Olympic College Topic 12 Factorisation
Definition: A factor of a number is an integer that divides the number exactly.
So for example, the factors of 12 are 1,2,3,4,6 and 12.
If we write down the factors of two numbers then the factors which belong to both groups are called the common factors.
For example: The factors of 12 are 1 , 2 ,3, 4 ,6, and 12
The factors of 20 are 1 , 2 , 4 ,5,10 and 20
The common factors of 12 and 20 are 1,2 and 4
The Greatest Common Factor (GCF) of two numbers is the largest of the common factors. For example, the greatest common factor of 12 and 20 is 4.
It is often more convenient to simplify the process of finding the GCF mentally by going through the factors of each and finding the largest factor that belongs to both.
If the only common factor of two numbers is 1 then that we say that the two numbers are mutually prime or prime for short.
Example 1: Find the GCF of the following numbers.
(a) 20 and 30 (b) 12 and 24 (c) 16 and 44
(d) 4 and 9 (e) 60 and 32 (f) 8,12 and 24
Solution(a): Factors of 20 = 1,2,4,5, 10 and 20
Factors of 30 = 1,2,3,5,6, 10 ,15 and 30 GCF of 20 and 30 = 10
Solution(b): Factors of 12 = 1,2,3,4,6 and 12
Factors of 24 = 1,2,3,4,6,8, 12 and 24 GCF of 12 and 24 = 12
Solution(c): Factors of 16 = 1,2, 4 ,8,and 16
Factors of 44 = 1,2, 4 ,11,22 and 44 GCF of 16 and 44 = 4
Solution(d): Factors of 4 = 1,2 and 4
Factors of 9 = 1,3 and 9 GCF of 4 and 9 = 1 (Prime)
Solution(e): Factors of 60 = 1,2,3, 4 ,5,6,10,12,15,20,30 and 60
Factors of 32 = 1,2, 4 ,8,16 and 32 GCF of 60 and 32 = 4
Solution(f): Factors of 8 = 1,2, 4 ,and 8
Factors of 12 = 1,2,3, 4 ,6 and 12
Factors of 24 = 1,2,3, 4 ,6,8,12 and 24 GCF of 8,12 and 24 = 4
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Olympic College Topic 12 Factorisation
To find the factors of an exponential we simply write down all its powers starting at 1.
Example 2: Find all the factors of x 4 .
Solution: The factors of x
4
are 1,x,x
2
x
3
and x
4
Example 3: Find all the factors of y
3
.
Solution: The factors of y
3
are 1,y,y
2 and y
3
To find the GCF of two exponentials it will be the smaller of the two powers.
Example 3: Find the GCF of x
3
and x
7
Solution: The factors of x
3
are 1,x,x
2
and x
3
The factors of x
7
are 1,x,x
2 x
3
,x
4
,x
5
,x
6
and x
7
The GCF of x
3
and x
7
= x
3
(Notice that this is the smaller of the two powers)
Example 4: Find the GCF of y 5 and y 6
Solution: The factors of y
5
are 1,y,y
2
,y
4
and y
5
The factors of y
6
are 1,y,y
2
y
3
,y
4
, y
5 and y
6
The GCF of y
5
and y
6
= y
6
(Notice that this is the smaller of the two powers)
When the expression is a product of two or more exponentials then the factors are the products of all the permutations of each exponential.
Example 5: Find the factors of x 3 y 2
Solution: The factors of x
3 y
2 are 1,x,x
2
, x
3 y,xy,x
2 y,x
3 y y
2
,xy
2
,x
2 y
2
,x
3 y
3
Example 6: Find the factors of b
2 d
5
Solution: The factors of b 2 d 5 are 1,b,b 2 d,bd,b 2 d d 2 ,bd 2 ,b 2 d 2 d
3
,bd
3
,b
2 d
3 d
4
,bd
4
,b
2 d
4 d
5
,bd
5
,b
2 d
5
When the expression is a multiple of an exponential then its factors are all the permutations of the factors of the number with the powers of the exponential.
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Olympic College Topic 12 Factorisation
Example 7: Find all the factors of 6x
3
Solution: The factors of 6x 3 are 1,x,x 2 , x 3 and 2,2x,2x 2 ,2x 3 and 3,3x,3x 2 ,3x 3 and 6,6x,6x 2 ,6x 3
Example 8: Find all the factors of 4y
2
Solution: The factors of 4y
2
are 1,y,y
2
and 2,2y,2y
2
and 4,4y,4y
2
The greatest common factor (GCF) of two expressions is the largest factor they both share.
There are two methods to finding the GCF of two expressions the first is by writing out all the factors of each expression and then finding the largest result that they both share. This method is very time consuming as the following worked example shows.
Example 9: Find the GCF of 12x 3 and 16x 2
Solution: The factors of 12x
3 are 1 , x , x
2
, x
3
2,2x,2x
2
,2x
3
3,3x,3x
2
,3x
3
4,4x,4x
2
,4x
3
6,6x,6x
2
,6x
3
12,12x,12x
2
,12x
3
The factors of 16x
2
are 1 , x , x
2
2,2x,2x
2
4,4x,4x
2
8,8x,8x
2
16,16x,16x
2
The common factors are 1,x,x
2
2,2x,2x
2
4,4x, 4x
2
The GCF of 12x
3
and 16x
2
is 4x
2
The second method for finding the GCF of two expressions is to find the GCF of the constants and the exponentials separately and then combine the result.
