Olympic College Topic 12 Factorisation Topic 12 Factorisation 1. How to find the greatest common factors of an algebraic expression. Definition:A factor

Olympic College Topic 12 Factorisation

Topic 12 Factorisation

1. How to find the greatest common factors of an algebraic expression.

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 bothgroups are called the common factors.

For example: The factors of 12 are 1,2,3,4,6, and 12The factors of 20 are 1,2,4,5,10 and 20The common factors of 12 and 20 are 1,2 and 4

The Greatest Common Factor (GCF) of two numbers is the largest of the commonfactors. 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 bygoing 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 twonumbers are mutually prime or prime for short.

Example 1: Find the GCF of the following numbers.

(a) 20 and 30

(d) 4 and 9

Solution(a): Factors of 20 =Factors of 30 =

Solution(b): Factors of 12 =Factors of 24 =

Solution(c): Factors of 16 =Factors of 44 =

Solution(d): Factors of 4 =Factors of 9 =

Solution(e): Factors of 60 =Factors of 32 =

Solution(f): Factors of 8 =Factors of 12 =Factors of 24 =

(b) 12 and 24

(e) 60 and 32

1,2,4,5,10 and 201,2,3,5,6,10,15 and 30

1,2,3,4,6 and 121,2,3,4,6,8,12 and 24

1,2,4,8,and 161,2,4,11,22 and 44

1,2 and 41,3 and 9

1,2,3,4,5,6,10,12,15,20,30 and 601,2,4,8,16 and 32

1,2,4,and 81,2,3,4,6 and 121,2,3,4,6,8,12 and 24

(c) 16 and 44

(f) 8,12 and 24

GCF of 20 and 30 = 10

GCF of 12 and 24 = 12

GCF of 16 and 44 = 4

GCF of 4 and 9 = 1 (Prime)

GCF of 60 and 32 = 4

GCF of 8,12 and 24 = 4

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To find the factors of an exponential we simply write down all its powers starting at 1.

Example 2: Find all the factors of x4.

Solution: The factors of x4 are 1,x,x2 x3 and x4

Example 3: Find all the factors of y3.

Solution: The factors of y3 are 1,y,y2and y3

To find the GCF of two exponentials it will be the smaller of the two powers.

Example 3: Find the GCF of x3 and x7

Solution: The factors of x3 are 1,x,x2 and x3

The factors of x7 are 1,x,x2 x3,x4,x5,x6 and x7

(Notice that this is the smaller of the two powers)The GCF of x3 and x7 = x3

Example 4: Find the GCF of y5 and y6

Solution: The factors of y5 are 1,y,y2,y4 and y5

The factors of y6 are 1,y,y2 y3,y4,y5and y6

The GCF of y5 and y6 = y6 (Notice that this is the smaller of the two powers)

When the expression is a product of two or more exponentials then the factors arethe products of all the permutations of each exponential.

Example 5: Find the factors of x3y2

Solution: The factors of x3y2 are 1,x,x2, x3 y,xy,x2y,x3y y2,xy2,x2y2,x3y3

Example 6: Find the factors of b2d5

Solution: The factors of b2d5 are 1,b,b2

d3,bd3,b2d3

d,bd,b2dd4,bd4,b2d4

d2,bd2,b2d2

d5,bd5,b2d5

When the expression is a multiple of an exponential then its factors are all the permutations ofthe factors of the number with the powers of the exponential.

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Example 7: Find all the factors of 6x3

Solution: The factors of 6x3 are 1,x,x2, x3 and 2,2x,2x2,2x3 and 3,3x,3x2,3x3 and 6,6x,6x2,6x3

Example 8: Find all the factors of 4y2

Solution: The factors of 4y2 are 1,y,y2 and 2,2y,2y2 and 4,4y,4y2

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 thefactors of each expression and then finding the largest result that they both share. This method isvery time consuming as the following worked example shows.

Example 9: Find the GCF of 12x3 and 16x2

Solution: The factors of 12x3are

The factors of 16x2 are

1,x,x2, x3

4,4x,4x2,4x3

1,x,x2

2,2x,2x2,2x3

6,6x,6x2,6x3

2,2x,2x2

3,3x,3x2,3x3

12,12x,12x2,12x3

4,4x,4x2

8,8x,8x2 16,16x,16x2

The common factors are 1,x,x2 2,2x,2x2 4,4x,4x2

The GCF of 12x3 and 16x2 is 4x2

The second method for finding the GCF of two expressions is to find the GCF of the constantsand the exponentials separately and then combine the result.

