge of exponents is required to determine the common factor. In the
example below the common factor is a power. Remember you can always expand to see if you are
correct.
Factorise m7 + m4
m7 + m 4
= m 4 (m3 + 1)
Factorise ma + ma+2
ma + ma+2
= m a (1 + m 2 )
Factorise e2 x + e x
e2 x + e x
= e x (e 2 + 1)
Factorise e x + e− x
e x + e− x
e x + e− x
= e − x (e 2 x + 1)
or
= e x (1 + e −2 x )
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Factorising trinomials
In this section we will practise factorising quadratic trinomials. Trinomials are polynomials that
contain three terms. A quadratic trinomial is a trinomial in the form: ax 2 + bx + c where a, b and c
(a≠0) are coefficients, for example: 3 x 2 − 4 x + 1 .
This is an important skill to have for simplifying algebraic expressions and graphing functions.
Two methods are shown. The first method is logical where thinking about possible solutions is
required. The second method is a sequence of steps, so following the method to achieve an
answer just involves remembering the method.
Example: Factorise the trinomial x 2 + 7 x + 10
Mental Strategy
Sequence of Steps Strategy
You know that a trinomial may produce 2
monomial factors.
Using ideas from the FOIL method of
2
expanding, the first term of the trinomial, x ,
came from x × x ;
the last term 10 could come from the factor
pair 2 and 5, or 1 and 10.
Construct a t diagram to help.
The general form of a trinomial is
ax 2 + bx + c . So for the trinomial being
factorised x 2 + 7 x + 10 ;
ac =×
1 10 =10 and b =7 .
Think of two numbers that multiply to 10 and
add to 7? 2 and 5. If there are no such
numbers, then it means it cannot be
factorised.
2
On the left, list the factors of the first term x .
On the right list the factors of the last term 10.
(In this case: 2 & 5 and 1 & 10)
The arrows represent the two middle terms
(the OI from FOIL). Under each pair write sum
of the two middle terms indicated by the
arrow.
Now the trinomial becomes:
x 2 + 7 x + 10
= x 2 + 2 x + 5 x + 10
Using our knowledge of grouping factors:
(x+2)
x
x
This gives us the breakdown of the middle
term 7x into two separate terms 2 x and 5 x .
x 2 + 7 x + 10
2
5
1
10
2x+5x = 7x
1x+10 = 11x
= x 2 + 2 x + 5 x + 10
= x( x + 2) + 5( x + 2)
=
( x + 2)( x + 5)
(x+5)
The middle term in the question is 7x, so the
result is:
If no grouping factor is found – a mistake has
been made.
x 2 + 7 x + 10 =
( x + 2)( x + 5)
After some practice, you may use the Mental Strategy for easier questions and the Sequence
Strategy for more difficult questions.
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Example: Factorise the trinomial x 2 + 2 x − 15
Mental Strategy
Sequence of Steps Strategy
2
The first term of the trinomial, x , came from
x × x . To obtain (-15) from two numbers, one
will be positive, the other negative; so the
factor pairs of (-15) are 1 and -15, -1 and 15, 3
and -5 or -3 and 5. Note factors that differ
only by sign must be included.
x
x
1
-1
3
-3
-15
15
-5
5
1x-15x
=-14x
x+15x
= 14x
3x-5x
=-2x
3x+5x
= 2x
For the trinomial being factorised
x 2 + 2 x − 15 ;
−
ac =
1× −15 =
15 and b =
2.
Two numbers that multiply to -15 and add to 2
are -3 and +5
(-3 x 5 = -15, -3 + 5 = 2)
This gives us the breakdown of the middle term
2x into two separate terms −3 x and 5 x .
Now the trinomial becomes:
x 2 + 2 x − 15
= x 2 − 3 x + 5 x − 15
= x( x − 3) + 5( x − 3)
= ( x − 3)( x + 5)
This indicates the correct option is:
x 2 + 2 x − 15 =
( x − 3)( x + 5)
Example: Factorise the trinomial 2 x 2 + 3 x − 2
Mental Strategy
Sequence of Steps Strategy
2
The first term of the trinomial, 2x , came
from 2x × x . To obtain (-2) from two
numbers, one will be positive, the other
negative; so the factor pairs of (-2) are 1 and 2, -1 and 2, -2 and 1 or 2 and -1.
For the trinomial being factorised
2 x 2 + 3x − 2 ;
ac =2 × − 2 =− 4 and b =3 .