Example 10: Find the GCF of 12x
3
and 16x
2
Solution: The GCF of 12 and 16 is 4
The GCF of x
2
and x
3
is x
2
(The smaller of the two powers)
The GCF of 12x
2
and 16x
3
is 4x
2
Example 11: Find the GCF of 10y
4 x
2
and 25y
3 x
3
Solution: The GCF of 10 and 25 is 5
The GCF of x
2
and x
3
is x
2
(The smaller of the two powers)
The GCF of y
4
and y
3
is y
3
(The smaller of the two powers)
The GCF of 10y
4 x
2
and 25y
3 x
3
is 5x
2 y
3
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Olympic College Topic 12 Factorisation
Exercise 1:
1. Find all the factors of the following numbers.
(a) 26 (b) 60 (c) 82 (d) 144 (e) 100
2. Find all the factors of the following exponentials.
(a) x
6
(b) x
4
(c) b
3
x
2
(d) c
2
d
2
(e) a
5
b
3. Find all the factors of the following expressions.
(a) 6x
3
(b) 4b
4
(c) 20x
2
(d) 10d
3
(e) 12b
4. Find the greatest Common Factors GCF for each of the following.
(a) 12 and 20 (b) 16 and 60 (c) 43 and 57
(d) 100 and 45 (e) 16,24 and 30 (f) 18,36 and 9
5. Find the greatest Common Factors GCF for each of the following.
(a) 6b 3 and 15b 2 (b) 30c 2 and 20c 6 (c) 2d 3 and 6d 3
(d) 12x
4
and 14x
5
(e) 7x
3
and 14x
4
(f) 18g
10
and 81g
22
(g) 14e
4
and 21e
8
(h) 2k
5
and 3k
5
(i) 8g
2
and 36g
12
6. Find the greatest Common Factors GCF for each of the following.
(a) 4y
3 x
5
and 20y
6 x
6
(b) 8b
4 c
2
and 40b
2 c
4
(c) y
5 x
5
and 2y
3 x
3
(d) 14a
3 b
2
and 21a
3 b
2
(e) 5x
4 y
9
and 9x
9 y
4
(f) 9r
2 t
11
and 21r
12 t
12
(g) 6c
7 d
9
and 12c
7 d
6
(h) 4c
2 d
2
and 40c
3 d
2
(i) 10c
4 d
2 e and 8c
2 d
2 e
4
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Olympic College Topic 12 Factorisation
When we expand parenthesis we can use the distributive law so that 5(2x – 7) = 10x – 35
Factorising an expression simplifies it by reversing the process of expanding parenthesis.
So we take 10x – 35 = 5(2x – 7) this process is called factoring by taking out a common factor
This process can be split into two steps.
Step 1: Find the common factors of each term in the expression
In the above example the common factors of 10x and 35 is 5.
Step 2: Write down the common factor and in the parenthesis write down what is left when you divide each term in the expression by the common factor.
In the above example we get 10x – 35 = 5(……….)
The first term is 10x when you divide it by the common factor 5 you get 2x
The second term is – 35 and when you divide it by the common factor 5 you get – 7
These two results are now put in the parenthesis to complete the process.
So 10x – 35 = 5(2x – 7)
Example 1: Factor 18x – 24y
Solution: The two terms of this expression are 18x and – 24y the GCF is 6
18x – 24y = 6(3x – 4y)
Example 2: Factor 3a – 12b + 24n
Solution: The three terms of this expression are 3a ,
– 12b and 24n the GCF is 3
3a – 12b + 24n = 3(a – 4b + 8n)
Example 3: Factor 8x + 7y + 14z
Solution: The three terms of this expression are 8x , 7y and 14z the GCF is 1 so these terms are prime so it is not possible to factor this expression except in a very trivial way as shown below so we typically just say Prime and leave the expression alone
8x + 7y + 14z = 1(8x + 7y + 14z)
Example 4: Factor 4t
3
– 6t
Solution: The two terms of this expression are 4t
3
and – 6t the common factors are 2 and t so combined the common factor is 2t.
So 4t
3
– 6t = 2t(2t
2
– 3)
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Olympic College Topic 12 Factorisation
Example 5: Factor 40x
2
– 25x
Solution: The two terms of this expression are 40x 2 and – 25x the common factors are 5 and x so combined the common factor is 5x.
So 40x
2
– 25x = 5x(8x – 5)
Example 6: Factor the expression 3x
3 y
2
+ 9x
2 y
3
Solution: The 2 terms of this expression are 3x
3 y
2
and 9x
2 y
3
The common factors are 3, x
2
and y
2
so combined the common factor is 3x
2 y
2
.
So 3x 3 y 2 + 9x 2 y 3 = 3x 2 y 2 (x +3y)
Example 7: Factor the expression 3ab
2
+ 6a
2 b
2
– 12a
3 b
3
Solution: The 3 terms of this expression are 3ab
2
, 6a
2 b
2
and – 12a
3 b
3
The common factors are 3 , a and b
2
so combined the common factor is 3ab
2
.
So 3ab
2
+ 6a
2 b
2
– 12a
3 b
3
= 3ab
2
(1 + 2a – 4ab)
Example 8: Factor the expression 4x
2 y – 8xy + 10x
Solution: The 3 terms of this expression are 4x 2 y , – 8xy and 10x
The common factors are 2 , and x so combined the common factor is 2x.
So 4x
2 y – 8xy + 10x = 2x(2xy – 4y + 5)
Example 9: Factor the expression 4a(x + y) + 6b(x + y)
Solution: The 2 terms of this expression are 4a(x + y) + 6b(x + y)
The common factors are 2 , and (x + y) and when we combine them we get a GCF of 2(x + y)
So 4a(x + y) + 6b(x + y) = 2(x + y)(2a + 3b)
It may seem a little strange that (x + y) is a factor but mathematically we can just consider it a single object that both terms contain.