Example 10: Find the GCF of 12x3 and 16x2

Solution: The GCF of 12 and 16 is 4(The smaller of the two powers)The GCF of x2 and x3 is x2

The GCF of 12x2 and 16x3 is 4x2

Example 11: Find the GCF of 10y4x2 and 25y3x3

Solution: The GCF of 10 and 25 is 5The GCF of x2 and x3 is x2

The GCF of y4 and y3 is y3

(The smaller of the two powers)(The smaller of the two powers)

The GCF of 10y4x2 and 25y3x3 is 5x2y3

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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) x6 (b) x4 (c) b3 x2 (d) c2 d2 (e) a5 b

3. Find all the factors of the following expressions.

(a) 6x3 (b) 4b4 (c) 20x2 (d) 10d3 (e) 12b

4. Find the greatest Common Factors GCF for each of the following.

(a) 12 and 20

(d) 100 and 45

(b) 16 and 60

(e) 16,24 and 30

(c) 43 and 57

(f) 18,36 and 9


(a) 6b3 and 15b2

(d) 12x4 and 14x5

(g) 14e4 and 21e8

(b) 30c2 and 20c6

(e) 7x3 and 14x4

(h) 2k5 and 3k5

(c) 2d3 and 6d3

(f) 18g10 and 81g22

(i) 8g2 and 36g12


(a) 4y3x5 and 20y6x6

(d) 14a3b2 and 21a3b2

(g) 6c7d9 and 12c7d6

(b) 8b4c2 and 40b2c4

(e) 5x4y9 and 9x9y4

(h) 4c2d2 and 40c3d2

(c) y5x5 and 2y3x3

(f) 9r2t11 and 21r12t12

(i) 10c4d2e and 8c2d2e4

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2. Factoring by taking out a Common factors.

When we expand parenthesis we can use the distributive law so that 5(2x – 7) = 10x – 35Factorising 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 expressionIn 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 youdivide 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 2xThe second term is – 35 and when you divide it by the common factor 5 you get – 7These two results are now put in the parenthesis to complete the process.

= 5(2x – 7)So

Example 1: Factor

10x – 35

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 termsare prime so it is not possible to factor this expression except in a very trivial wayas shown below so we typically just say Prime and leave the expression alone

= 1(8x + 7y + 14z)8x + 7y + 14z

4t3 – 6t

Example 4: Factor

Solution: The two terms of this expression are 4t3 and – 6t the common factors are 2 and t socombined the common factor is 2t.

So 4t3 – 6t = 2t(2t2 – 3)


Example 5: Factor 40x2 – 25x

Solution: The two terms of this expression are 40x2 and – 25x the common factors are 5 andx so combined the common factor is 5x.

So 40x2 – 25x = 5x(8x – 5)

Example 6: Factor the expression 3x3y2 + 9x2y3

Solution: The 2 terms of this expression are 3x3y2 and 9x2y3

The common factors are 3, x2 and y2 so combined the common factor is 3x2y2.

So 3x3y2 + 9x2y3 = 3x2y2(x +3y)

Example 7: Factor the expression 3ab2 + 6a2b2 – 12a3b3

Solution: The 3 terms of this expression are 3ab2 , 6a2b2 and – 12a3b3

The common factors are 3 , a and b2 so combined the common factor is 3ab2.

So 3ab2 + 6a2b2 – 12a3b3 = 3ab2(1 + 2a – 4ab)

Example 8: Factor the expression 4x2y – 8xy + 10x

Solution: The 3 terms of this expression are 4x2y , – 8xy and 10x

The common factors are 2 , and x so combined the common factor is 2x.

So 4x2y – 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 GCFof 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 justconsider it a single object that both terms contain.

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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 aGCF 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 justconsider 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 toexpand the expression to make sure it is equal to the one given at the start.

So in Example 8 we were asked to factor 4x2y – 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)

4x2y – 8xy + 10x

This is the same expression so we have shown that our factoring part of the process was donecorrectly.