2x
-1
1
-2
2
Two numbers that multiply to -4 and add to 3 are
-1 and +4
(-1 x 4 = -4, -1 + 4 = 3)
This gives us the breakdown of the middle term
3x into two separate terms − x and 4x .
x
2
-2
1
-1
Now the trinomial becomes:
1x-4x
= -3x
2x+2x
=0
2x-2x
=0
1x+4x
=3x
This indicates the correct option is:
2 x 2 + 3x − 2
= 2x2 − x + 4x − 2
= x(2 x − 1) + 2(2 x − 1)
= (2 x − 1)( x + 2)
2 x 2 + 3x − 2 =
(2 x − 1)( x + 2)
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Example: Factorise the trinomial 3 x 2 − 11x + 6
Mental Strategy
Sequence of Steps Strategy
2
The first term of the trinomial, 3x , came from
3x × x . To obtain 6 from two numbers, both
will be positive or both negative. With the
middle term negative, the factor pairs of 6 are
-1 and -6, or -2 and -3.
So for the trinomial being factorised
3 x 2 − 11x + 6 ; ac =3 × 6 =18 and b =−11 .
3x
-1
-2
Two numbers that multiply to 18 and add to -11
are -2 and -9
(-2 x -9 = 18, -2 + -9 = -11)
This gives us the breakdown of the middle term
−11x into two separate terms -2 x and -9x .
x
-6
-3
Now the trinomial becomes:
-1x-18x =19x
-2x-9x = 11x
3 x 2 − 11x + 6
= 3x 2 − 2 x − 9 x + 6
= x(3 x − 2) − 3(3 x − 2)
This indicates the correct option is:
=(3 x − 2)( x − 3)
3 x 2 − 11x + 6= (3 x − 2)( x − 3)
Example: Factorise the trinomial 3 x 2 − 3 x − 90
Looking at this trinomial, you will notice that there is a common factor of 3 in each term, so the first
step in this factorisation is taking out the common factor. This makes the remaining factorising
much easier
3 x 2 − 3 x − 90
= 3( x 2 − x − 30)
Now the trinomial: x 2 − x − 30 is to be factorised
Mental Strategy
Sequence of Steps Strategy
2
The first term of the trinomial, x , came from
x × x . To obtain -30 from two numbers, one
will be positive and one will be negative. The
factor pairs of -30 are -1 and 30, 1 and -30, 5
and -6 or -5 and 6.
So for the trinomial being factorised
ac = 1× −30 = −30 and b = −1 .
x 2 − x − 30 ;
x
-1
1
5
-5
Two numbers that multiply to -30 and add to -1
are 5 and -6
(5 x -6 = -30, 5 + -6 = -1)
This gives us the breakdown of the middle term
−1x into two separate terms 5 x and -6x .
x
30
-30
-6
6
Now the trinomial becomes:
-1x+30x
=29x
1x-30x
=-29x
5x-6x
= -x
5x+6x
=x
This indicates the correct option is:
x 2 − x − 30 = ( x + 5)( x − 6)
x 2 − x − 30
= x 2 + 5 x − 6 x − 30
= x( x + 5) − 6( x + 5)
=
( x + 5)( x − 6)
The final result is 3 x 2 − 3 x − 90= 3( x 2 − x − 30)= 3 ( x + 5 )( x − 6 )
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Example: Factorise the trinomial 6 x 2 + 11x + 4
Mental Strategy
Sequence of Steps Strategy
2
The first term of the trinomial, 6x , came
from x × 6 x or 2 x × 3 x . To obtain 4 from two
numbers, both will be positive. The factor
pairs of 4 are 1 and 4, or 2 and 2.
This time there is two options for the first
term.
x
6x
2x
3x
2
1
2
4
2x+6x
= 8x
6x+4x
=10x
So for the trinomial being factorised
6 x 2 + 11x + 4 ; ac = 6 × 4 = 24 and b = 11 .
Two numbers that multiply to 24 and add to 11
are 3 and 8
(3 x 8 = 24, 3 + 8 = 11)
This gives us the breakdown of the middle term
11x into two separate terms 3 x and 8x .
Now the trinomial becomes:
6x+4x
=10x
3x+8x
=11x
6 x 2 + 11x + 4
= 6 x 2 + 3x + 8 x + 4
= 3 x(2 x + 1) + 4(2 x + 1)
=
(3 x + 4)(2 x + 1)
This indicates the correct option is:
6 x 2 + 11x + 4= (2 x + 1)(3 x + 4)
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Difference of squares
A difference of squares is written as the identity a 2 − b 2 = (a + b)(a − b)
You may remember seeing this, reversed around, as a useful identity for expanding.
When do you know if you have a Difference of Squares?