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Olympic College Topic 12 Factorisation
Example 10: Factor the expression 4(x + 4)
2
+ 12(x + 4)
Solution: The 2 terms of this expression are 4(x + 4) 2 + 12(x + 4)
The common factors are 4 , and (x + 4)
2
and when we combine them we get a
GCF of 4(x + 4) 2
So 4(x + 4)
2
+ 12(x + 4) = 4(x + 4)( (x + 4) + 3)
= 4(x + 4)( (x + 7)
It may seem a little strange that (x + y) is a factor but mathematically we can just consider it a single object that both terms contain.
There are two checks that you can make to be sure that you have factored properly. The first is to expand the expression to make sure it is equal to the one given at the start.
So in Example 8 we were asked to factor 4x
2 y – 8xy + 10x
The solution was 2x(2xy – 4y + 5) to check that this is correct we expand the parenthesis.
2x(2xy – 4y + 5) = 2x(2xy) + 2x(– 4y) + 2x(5)
= 4x
2 y – 8xy + 10x
This is the same expression so we have shown that our factoring part of the process was done correctly.
The second check is to see if we have used the correct Greatest Common Factor (GCF)
For example suppose we were asked to factor the expression 8x 3 y 2 + 12x 4 y 3
Lets assume that we wrongly thought that the GCF was 2x
3 y in which case we would get as our factoring
8x
3 y
2
+ 12x
4 y
3
= 2x
3 y(4y + 6xy)
We can tell this would not be the full factoring by looking at what is left in the parenthesis in this case we have 4y + 6xy and since this expression can still be factored more as it has a common factor of 2y we know we have factored the original expression wrongly and we should have used the GCF of 4x 3 y 2 this would then give us \the correct factorization is
8x
3 y
2
+ 12x
4 y
3
= 4x
3 y
2
(2 + 3x)
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Olympic College Topic 12 Factorisation
Exercise 2:
1. Find the GCF of the terms of the polynomial 8 x
6
+ 12 x
3
A. x
3
B. 2x
3
C. 4x
3
D 8x
6
2.
Factor each of the following expressions.
(a) 8x + 8 (b) 6d + 12e (c) 4x + 9 (d) 12 + 24y
(e) 12c – 24 (f) 10y – 15x (g) 12x – 10b (h) 15e – 20g
(i) 6x – 16 (j) 5 – 25g (k) – 2x + 4y (l) – 4x – 12d
3.
Factor each of the following expressions.
(a) 4x + 12y + 6c (b) 8b + 4c + 8d (c) 42d + 12 – 8c
(d) 12 + 8d – 12c (e) 30c – 12d + 15e (f) 25x – 5c – 15
(g) 2x – 18 + 24y (h) 50 – 20g – 15h (i) 6x + 3y + 3t
(j) 16b – 20h – 10j (k) 20t – 15 + 50p (l) 6x – 3 + 9y
4.
Factor each of the following expressions.
(a) 2ab + 3ac (b) 21b + 3bd (c) 5c + tc
(d) 22k + 11gk (e) 30c – 9cd (f) 20x – 15xc
(g) 20xd – 10xg (h) 15rt – 20ft (i) 30xc + 3yc
(j) 15by + 12b – 9y (k) 4ab – 6ac + 12ad (l) 12x + 6x
2
5.
Factor each of the following expressions.
(a) 4b 4 – 8b 3 (b) 6ab + 9ab 3 – 3a 2 b (c) 2cy 2 + 20c 2 y 3
(d) 12by
2
+ 18b
2 y
3
– 2by (e) 12ax
3 y
2
+ 8a
2 x
3
y
2
(f) 6b
3 y
2
+ b
3 y
3
– 9b
3 y
(g) 30xyz + 9x
2 y
2 z
2
(h) 4x
2
+12x (i) 15x
3 c – 40x
3 c
4
(j) 24t 4 + 4t 3 + 32t (k) 3c 4 d – 9c 5 d 5 (l) 5x 4 c – 10x 2 c 7
(m) 3cx
2
– 18c
2 x (n) 5mnp
5
– 20m
5 n
6
(o) 3x
2 y
4
+ 3x
2 y
6
(p) 3a(x + 1) – 6b(x + 1) (q) 3a(x + y) – 9a
2
(x + y) (r) (x + 7)
2
– 2(x + 7)
Page | 8

Olympic College Topic 12 Factorisation
When you look at the following expansions you will notice a pattern.
(x +3)(x – 3) = x 2 – 3x + 3x – 9 = x 2 – 9
(b +10)(b – 10) = b
2
– 10b + 10b – 100 = b
2
– 100
(x +3)(x – 3) = x
2
– 3x + 3x – 9 = x
2
– 9
(a +b)(a – b) = a 2 – ab + ab – b 2 = a 2 – b 2
The general pattern gives us a way to factor the difference of two squares by simply finding the values of a and b.
Definition: A difference of two squares is an expression that contain two terms that are subtracted. The two terms will in turn be squares of some other term.
When you factor such expressions the answer is always in the form
(a + b)(a – b) where a and b are the square roots of the first and second terms. a 2 – b 2 = (a + b)(a – b)
Example 1: Factor the expression t
2
– 4
Solution: The 2 terms of this expression are t
2
and 4, their square roots are t and 2.
So t 2 – 4 = (t + 2)(t – 2)
Example 2: Factor the expression x
2
– 36
Solution: The 2 terms of this expression are x
2
and 36, their square roots are x and 6.
So x
2
– 36 = (x + 6)(x – 6)
Example 3: Factor the expression 25x
2
– 16
Solution: The 2 terms of this expression are 25x 2 and 16, their square roots are 5x and 4.