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 8x3y2 + 12x4y3

Lets assume that we wrongly thought that the GCF was 2x3y in which case we would get as ourfactoring

8x3y2 + 12x4y3 = 2x3y(4y + 6xy)

We can tell this would not be the full factoring by looking at what is left in the parenthesis inthis case we have 4y + 6xy and since this expression can still be factored more as it has acommon factor of 2y we know we have factored the original expression wrongly and we shouldhave used the GCF of 4x3y2 this would then give us \the correct factorization is

8x3y2 + 12x4y3 = 4x3y2(2 + 3x)


Exercise 2:

1. Find the GCF of the terms of the polynomial 8x6 + 12x3

A. x3 B. 2x3 C. 4x3 D 8x6

2. Factor each of the following expressions.

(a) 8x + 8

(e) 12c – 24

(i) 6x – 16

(b) 6d + 12e

(f) 10y – 15x

(j) 5 – 25g

(c) 4x + 9

(g) 12x – 10b

(k) – 2x + 4y

(d) 12 + 24y

(h) 15e – 20g

(l) – 4x – 12d


(a) 4x + 12y + 6c

(d) 12 + 8d – 12c

(g) 2x – 18 + 24y

(j) 16b – 20h – 10j

(b) 8b + 4c + 8d

(e) 30c – 12d + 15e

(h) 50 – 20g – 15h

(k) 20t – 15 + 50p

(c) 42d + 12 – 8c

(f) 25x – 5c – 15

(i) 6x + 3y + 3t

(l) 6x – 3 + 9y


(a) 2ab + 3ac

(d) 22k + 11gk

(g) 20xd – 10xg

(j) 15by + 12b – 9y

(b) 21b + 3bd

(e) 30c – 9cd

(h) 15rt – 20ft

(k) 4ab – 6ac + 12ad

(c) 5c + tc

(f) 20x – 15xc

(i) 30xc + 3yc

(l) 12x + 6x2


(a) 4b4 – 8b3

(d) 12by2 + 18b2y3 – 2by

(b) 6ab + 9ab3 – 3a2b

(e) 12ax3y2 + 8a2x3 y2

(c) 2cy2 + 20c2y3

(f) 6b3y2 + b3y3 – 9b3y

(g) 30xyz + 9x2y2z2 (h) 4x2 +12x (i) 15x3c – 40x3c4

(j) 24t4 + 4t3 + 32t

(m) 3cx2 – 18c2x

(p) 3a(x + 1) – 6b(x + 1)

(k) 3c4d – 9c5d5

(n) 5mnp5 – 20m5n6

(q) 3a(x + y) – 9a2(x + y)

(l) 5x4c – 10x2c7

(o) 3x2y4 + 3x2y6

(r) (x + 7)2 – 2(x + 7)

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3. Factoring the Difference of Two Squares

When you look at the following expansions you will notice a pattern.

(x +3)(x – 3) = x2 – 3x + 3x – 9 = x2 – 9

(b +10)(b – 10) = b2 – 10b + 10b – 100 = b2 – 100

(x +3)(x – 3)

(a +b)(a – b)

=

=

x2 – 3x + 3x – 9

a2 – ab + ab – b2

=

=

x2 – 9

a2 – b2

The general pattern gives us a way to factor the difference of two squares by simply finding thevalues of a and b.

Definition: A difference of two squares is an expression that contain two terms that aresubtracted. 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.

a2 – b2 = (a + b)(a – b)Example 1: Factor the expression t2 – 4

Solution: The 2 terms of this expression are t2 and 4, their square roots are t and 2.

So t2 – 4 = (t + 2)(t – 2)

Example 2: Factor the expression x2 – 36

Solution: The 2 terms of this expression are x2 and 36, their square roots are x and 6.

So x2 – 36 = (x + 6)(x – 6)

Example 3: Factor the expression 25x2 – 16

Solution: The 2 terms of this expression are 25x2 and 16, their square roots are 5x and 4.

So 25x2 – 16 = (5x + 4)(5x – 4)

Example 4: Factor the expression 4y6 – 9a4

Solution: The 2 terms of this expression are 4y6 and 9a4, their square roots are 2y3 and 3a2.

So 4y2 – 9a2 = (2y3 + 3a2)(2y3 – 3a2)

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Exercise 3

1. Factor the following expressions.

(a) x2 – 25

(d) 144 – b2

(g) 9x2 – 4

(j) 9x2 – 4y2

(b) x2 – 49

(e) x2 – b2

(h) 16a2 – 25b2

(k) b6 – y6

(c) b2 – 100

(f) 4x2 – 81

(i) 1 – 4a2b4

(l) 4x2 – 9b2


4. Factoring trinomials of the form x2 + bx +c

These expressions contain three terms (hence the name trinomial).