The example below is a Difference of Squares
− 4
x2
x 2 is x squared
Difference
( subtract )
4 is 2 squared (22 )
Note: There is no middle term
Example: Factorise x 2 − 4
x2 − 4
= x 2 − 22
=
( x − 2)( x + 2)
Example: Factorise x 2 + 16
This is not a difference of squares
Example: Factorise x 2 − 225
x 2 − 225
= x 2 − 152
=+
( x 15)( x − 15)
Example: Factorise 3 x 2 − 75
3 x 2 − 75
• Common factor of 3
= 3( x 2 − 25)
= 3( x 2 − 52 )
=3( x − 5)( x + 5)
Example: Factorise 4 x 2 − 49
4 x 2 − 49
= (2 x) 2 − 7 2
=
(2 x − 7)(2 x + 7)
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The question below uses our knowledge of square and square roots.
Example: Factorise x 2 − 5
x2 − 5
( )
=
x2 − 5
2
• 5 can be rewritten as
( 5)
2
=
( x − 5)( x + 5)
Perfect squares
The identities for perfect squares are:
a 2 + 2ab + b 2 = (a + b) 2
and a 2 − 2ab + b 2 = (a − b) 2
You may remember seeing these, reversed around, as useful identities for expanding. However,
for factorising you need to be able to see the how the middle term relates to 2ab.
Look at the trinomial. If the first and last terms are perfect squares, then examine the middle more
closely. If the middle term is 2ab where a is from the first term and b is from the last term, then it is
a perfect square. Remember you can always factorise as a trinomial, you will get the same result.
2
2
Example: x2 + 6 x + 92 = ( x + 3)
x
2×3× x
3
2
= ( x − 5) 2
Example: x2 − 10− x + 25
2
x
2× 5× x
5
Example: 4 x 22 + 4 x + 12= (2 x + 1) 2
(2 x )
2×1×2 x
1
Example: 4 x 22 − 20
x + 25
= (2 x − 5) 2
−
2
(2 x )
2× 5×2 x
5
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Sum and difference of cubes
Two other identities worth knowing are the sum and difference of cubes. The appearance of these
is below:
Sum of Cubes
Difference of Cubes
Identity:
Identity:
a 3 + b3 = (a + b)(a 2 − ab + b 2 )
a 3 − b3 = (a − b)(a 2 + ab + b 2 )
Examples:
Examples:
(i) x3 + 8
(i) x3 − 125
= x33 + 233
= x33 − 533
b
a
a
b
= ( x + 2)( x −2 2 x +2 4)
= ( x − 5)( x 2 +2 5 x +2 25)
(ii) 108 + 4 x 3 • common factor of 4
(ii) -1 + 8 x3
2
a +b
a − ab + b
a +b
= 4(27 + x 3 )
= 8 x3 - 1
= 4(33 + x 3 )
= (2 x)3 - 13
a − ab + b
= 4(3 + x)(9 + 3 x + x 2 )
= (2 x - 1)(4 x 2 + 2 x + 1)
(iii) 64 x 3 + 1
(iii) 24x3 − 81 y 3 • common factor of 3
( 4x)
=
3
+ 13
= (4 x + 1)(16 x 2 − 4 x + 1)
= 3 8 x3 − 27 y 3 
= 3 (2 x)3 − (3 y )3 
= 3 ( 2 x − 3 y ) ( 4 x 2 + 6 xy + 9 y 2 )
Factorising a sum or difference of cubes results in the product of a binomial factor and a trinomial
factor, which can be written as a 3 ± b3 = (a ± b)(a 2  ab + b 2 ) . The trinomial factor will not factorise
further.
The video below covers some examples similar to those above. Click on the image below to start
downloading the video. Depending on your internet connection, this may take a minute or two.
Video ‘Factorisation’
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Activity
1. Factorise the following.
(a)
8a + 12
(b)
6 x − 12
(c)
12 p − 36
(d)
7 a − 21b
(e)
bg + 2bf
(f)
14a + 24b
(g)
a 2 + 2a
(h)
6 x − 6 x2
(i)
12 p 2 − 6 p
(j)
14a − 21a 2
(k)
3g 2 + 2 g
(l)
12a − 20a 2
(m)
8a 2 + 12ab
(n)
9ax 2 − 12a 2 x
(o)
a 4b3 − a 3b3
(p)
7 a 2b − 49ab 2
(r)
−
(q)
−
2g − 6g 2
4 x3 − 24 x
2. Factorise the following.(Grouping factors)
(a)
a ( x + y) + 2( x + y)
(b)
3( x + y ) + 2 ( x + y )
(c)
a ( x − y) − x + y
(d)
3a ( x − y ) − 2 ( y − x )
(e)
4a 4 ( y − 3) − 6a 3 ( y − 3)
(f)
3( p + q ) − 6 ( p + q )
(g)
3a + ab + 3c + bc
(h)
a 2 + 3a − ac − 3c
(i)
a 3 − 3a 2 + ab − 3b
(j)
x2 y 2 + 1 − x2 y − y
(k)
2 xy 2 + 2 + y 2 + 4 x
(l)
12 x3 + x 2 y − 4 x3 y − 3 x 2
(m)
ax + ay + bx + by
(n)
4 x 2 + x − 12ax − 3a
(o)
6 x3 − 9 x 2 − 2 x + 3
(p)
2a 2b − 2a − b + ab 2
2
2
3. Factorise the following.(Common factor - exponent)
(a)
(b)
4m 4 − 2m 2
x n+2 + x n−2
(c)
e x +1 + e x
(e)
e4m + e2m
(d)