So 25x
2
– 16 = (5x + 4)(5x – 4)
Example 4: Factor the expression 4y
6
– 9a
4
Solution: The 2 terms of this expression are 4y
6
and 9a
4
, their square roots are 2y
3
and 3a
2
.
So 4y
2
– 9a
2
= (2y
3
+ 3a
2
)(2y
3
– 3a
2
)
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Olympic College Topic 12 Factorisation
Exercise 3
1. Factor the following expressions.
(a) x
2
– 25 (b) x
2
– 49 (c) b
2
– 100
(d) 144 – b
2
(e) x
2
– b
2
(f) 4x
2
– 81
(g) 9x
2
– 4 (h) 16a
2
– 25b
2
(i) 1 – 4a
2 b
4
(j) 9x
2
– 4y
2
(k) b
6
– y
6
(l) 4x
2
– 9b
2
Page | 10

Olympic College Topic 12 Factorisation
2
These expressions contain three terms (hence the name trinomial).
The most common type are simple trinomials – these are expressions where there is only an x
2 term and have the following general form x
2
+ bx + c.
When you factor x
2
+ bx + c it will always take the form (x …..)(x ……)
The missing terms in the parenthesis are the two numbers which are the factors of the constant term c and whose sum equals the coefficient of x called b.
2




The first situation is when a and c are both positive. In this situation the two factors of c will both be positive.
Example 1: Factor the expression x
2
+ 5x + 6
Solution : In this situation we are looking for two factors of 6 that add up to 5. Since both b and c are positive we need only look at positive values of the factors of 6.
The Positive factors of 6 are:-
+1 and + 6 Add to 7 so not the correct combination
+2 and + 3 Add to 5 so this is correct combination
Using these two factors of 6 we get x
2
+ 5x + 6 = (x + 2)(x + 3)
Example 2: Factor the expression x
2
+ 7x + 10
Solution : In this situation we are looking for two factors of 10 that add up to 7. Since both b and c are positive we need only look at positive values of the factors of 10.
The Positive factors of 10 are:-
+1 and + 10 Add to 11 so this is not the correct combination
+2 and + 5 Add to 7 so this is correct combination
Using these two factors of 10 we get x
2
+ 7x + 10 = (x + 2)(x + 5)
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Olympic College Topic 12 Factorisation
The second situation is when b is negative and c is positive. In this situation the two factors will both be negative.
Example 3: Factor the expression x 2 – 6x + 8
Solution : In this situation we are looking for two factors of 8 that add up to – 6 . Since b is negative and c are positive we need only look at the negative factors of 8.
The negative factors of 8 are:-
– 1 and – 8
Add to – 9 so this is not the correct combination
– 2 and – 4
Add to – 6 so this is correct combination
Using these two factors of 8 we get x
2
– 6x + 8 = (x – 2)(x – 4)
Example 4: Factor the expression x
2
– 7x + 12
Solution : In this situation we are looking for two factors of 12 that add up to – 7 . Since b is negative and c are positive we need only look at the negative factors of 12.
The negative factors of 12 are:-
– 1 and – 12 Add to – 9 so this is not the correct combination
– 2 and – 6
Add to – 8 so this is not the correct combination
– 3 and – 4
Add to – 7 so this is not the correct combination
Using these two factors of 12 we get x
2
– 7x + 12 = (x – 3)(x – 4)
Example 5: Factor the expression b
2
– 20b + 100
Solution : In this situation we are looking for two factors of 100 that add up to – 20 . Since b is negative and c are positive we need only look at the negative factors of 100.
The negative factors of 100 are:-
– 1 and – 100
Add to – 101 so this is not the correct combination
– 2 and – 50
Add to – 52 so this is not the correct combination
– 4 and – 25
Add to – 29 so this is not the correct combination
– 5 and – 20
Add to – 25 so this is not the correct combination
– 10 and – 10 Add to – 20 so this is not the correct combination
Using these two factors of 100 we get b
2
– 20b + 100 = (b – 10)(b – 10) = (b – 10)
2
Note: This process does not need to be as laborious as it seems – since the factoring is unique once you find a combination that works you can stop.
Also with practice you will get an insight into which combinations are more likely to succeed and so you will try these ones first and hopefully reduce the number of failed combinations to a minimum.
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Olympic College Topic 12 Factorisation
The third situation is when b is positive and c is negative. In this situation the larger factor will be positive and the smaller factor will be negative.
Example 6: Factor the expression x 2 + 2x – 15
Solution : In this situation we are looking for factors of – 15 that add up to 2. Since the large factor is positive and the smaller is factor is negative we only need to look at the following combinations.
Factors of – 15 are:-
15 and – 1 Add to 14 so this is not the correct combination
5 and – 3 Add to 2 so this is the correct combination
Using these two factors of – 15 we get x
2
+ 2x – 15 = (x + 5)(x – 3)
Example 7: Factor the expression x
2
+ 10x – 24
Solution : In this situation we are looking for factors of – 24 that add up to 10. Since the large factor is positive and the smaller is factor is negative we only need to look at the following combinations.
Factors of – 24 are:-
24 and – 1 Add to 23 so this is not the correct combination
12 and – 2 Add to 10 so this is the correct combination
Using these two factors of – 24 we get x
2
+ 10x – 24 = (x + 12)(x – 2)
Example 8: Factor the expression y 2 + 13y – 60
Solution : In this situation we are looking for factors of – 60 that add up to 13. Since the large factor is positive and the smaller is factor is negative we only need to look at the following combinations.