The most common type are simple trinomials – these are expressions where there is only an x2

term and have the following general form x2 + bx + c.

When you factor x2 + 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 constantterm c and whose sum equals the coefficient of x called b.

There are 4 basic situations in factoring x2 + bx + c

b and c are both positiveb is negative and c is positiveb is positive and c is negativeb is negative and c is negative

The first situation is when a and c are both positive. In this situation the two factors of c will bothbe positive.

Example 1: Factor the expression x2 + 5x + 6

Solution : In this situation we are looking for two factors of 6 that add up to 5. Since both band c are positive we need only look at positive values of the factors of 6.The Positive factors of 6 are:-

+1 and + 6+2 and + 3

Add to 7 so not the correct combinationAdd to 5 so this is correct combination

x2 + 5x + 6 = (x + 2)(x + 3)Using these two factors of 6 we get

Example 2: Factor the expression x2 + 7x + 10

Solution : In this situation we are looking for two factors of 10 that add up to 7. Since both band c are positive we need only look at positive values of the factors of 10.The Positive factors of 10 are:-

+1 and + 10+2 and + 5

Add to 11 so this is not the correct combinationAdd to 7 so this is correct combination

Using these two factors of 10 we get x2 + 7x + 10 = (x + 2)(x + 5)

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The second situation is when b is negative and c is positive. In this situation the two factors willboth be negative.

Example 3: Factor the expression x2 – 6x + 8

Solution : In this situation we are looking for two factors of 8 that add up to – 6 . Since b isnegative and c are positive we need only look at the negative factors of 8.The negative factors of 8 are:-

– 1 and – 8– 2 and – 4

Add to – 9 so this is not the correct combinationAdd to – 6 so this is correct combination

Using these two factors of 8 we get x2 – 6x + 8 = (x – 2)(x – 4)

Example 4: Factor the expression x2 – 7x + 12

Solution : In this situation we are looking for two factors of 12 that add up to – 7 . Since b isnegative and c are positive we need only look at the negative factors of 12.The negative factors of 12 are:-

– 1 and – 12– 2 and – 6– 3 and – 4

Add to – 9 so this is not the correct combinationAdd to – 8 so this is not the correct combinationAdd to – 7 so this is not the correct combination

Using these two factors of 12 we get x2 – 7x + 12= (x – 3)(x – 4)

Example 5: Factor the expression b2 – 20b + 100

Solution : In this situation we are looking for two factors of 100 that add up to – 20 . Since bis negative and c are positive we need only look at the negative factors of 100.The negative factors of 100 are:-

– 1 and – 100– 2 and – 50– 4 and – 25– 5 and – 20– 10 and – 10

Add to – 101 so this is not the correct combinationAdd to – 52 so this is not the correct combinationAdd to – 29 so this is not the correct combinationAdd to – 25 so this is not the correct combinationAdd to – 20 so this is not the correct combination

Using these two factors of 100 we get b2 – 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 uniqueonce you find a combination that works you can stop.

Also with practice you will get an insight into which combinations are more likely tosucceed and so you will try these ones first and hopefully reduce the number of failedcombinations to a minimum.

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The third situation is when b is positive and c is negative. In this situation the larger factor willbe positive and the smaller factor will be negative.

Example 6: Factor the expression x2 + 2x – 15

Solution : In this situation we are looking for factors of – 15 that add up to 2. Since the largefactor is positive and the smaller is factor is negative we only need to look at thefollowing combinations.

Factors of – 15 are:-

15 and – 15 and – 3

Add to 14 so this is not the correct combinationAdd to 2 so this is the correct combination

x2 + 2x – 15 = (x + 5)(x – 3)Using these two factors of – 15 we get

Example 7: Factor the expression x2 + 10x – 24



24 and – 112 and – 2

Add to 23 so this is not the correct combinationAdd to 10 so this is the correct combination

Using these two factors of – 24 we get x2 + 10x – 24 = (x + 12)(x – 2)

Example 8: Factor the expression y2 + 13y – 60



60 and – 130 and – 215 and – 2

Add to 59 so this is not the correct combinationAdd to 28 so this is not the correct combinationAdd to 13 so this is the correct combination

Using these two factors of – 60 we get y2 + 13y – 60 = (y + 15)(y – 2)

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The fourth situation is when b is negative and c is positive. In this situation the larger factor willbe negative and the smaller factor will be positive.