e 2 x − e −2 x
4. The following questions are partially factorised, complete the missing factor.
x 2 + 5 x + 6 = ( x + 2 )( x + ...)
x 2 + 3 x + 2 = ( x + 2 )( x + ...)
(a)
(b)
(c)
x 2 + 7 x + 12 = ( x + 3)( x + ...)
(d)
(e)
x 2 − 7 x + 10 = ( x − 5 )( x − ...)
(f)
(g)
x + 2 x − 35 = ( x + 7 )( x − ...)
(h)
(i)
x − 2 x − 15 = ( x + 3)(... ... ...)
(j)
(k)
x − 10 x + 9 =
( x − 1)(... ... ...)
(l)
2
2
2
( x − 2 )( x − ...)
x 2 − 2 x − 8 = ( x + 2 )( x − ...)
x 2 + 4 x − 12 = ( x − 2 )( x + ...)
x 2 − x − 20 = ( x + 4 )(... ... ...)
x 2 + 15 x + 56 = ( x + 7 )(... ... ...)
x 2 − 3x + 2 =
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5. Factorise these trinomials.
(a)
a 2 + 7a + 6
(b)
p 2 + 9 p + 20
(c)
a 2 − 10a + 24
(d)
a 2 + 11a + 24
(e)
x 2 − 15 x + 36
(f)
x 2 + 13 x + 36
(g)
x 2 + 16 x − 36
(h)
2 x2 + 5x − 3
(i)
3 x 2 − 11x + 6
(j)
4 x 2 + 25 x − 21
(k)
6 x2 − x −1
(l)
4a 2 − 12a + 5
(m)
5 y 2 + 13 y − 6
(n)
3b 2 + 14b − 5
(o)
10 x 2 + 9 x − 9
(p)
30m 2 − 37 m + 10
(q)
6a 2 − 47 a + 77
(r)
8 x 2 + 2 x − 15
6. Factorise the following by using the difference of squares method.
(a)
(b)
b 2 − 49
81 − x 2
(c)
4b 2 − 49
(d)
9x2 − 4 y 2
(e)
4a 2 − 25b 2
(f)
( x + 1)
2
− 49
7. The following questions can be factorised by the perfect square method. Factorise the following
by using the perfect square method or alternatively, factorise as trinomials.
(a)
(b)
a 2 + 14a + 49
x 2 − 12 x + 36
(c)
b 2 − 22b + 121
(d)
4 x2 + 4 x + 1
(e)
25a 2 − 20a + 4
(f)
4 x 2 + 28 x + 49
8. Factorise the following by using the sum or difference of cubes method.
(a)
(b)
b3 − 8
64 − x 3
(c)
8b3 − 125
(d)
x3 − 8 y 3
(e)
27 x 3 − 125b3
(f)
64 x3 − 1000
9. Factorise the questions below, one or more of the skills above is required.
(a)
(b)
48 x 2 − 3
6 x 2 − 33 x + 36
(c)
x4 + 2x2 − 3
(d)
( x − y) − ( x + y)
(e)
x2 + 6 x + 9 − y 2
(f)
x 3 + 3 x 2 − xb 2 − 3b 2
(g)
( x + y)
(h)
a 2 − 9 − 4ac + 12c
(i)
5m3 − m 2 − 20m + 4
(j)
x6 − 1
(k)
4 x 2 y 2 − x 2 − 4 xy − 4 y 2
(l)
2 ( 2a + 1) − 8 ( a − 2 )
2
+ 2( x + y) +1
2
2
2
2
The answers to the activity questions are found in the pdf named:
‘Answers to Activity Questions (PCA)’
Centre for Teaching and Learning | Academic Practice | Academic Skills
T +61 2 6626 9262 E [email protected] W www.scu.edu.au/teachinglearning
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[last edited on] CRICOS Provider: 01241G