Factors of – 60 are:-
60 and – 1 Add to 59 so this is not the correct combination
30 and – 2 Add to 28 so this is not the correct combination
15 and – 2 Add to 13 so this is the correct combination
Using these two factors of – 60 we get y
2
+ 13y – 60 = (y + 15)(y – 2)
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Olympic College Topic 12 Factorisation
The fourth situation is when b is negative and c is positive. In this situation the larger factor will be negative and the smaller factor will be positive.
Example 9: Factor the expression y 2 – y – 12
Solution: In this situation we are looking for factors of– 12 that add up to – 1. Since the large factor is negative and the smaller is factor is positive we only need to look at the following combinations.
Factors of – 60 are:-
1 and – 12 Add to – 11 so this is not the correct combination
2 and – 6 Add to – 4 so this is not the correct combination
3 and – 4 Add to – 1 so this is the correct combination
Using these two factors of – 12 we get y
2
– y – 12 = (y + 3)(y – 4)
Example 10: Factor the expression k
2
– 3k – 40
Solution: In this situation we are looking for factors of– 40 that add up to – 3. Since the large factor is negative and the smaller is factor is positive we only need to look at the following combinations.
Factors of – 40 are:-
1 and – 40 Add to – 41 so this is not the correct combination
2 and – 20 Add to – 18 so this is not the correct combination
4 and – 10 Add to – 6 so this is not the correct combination
5 and – 8 Add to – 3 so this is not the correct combination
Using these two factors of – 40 we get k
2
– 3k – 40 = (k + 5)(k – 8)
Exercise 4
1. Factor the following expressions.
(a) x
2
+ 8x + 7 (b) x
2
– 7x + 10 (c) x
2
– 6x + 8
(d) x
2
+ 5x + 6 (e) x
2
+ 4x + 3 (f) x
2
– 7x – 30
(g) x 2 + 2x – 80 (h) x 2 – 8x + 12 (i) x 2 – x – 20
(j) x
2
+ 2x – 63 (k) x
2
+ 3x – 88 (l) x
2
– 4x – 21
(m) x
2
– 5x – 150 (n) b
2
+ 2b – 15 (o) a
2
– 10a + 16
(p) x
2
+ 2x – 63 (q) x
2
+ 3x – 88 (r) x
2
– 4x – 21
(s) x 2 – 5x – 150 (t) b 2 + 8t + 15 (u) a 2 – 10a + 16
(v) x
2
+ 2x – 63 (w) x
2
+ 3x – 88 (x) x
2
– 4x – 21
Page | 14

Olympic College Topic 12 Factorisation
2
A trinomial of the type ax 2 + bx + c in which typically there is not single x 2 term but some multiple instead are normally the most difficult to factor as they do not fall neatly into categories as they did for situations like factor x
2
+ bx + c.
When you factor a complex trinomial of the form ax
2
+ bx + c there are 2 main stages these are.
Example 1: Factor the expression 2x 2 + 5x – 3
Solution: Step 1 Get the x term part of the factor.
There are a set of rules you can follow that will help you to factorise a complex trinomial expression the first is look at the number of x 2 the expression contains , this will tell you the possible number of x's that go into the factor , so for example if the expression has 2x
2
can only be formed by 2x times x. So the solution must take the form on the other hand if it were 4x
2
it could be one of two possible situations either ( 4x .....)(x ......) or (2x .....)(2x ......).
In this example we have 2x
2
+ 5x – 3 and we get.
2x
2
+ 5x – 3 = (2x .....)(x ......)
Step 2 Get the constant term part of the factor.
The second stage in the process is to look at the constant term, in this case its value is – 3 and try all its factor pairs in order until we find the combination that works.
So in this example we can use (1 & – 3) and (– 3 & 1) and (– 1 & 3) and (3 & – 1 )
These 4 combinations as shown below.
We now use FOIL on each one until we get the one that works. Notice that in the factors of – 3 the order is also important so we reverse each possible factor pair.
(2x + 1)(x – 3) this gives 2x 2 – 5x – 3
(2x – 1)(x + 3) this gives 2x
2
+ 5x – 3 This is the correct one.
(2x + 3)(x – 1) this gives 2x
2
+ x – 3
(2x – 3)(x + 1) this gives 2x
2
– x – 3
Only one combination actually works and this is our solution.
So 2x
2
+ 5x – 3 = (2x – 1)(x + 3)
Page | 15

Olympic College Topic 12 Factorisation
Example 2: Factor the expression 5y
2
– 12y + 4
Solution: Step 1 Get the y term part of the factor.
There are a set of rules you can follow that will help you to factorise a complex trinomial expression the first is look at the number of y
2
the expression contains , this will tell you the possible number of y's that go into the factor.
So for example since 5y 2 can only be formed by 5y times y. So the solution must take the form 5y 2 – 12y + 4 = (5y .....)(y ......)
Step 2 Get the constant term part of the factor.
The second stage in the process is to look at the constant term, in this case its value is + 4 and try all its factor pairs in order until we find the combination that works.
So in this example we can use (1& 4) and (4 & 1) and (2 & 2) and (– 1& – 4) and
(– 4 &– 1) and (– 2 & –2)
This in turn gives rise to 6 combinations as shown below. We now use FOIL on each one until we get the one that works. Notice that in the factors of 4 the order is also important so we reverse each possible factor pair.
(5y + 4)(y + 1) this gives 5y
2
+ 9y + 4
(5y + 2)(y + 2) this gives 5y
2
+ 12y + 4
(5y – 1)(y – 4) this gives 5y 2 – 21y + 4
(5y – 4)(y – 1) this gives 5y
2
– 9y + 4
(5y – 2)(y – 2) this gives 5y
2
– 12y + 4 This is the correct one.
Only one combination actually works and this is our solution.