Example 9: Factor the expression y2 – y – 12

Solution: In this situation we are looking for factors of– 12 that add up to – 1. Since the largefactor is negative and the smaller is factor is positive we only need to look at thefollowing combinations.


1 and – 122 and – 63 and – 4

Add to – 11 so this is not the correct combinationAdd to – 4 so this is not the correct combinationAdd to – 1 so this is the correct combination

y2 – y – 12 = (y + 3)(y – 4)Using these two factors of – 12 we get

Example 10: Factor the expression k2 – 3k – 40

Solution: In this situation we are looking for factors of– 40 that add up to – 3. Since the largefactor is negative and the smaller is factor is positive we only need to look at thefollowing combinations.


1245

and – 40and – 20and – 10and – 8

Add to – 41 so this is not the correct combinationAdd to – 18 so this is not the correct combinationAdd to – 6 so this is not the correct combinationAdd to – 3 so this is not the correct combination

k2 – 3k – 40 = (k + 5)(k – 8)Using these two factors of – 40 we get

Exercise 4

1.

Factor the following expressions.

(a) x2 + 8x + 7

(d) x2 + 5x + 6

(g) x2 + 2x – 80

(j) x2 + 2x – 63

(m) x2 – 5x – 150

(p) x2 + 2x – 63

(s) x2 – 5x – 150

(v) x2 + 2x – 63

(b) x2 – 7x + 10

(e) x2 + 4x + 3

(h) x2 – 8x + 12

(k) x2 + 3x – 88

(n) b2 + 2b – 15

(q) x2 + 3x – 88

(t) b2 + 8t + 15

(w) x2 + 3x – 88

(c) x2 – 6x + 8

(f) x2 – 7x – 30

(i) x2 – x – 20

(l) x2 – 4x – 21

(o) a2 – 10a + 16

(r) x2 – 4x – 21

(u) a2 – 10a + 16

(x) x2 – 4x – 21

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5. Factoring trinomials of the form ax2 + bx + c

A trinomial of the type ax2 + bx + c in which typically there is not single x2 term but somemultiple instead are normally the most difficult to factor as they do not fall neatly into categoriesas they did for situations like factor x2 + bx + c.

When you factor a complex trinomial of the form ax2 + bx + c there are 2 main stages these are.

1. Get the x term part of the factor

2. Get the constant term part of the factor.Example 1: Factor the expression 2x2 + 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 complextrinomial expression the first is look at the number of x2 the expression contains ,this will tell you the possible number of x's that go into the factor , so for exampleif the expression has 2x2 can only be formed by 2x times x. So the solution musttake the form on the other hand if it were 4x2 it could be one of two possiblesituations either(4x .....)(x ......) or (2x .....)(2x ......).

In this example we have 2x2 + 5x – 3 and we get.

2x2 + 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 valueis – 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 thefactors of – 3 the order is also important so we reverse each possible factor pair.

(2x + 1)(x – 3) this gives 2x2 – 5x – 3

This is the correct one.(2x – 1)(x + 3)

(2x + 3)(x – 1)

(2x – 3)(x + 1)

this gives 2x2 + 5x – 3

this gives 2x2 + x – 3

this gives 2x2 – x – 3

Only one combination actually works and this is our solution.

So 2x2 + 5x – 3 = (2x – 1)(x + 3)


Example 2: Factor the expression 5y2 – 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 complextrinomial expression the first is look at the number of y2 the expression contains ,this will tell you the possible number of y's that go into the factor.

So for example since 5y2 can only be formed by 5y times y. So the solution musttake the form 5y2 – 12y + 4 = (5y .....)(y ......)


The second stage in the process is to look at the constant term, in this case its valueis + 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 oneach one until we get the one that works. Notice that in the factors of 4 the order isalso important so we reverse each possible factor pair.

(5y + 4)(y + 1)

(5y + 2)(y + 2)

(5y – 1)(y – 4)

(5y – 4)(y – 1)

this gives 5y2 + 9y + 4

this gives 5y2 + 12y + 4

this gives 5y2 – 21y + 4

this gives 5y2 – 9y + 4

(5y – 2)(y – 2) this gives 5y2 – 12y + 4 This is the correct one.


So 5y2 – 12y + 4 = (5y – 2)(y – 2)

Notice that the first three choices were bound to give positive values for the y–termand so we could have saved ourselves some time by not bothering to check them asthey clearly cannot get us the – 12y term that we need.

Another concept that can reduce your workload is when you get a solution that isalmost correct one except for the sign such as when you tested

(5y + 2)(y + 2) = 5y2 + 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 combinationssimilar 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) = 5y2 – 12y + 4 the correct factorisation.

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Example 3: Factor the expression 4x2 – 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 complextrinomial expression the first is look at the number of x2 the expression contains ,this will tell you the possible number of x's that go into the factor.

So for example since 5x2 can only be formed by 4x times x or by 2x times 2x.

So the solution must take the one of these forms

4x2 – 17x – 15 = (4x .....)(x ......)

4x2 – 17x – 15 = (2x .....)(2x ......)


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 atthe constant term, in this case its value – 15 and try all its factor pairs in order untilwe find the combination that works. Since both terms have 2x we do not need toreverse 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 oneach one until we get the one that works.

(2x + 1)(2x – 15)(2x – 1)(2x + 15)(2x + 3)(2x – 5)(2x – 3)(2x + 5)

this gives 4x2 – 27x – 15this gives 4x2 + 27x – 15this gives 4x2 – 4x – 15this gives 4x2 + 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 oneach one until we get the one that works.

(4x + 1)(x – 15)(4x – 1)(2x + 15)(4x + 3)(x – 5)(4x – 3)(x + 5)

this gives 4x2 – 59x – 15this gives 4x2 + 59x – 15this gives 4x2 – 17x – 15 This is the correct one.this gives 4x2 + 17x – 15


So 4x2 – 17x – 15 = (4x + 3)(x – 5)

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Exercise 5:

1. Factor the following expressions.

(a) 2x2 + 15x + 7

(d) 2x2 – 11x + 15

(g) 5x² + 3x – 8

(j) 16x² – 40x + 25

(m) (4x + 1)(x – 5)

(p) 3x² – 26x – 9

(b) 7y2 – 2y – 24

(e) 3p2 – 13p + 10

(h) 2x² + 3x – 2

(k) 4x² + 40x + 25

(n) (4b + 3)(4b + 5)

(q) 25x² + 20x + 4

(c) 16x2 + 24x + 9

(f) 4x2 – 12x – 40

(i) 2x² – 11x + 5

(l) 4c² – 13c + 9

(o) (3y – 1)(2y – 9)

(r) 9d² + 24d + 16

2. Which of the following is the factorization of 3x2 + 7x – 6

A. (3x – 2)(x – 3) B. (x + 3)(3x + 2)

C. (3x – 2)(x + 3) D. (3x + 2)(x – 3)

3. Which of the following is the factorization of 40p2 – 13p – 36

A. (8p + 9)(5p + 4)

C. (8p – 9)(5p – 4)

B. ( 8p – 9)(5p + 4)

D. (8p + 9)(5p – 4)

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6. Factoring - Special Cases

There are 3 special cases of factoring we will deal with in this section.

Case 1: Trinomials with a negative x2 coefficient such as – x2 + 10x – 9

Case 2: Trinomials in x and y such as x2 + 6xy + 5y2

Case 3: Combined factoring where a common factor and another factoring method arecombined such as ax2 + 4ax + 3a or ax2 – 9a

Case 1: Trinomials with a negative x2 coefficient

In these situations we take out the negative as a common factor of – 1 and the resulting trinomialcan be solved in the usual way.

Example 1: Factor – x2 + 10x – 9

Solution: – x2 + 10x – 9 =

=

– (x2 – 10x + 9)

– (x – 9)(x – 1)

Take out – 1 as a common factor

Factoring x2 – 10x + 9

Example 2: Factor – 2x2 – 5x + 18

Solution: – 2x2 – 5x + 18 =

=

– (2x2 + 5x – 9)

– (2x – 9)(x – 1)

Take out – 1 as a common factor

Factoring 2x2 + 5x – 9

Case 2: Trinomials in x and y

In these situations there will be a trinomial in x and y such as x2 + 6xy + 5y2 in order to factorthese we use a method similar to those used for other trinomials.

Example 3: Factor x2 + 6xy + 5y2

Solution: in order to factor x2 + 6xy + 5y2 we need to look for factors of 5 that add up to 6 inthis case we get + 5 and + 1 this represents + 5y and + y and so we get the solution:

x2 + 6xy + 5y2 = (x + 5y)(x + y)

Example 4: Factor x2 + 3xy – 10y2

Solution: in order to factor x2 + 3xy – 10y2 we need to look for factors of0 – 10 that add up to3 in this case we get + 2 and – 5 this represents + 2y and – 5 and so we get thesolution:

x2 + 3xy – 10y2 = (x + 2y)(x + – 5y)

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Case 3: Combined Factoring

In this situation we combine two or more of the methods where the first process involves takingout a common factor followed by one ot the processes described earlier.