So 5y
2
– 12y + 4 = (5y – 2)(y – 2)
Notice that the first three choices were bound to give positive values for the y–term and so we could have saved ourselves some time by not bothering to check them as they clearly cannot get us the – 12y term that we need.
Another concept that can reduce your workload is when you get a solution that is almost correct one except for the sign such as when you tested
(5y + 2)(y + 2) = 5y 2 + 12y + 4 we got + 12y instead of the – 12y we were seeking
In these situations it is often a good idea to look at other factor combinations similar to those ones but with different signs so instead of using the factors (2 & 2) we used (– 2 & –2) to get (5y – 2)(y – 2) = 5y
2
– 12y + 4 the correct factorisation.
Page | 16

Olympic College Topic 12 Factorisation
Example 3: Factor the expression 4x
2
– 17x – 15
Solution: Step 1 Get the x term part of the factor.
There are a set of rules you can follow that will help you to factorise a complex trinomial expression the first is look at the number of x
2
the expression contains , this will tell you the possible number of x's that go into the factor.
So for example since 5x
2
can only be formed by 4x times x or by 2x times 2x.
So the solution must take the one of these forms
4x
2
– 17x – 15 = (4x .....)(x ......)
4x
2
– 17x – 15 = (2x .....)(2x ......)
Step 2 Get the constant term part of the factor.
Since we do not know which of the above 2 forms (4x ....)(x .....) or (2x ....)(2x .....)
Will yield the correct factorization two must pick one and test it first.
If we choose to test (2x….)(2x….) then the second stage of the process is to look at the constant term, in this case its value – 15 and try all its factor pairs in order until we find the combination that works. Since both terms have 2x we do not need to reverse the factors.
So we can use (1& – 15) and (–1 & 15) and (3 & –5) and (–3 & 5)
This in turn gives rise to 4 combinations as shown below. We now use FOIL on each one until we get the one that works.
(2x + 1)(2x – 15) this gives 4x
2
– 27x – 15
(2x – 1)(2x + 15) this gives 4x
2
+ 27x – 15
(2x + 3)(2x – 5) this gives 4x 2 – 4x – 15
(2x – 3)(2x + 5) this gives 4x
2
+ 4x – 15
None of these combination work so we must try the combination (4x ....)(x .....)
So we can use (1& – 15) and (–1 & 15) and (3 & –5) and (–3 & 5)
This in turn gives rise to 4 combinations as shown below. We now use FOIL on each one until we get the one that works.
(4x + 1)(x – 15) this gives 4x
2
– 59x – 15
(4x – 1)(2x + 15) this gives 4x
2
+ 59x – 15
(4x + 3)(x – 5) this gives 4x
2
– 17x – 15 This is the correct one.
(4x – 3)(x + 5) this gives 4x 2 + 17x – 15
Only one combination actually works and this is our solution.
So 4x
2
– 17x – 15 = (4x + 3)(x – 5)
Page | 17

Olympic College Topic 12 Factorisation
Exercise 5:
1. Factor the following expressions.
(a) 2x
2
+ 15x + 7 (b) 7y
2
– 2y – 24 (c) 16x
2
+ 24x + 9
(d) 2x
2
– 11x + 15 (e) 3p
2
– 13p + 10 (f) 4x
2
– 12x – 40
(g) 5x² + 3x – 8 (h) 2x² + 3x – 2 (i) 2x² – 11x + 5
(j) 16x² – 40x + 25 (k) 4x² + 40x + 25 (l) 4c² – 13c + 9
(m) (4x + 1)(x – 5) (n) (4b + 3)(4b + 5) (o) (3y – 1)(2y – 9)
(p) 3x² – 26x – 9 (q) 25x² + 20x + 4 (r) 9d² + 24d + 16
2. Which of the following is the factorization of 3x
2
+ 7x – 6
A. (3 x – 2)( x – 3) B. ( x + 3)(3 x + 2)
C. (3 x – 2)( x + 3) D. (3 x + 2)( x – 3)
3. Which of the following is the factorization of 40 p 2 – 13 p – 36
A. (8 p + 9)(5 p + 4) B. ( 8 p – 9)(5 p + 4)
C. (8 p – 9)(5 p – 4) D. (8 p + 9)(5 p – 4)
Page | 18

Olympic College Topic 12 Factorisation
There are 3 special cases of factoring we will deal with in this section.
Case 1: Trinomials with a negative x 2 coefficient such as – x 2 + 10x – 9
Case 2: Trinomials in x and y such as x
2
+ 6xy + 5y
2
Case 3: Combined factoring where a common factor and another factoring method are combined such as ax 2 + 4ax + 3a or ax 2 – 9a
2
In these situations we take out the negative as a common factor of – 1 and the resulting trinomial can be solved in the usual way.
Example 1: Factor
– x 2
+ 10x – 9
Solution:
– x 2
+ 10x – 9 =
– (x 2
– 10x + 9) Take out – 1 as a common factor
= – (x – 9)(x – 1) Factoring x 2 – 10x + 9
Example 2: Factor
– 2x 2
– 5x + 18
Solution:
– 2x 2
– 5x + 18 =
– (2x 2
+ 5x – 9) Take out – 1 as a common factor
= – (2x – 9)(x – 1) Factoring 2x 2 + 5x – 9
In these situations there will be a trinomial in x and y such as x
2
+ 6xy + 5y
2
in order to factor these we use a method similar to those used for other trinomials.