Example 1: Factor the expression 5x2 – 180

Solution: First we take a common factor of 5 from the two terms then we factor thedifference of two squares that remains.

Take out a common factor of 5

Factor the difference of two squares

So 5x2 – 180

5(x2 – 36)

5x2 – 180

=

=

=

5(x2 – 36)

5(x + 6)(x – 6)

5(x + 6)(x – 6)

Example 2: Factor the expression 2ax2 + 12ax + 18a

Solution: First we take a common factor of 2a from the 3 terms then we factor what remains.

2ax2 + 12ax + 18a Take out a C.F. of 2a

Factor the trinomial x2 + 6x + 9

=

=

=

2a(x2 + 6x + 9)

2a(x + 3)(x + 3)

2a(x + 3)2

Example 3: Factor the expression 2ab4x2 + 8ab4x + 6ab4

Solution: First we take a common factor of 2ab4 from the 3 terms then we factor whatremains.

2ab4x2 + 8ab4x + 6ab4=

=

2a b4 (x2 + 4x + 3)

2a b4 (x + 3)(x + 1)

Take out a C.F. of 2ab4

Factor the trinomial x2 + 4x + 3

Example 4: Factor the expression x4 + 10x3 – 11x2

Solution: First we take a common factor of x2 from the 3 terms then we factor what remains.

x4 + 10x3 – 11x2 =

=

x2 (x2 + 10x – 11)

2a(x + 11)(x – 1)

Take out a C.F. of x2

Factor the trinomial x2 + 10x – 11


Exercise 6

1. Factor the following.

2.

3.

(a) – x2 + 5x – 6

Factor the following.

(a) x2 + 26xy + 25y2

(d) x2 + 3xy – 10y2

Factor the following.

(a) 4x2 – 100

(d) 2 – 50d2

(g) 6abx2 – 6abx – 36ab

(j) 3xy2 – 48xy4

(b) – 2x2 – 17x + 19

(b) x2 + 8xy + 12y2

(e) x2 – 3xy – 4y2

(b) 2x2 – 50y2

(e) 4x2 – 4x – 8

(h) 2a2 x2 + 12a2 x + 18a2

(k) 6a3 + 3a2 – 30a

(c) – 4x2 – x + 39

(c) x2 – 9xy + 14y2

(f) x2 – 4xy – 12y2

(c) 11x2 – 99c2

(f) 2ax2 – 2ax – 12a

(i) abx2 – 6abx + 5ab

(l) 4x4 + 10x3 – 6x2

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c , c2d, c2d22.(d) 1,d,d (e) 1,a,a ,a ,a ,a b,ba,ba ,ba ,ba ,ba5

(k) 3c d(1 – 3cd4)

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SolutionsExercise 1:

(c) 1,2,41,811.(a) 1,2,13,261.(d) 1,2,3,4,6,8,12,16,24,36,48,72,144

(b) 1,2,3,4,5,6,10,12,15,20,30,60(e) 1,2,4,5,10,20,25,50,100

2.(a) 1,x,x2,x3,x4,x5,x6 (b) 1,x,x2,x3,x4 (c) 1,x,x2 b,bx,bx2 b2,b2x,b2 x2 b3,b3x,b3 x22 c,cd,cd2 2 2 3 4 5 2 3 4

3.(a) 1,x,x2,x3 2,2x,2x2,2x3 3,3x,3x2,3x3 6,6x,6x2,6x3

3.(b) 1,b,b2,b3 ,b4 2,2b,2b2,2b3,2b4 4,4b,4b2,4b3,4b4

3.(c) 1,x,x2 2,2x,2x2 4,4x,4x2 5,5x,5x2 10,10x,10x2 20,20x,20x2

5,5d,5d2,5d3 10,10d,10d2,10d33.(d) 1,d,d2,d3

3.(e) 1,2,3,4,6,122,2d,2d2,2d3

b,2b,3b,4b,6b,12b

4.(a) 4 (b) 4 (c) 1 (d) 5 (e) 2 (f) 9

5.(a) 3b2 (b) 10c2 (c) 2d3 (d) 2x4 (e) 7x3 (f) 9g10 (g) 7e4 (h) k5 (i) 4g2

6.(a) 4y3x5 (b) 8b2c2 (c) y3x3 (d) 7a3b2 (e) x4y4 (f) 3r2t11 (g) 6c7d6 (h) 4c2d2 (i) 2c2d2e