Example 3: Factor x
2
+ 6xy + 5y
2
Solution: in order to factor x
2
+ 6xy + 5y
2
we need to look for factors of 5 that add up to 6 in this case we get + 5 and + 1 this represents + 5y and + y and so we get the solution: x
2
+ 6xy + 5y
2
= (x + 5y)(x + y)
Example 4: Factor x
2
+ 3xy – 10y
2
Solution: in order to factor x
2
+ 3xy – 10y
2
we need to look for factors of0 – 10 that add up to
3 in this case we get + 2 and – 5 this represents + 2y and – 5 and so we get the solution: x 2 + 3xy – 10y 2 = (x + 2y)(x + – 5y)
Page | 19

Olympic College Topic 12 Factorisation
In this situation we combine two or more of the methods where the first process involves taking out a common factor followed by one ot the processes described earlier.
Example 1: Factor the expression 5x
2
– 180
Solution: First we take a common factor of 5 from the two terms then we factor the difference of two squares that remains.
So 5x
2
– 180 = 5(x
2
– 36) Take out a common factor of 5
5(x
2
– 36) = 5(x + 6)(x – 6) Factor the difference of two squares
5x
2
– 180 = 5(x + 6)(x – 6)
Example 2: Factor the expression 2ax
2
+ 12ax + 18a
Solution: First we take a common factor of 2a from the 3 terms then we factor what remains.
2ax 2 + 12ax + 18a = 2a(x 2 + 6x + 9) Take out a C.F. of 2a
= 2a(x + 3)(x + 3) Factor the trinomial x
2
+ 6x + 9
= 2a(x + 3)
2
Example 3: Factor the expression 2ab 4 x 2 + 8ab 4 x + 6ab 4
Solution: First we take a common factor of 2ab
4
from the 3 terms then we factor what remains.
2ab
4 x
2
+ 8ab
4 x + 6ab
4
= 2a b
4
(x
2
+ 4x + 3) Take out a C.F. of 2ab
4
= 2a b
4
(x + 3)(x + 1) Factor the trinomial x
2
+ 4x + 3
Example 4: Factor the expression x
4
+ 10x
3
– 11x
2
Solution: First we take a common factor of x
2
from the 3 terms then we factor what remains. x
4
+ 10x
3
– 11x
2
= x
2
(x
2
+ 10x – 11) Take out a C.F. of x
2
= 2a(x + 11)(x – 1) Factor the trinomial x
2
+ 10x – 11
Page | 20

Olympic College Topic 12 Factorisation
Exercise 6
1. Factor the following.
(a) – x
2
+ 5x – 6 (b) – 2x
2
– 17x + 19 (c) – 4x
2
– x + 39
2. Factor the following.
(a) x
2
+ 26xy + 25y
2
(b) x
2
+ 8xy + 12y
2
(c) x
2
– 9xy + 14y
2
(d) x 2 + 3xy – 10y 2 (e) x 2 – 3xy – 4y 2 (f) x 2 – 4xy – 12y 2
3. Factor the following.
(a) 4x 2 – 100 (b) 2x 2 – 50y 2 (c) 11x 2 – 99c 2
(d) 2 – 50d
2
(e) 4x
2
– 4x – 8 (f) 2ax
2
– 2ax – 12a
(g) 6abx
2
– 6abx – 36ab (h) 2a
2
x
2
+ 12a
2
x + 18a
2
(i) abx
2
– 6abx + 5ab
(j) 3xy
2
– 48xy
4
(k) 6a
3
+ 3a
2
– 30a (l) 4x
4
+ 10x
3
– 6x
2
Page | 21

Olympic College Topic 12 Factorisation
Exercise 1:
1.(a) 1,2,13,26 (b) 1,2,3,4,5,6,10,12,15,20,30,60 (c) 1,2,41,81
1.(d) 1,2,3,4,6,8,12,16,24,36,48,72,144 (e) 1,2,4,5,10,20,25,50,100
2.(a) 1,x,x
2
,x
3
,x
4
,x
5
,x
6
(b) 1,x,x
2
,x
3
,x
4
(c) 1,x,x
2
b,bx,bx
2
b
2
,b
2 x,b
2
x
2
b
3
,b
3 x,b
3
x
2
2.(d) 1,d,d
2
c,cd,cd
2
c
2
, c
2 d, c
2 d
2
(e) 1,a,a
2
,a
3
,a
4
,a
5
b,ba,ba
2
,ba
3
,ba
4
,ba
5
3.(a) 1,x,x 2 ,x 3
3.(b) 1,b,b
2
,b
3
,b
4
2,2x,2x 2 ,2x 3 3,3x,3x
2,2b,2b
2
,2b
3
,2b
4
2 ,3x 3 6,6x,6x 2 ,6x
4,4b,4b
2
,4b
3
,4b
4
3
3.(c) 1,x,x
2
2,2x,2x
2
4,4x,4x
2
5,5x,5x
2
10,10x,10x
2
20,20x,20x
2
3.(d) 1,d,d
2
,d
3
2,2d,2d
2
,2d
3
5,5d,5d
2
,5d
3
10,10d,10d
2
,10d
3
3.(e) 1,2,3,4,6,12 b,2b,3b,4b,6b,12b
4.(a) 4 (b) 4 (c) 1 (d) 5 (e) 2 (f) 9
5.(a) 3b
2
(b) 10c
2
(c) 2d
3
(d) 2x
4
(e) 7x
3
(f) 9g
10
(g) 7e
4
(h) k
5
(i) 4g
2
6.(a) 4y
3 x
5
(b) 8b
2 c
2
(c) y
3 x
3
(d) 7a
3 b
2
(e) x
4 y
4
(f) 3r
2 t
11
(g) 6c
7 d
6
(h) 4c
2 d
2
(i) 2c
2 d
2 e
Exercise 2:
1. C
2.(a) 8(x + 1) (b) 6(d + 2e) (c) 4x + 9 (d) 12(1 + 2y) (e) 12(c – 2)
2.(f) 5(2y – 3x) (g) 2(6x – 5b) (h) 5(3e – 4g) (i) 2(x – 8)4 (j) 5(1 – 5g)
(k) 2(–x + 2) or – 2(x – 2y) (l) 4(– x – 3d) or – 4(x + 3d)
3.(a) 2(2x + 6y + 3c) (b) 4(2b + c + 2d) (c) 2(21d + 6 – 4c)
3.(d) 4(3 + 2d – 3c) (e) 3(10c – 4d + 5e) (f) 5(5x – c – 3)
3.(g) 2(x – 9 + 12y) (h) 5(10 – 4g – 3h) (i) 3(2x + y + t)
3.(j) 2(8b – 10h – 5j) (k) 5(4t – 3 + 10p) (l) 3(2x – 1 + 3y)
4.(a) a(2b + 3c) (b) 3b(7 + d) (c) c(5 + t) (d) 11k(2 + 2g)
4.(e) 3c(10 – 3d) (f) 5x(4 – 3c) (g) 10x(2d – g) (h) 5t(3r – 4f)
4.(i) 3c(10x + y) (j) 3(5by + 4b – 3y) (k) 2a(2b – 3c + 6d) (l) 6x(2 + x)
5.(a) 4b
2
(1 – 2b) (b) 3ab(2 + 3b
2
– a) (c) 2cy
2
(1 + 10y)
5.(d) 2by(6y + 9by
2
– 1) (e) 4ax
3 y
2
(3 + 2a) (f) b
3 y(6y + y
2
– 9)
5.(g) 3xyz(10 + 9xyz) (h) 4x(x +3) (i) 5x 3 c(3 – 8c 3 )
5.(j) 4t(6t
3
+ t
2
+ 8) (k) 3c
4 d(1 – 3cd
4
) (l) 5x
2 c(1 – 2c
6
)
5.(m) 3cx(x – 6c) (n) 5mn(p
5
– 4 m
4 n
5
) (o) 3x
2 y
4
(1 + y
4
)
5.(p) 3(x + 1)(a – 2b) (q) 3a(x + y)(1 – 3a) (r) (x + 7)(x +5)
Page | 22

Olympic College Topic 12 Factorisation
Exercise 3
1.(a) (x + 5)(x – 5) (b) (x + 7)(x – 7) (c) (x + 10)(x – 10)
1.(d) (12 + b)(12 – b) (e) (x + b)(x – b ) (f) (2x + 9)(2x – 9)
1.(g) (3x + 2)(3x – 2) (h) (4a + 5b)(4a – 5b) (i) (1 + 2ab
2
)(1 – 2ab
2
)
1.(j) (3x + 2y)(3x – 2y) (k) (b
3
+ y
3
)( b
3
– y
3
) (l) (2x + 3b)(2x – 3b)
Exercise 4
1.(a) (x + 7)(x + 1) (b) (x – 5)(x – 2) (c) (x – 4)(x – 2) (d) (x + 3)(x + 2)
1.(e) (x + 3)(x + 1) (f) (x + 3)(x – 10) (g) (x + 10)(x – 8) (h) (x – 6)(x – 2)
1.(i) (x + 4)(x – 5) (j) (x + 9)(x – 7) (k) (x + 11)(x – 8) (l) (x + 3)(x – 7)
1.(m) (x + 10)(x – 15) (n) (b + 5)(b – 3) (o) (a – 2)(a – 8) (p) (x + 9)(x – 7)
1.(q) (x + 11)(x – 3) (r) (x + 3)(x – 7) (s) (x + 10)(x – 15) (t) (b + 3)(b + 5)
1.(u) (a – 8)(x – 2) (v) (x + 9)(x – 7) (s) (x + 11)(x – 8) (t) (x + 3)(x – 7)
Exercise 5:
1.(a) (2x + 1)(x + 7) (b) (7y + 12 )(y – 2) (c) (4x + 3)(4x + 3)
1.(d) (2x – 5)(x – 3) (e) (3p + 2)(p – 5) (f) (5x + 6)(x – 1)
1.(g) (5x + 8)(x – 1) (h) (2x – 1)(x + 2) (i) (2x – 1)(x – 5)
1.(j) (2x – 1)(x – 5) (k) (4x + 5)(4x + 5) (l) (4c – 9)(c – 1)
1.(m) (4x + 1)(x – 5) (n) (4b + 3)(4b + 5) (o) (3y – 1)(2y – 9)
1.(p) (3x + 1)(x – 9) (q) (5x + 2) 2 (r) (3d + 4) 2
2. C 3. B
Exercise 6
1.(a) – (x – 3)(x – 2) (b) – (2x + 19)(x + 1) (c) – (4x + 13)(x – 3)
2. (a) (x + 25y)(x + y) (b) (x + 6y)(x + 2y) (c) (x – 2y)(x – 7y)
2.(d) (x + 5y)(x – 2y) (e) (x + y)(x – 4y) (f) (x + 2y)(x – 6y)
3.(a) 4(x + 5)(x – 5) (b) 2(x + 5y)(x – 5y) (c) 11(x + 3y)(x – 3y)
3.(d) 2(1 =5d)(1 – 5d) (e) 4(x + 1)(x – 2) (f) 2a(x + 2)(x – 3)
3.(g) 6ab(x + 2)(x – 3) (h) 2a
2
(x + 3)
2
(i) ab(x – 5)(x – 1)
3.(j) 3xy 2 (1 + y)(1 – y) (k) 3a(2a + 5)(a – 2) (l) 2x 2 (2x – 1)(x + 3)
Page | 23