Exercise 2:

1. C

2.(a) 8(x + 1)2.(f) 5(2y – 3x)

(b) 6(d + 2e)(g) 2(6x – 5b)

(c) 4x + 9(h) 5(3e – 4g)

(d) 12(1 + 2y)(i) 2(x – 8)4

(e) 12(c – 2)(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)3.(d) 4(3 + 2d – 3c)3.(g) 2(x – 9 + 12y)3.(j) 2(8b – 10h – 5j)

(b) 4(2b + c + 2d)(e) 3(10c – 4d + 5e)(h) 5(10 – 4g – 3h)(k) 5(4t – 3 + 10p)

(c) 2(21d + 6 – 4c)(f) 5(5x – c – 3)(i) 3(2x + y + t)(l) 3(2x – 1 + 3y)

4.(a) a(2b + 3c)4.(e) 3c(10 – 3d)4.(i) 3c(10x + y)

(b) 3b(7 + d)(f) 5x(4 – 3c)(j) 3(5by + 4b – 3y)

(c) c(5 + t)(g) 10x(2d – g)(k) 2a(2b – 3c + 6d)

(d) 11k(2 + 2g)(h) 5t(3r – 4f)(l) 6x(2 + x)

5.(a) 4b2(1 – 2b)5.(d) 2by(6y + 9by2 – 1)

(b) 3ab(2 + 3b2 – a)(e) 4ax3y2(3 + 2a)

(c) 2cy2(1 + 10y)(f) b3y(6y + y2 – 9)

5.(g) 3xyz(10 + 9xyz) (h) 4x(x +3) (i) 5x3c(3 – 8c3)5.(j) 4t(6t3 + t2 + 8)5.(m) 3cx(x – 6c)5.(p) 3(x + 1)(a – 2b)

4

(n) 5mn(p5 – 4 m4n5)(q) 3a(x + y)(1 – 3a)

(l) 5x2c(1 – 2c6)(o) 3x2y4(1 + y4)(r) (x + 7)(x +5)

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Exercise 3

1.(a) (x + 5)(x – 5)1.(d) (12 + b)(12 – b)1.(g) (3x + 2)(3x – 2)1.(j) (3x + 2y)(3x – 2y)

(b)(e)(h)(k)

(x + 7)(x – 7)(x + b)(x – b )

(4a + 5b)(4a – 5b)(b3 + y3)( b3 – y3)

(c) (x + 10)(x – 10)(f) (2x + 9)(2x – 9)(i) (1 + 2ab2 )(1 – 2ab2)(l) (2x + 3b)(2x – 3b)

Exercise 4

1.(a) (x + 7)(x + 1)1.(e) (x + 3)(x + 1)1.(i) (x + 4)(x – 5)1.(m) (x + 10)(x – 15)1.(q) (x + 11)(x – 3)1.(u) (a – 8)(x – 2)

(b) (x – 5)(x – 2)(f) (x + 3)(x – 10)(j) (x + 9)(x – 7)(n) (b + 5)(b – 3)(r) (x + 3)(x – 7)(v) (x + 9)(x – 7)

(c) (x – 4)(x – 2)(g) (x + 10)(x – 8)(k) (x + 11)(x – 8)(o) (a – 2)(a – 8)(s) (x + 10)(x – 15)(s) (x + 11)(x – 8)

(d) (x + 3)(x + 2)(h) (x – 6)(x – 2)(l) (x + 3)(x – 7)(p) (x + 9)(x – 7)(t) (b + 3)(b + 5)(t) (x + 3)(x – 7)

Exercise 5:

1.(a) (2x + 1)(x + 7)1.(d) (2x – 5)(x – 3)1.(g) (5x + 8)(x – 1)1.(j) (2x – 1)(x – 5)1.(m) (4x + 1)(x – 5)1.(p) (3x + 1)(x – 9)

2. C

3. B

Exercise 6

1.(a) – (x – 3)(x – 2)

2. (a) (x + 25y)(x + y)2.(d) (x + 5y)(x – 2y)

3.(a) 4(x + 5)(x – 5)3.(d) 2(1 =5d)(1 – 5d)3.(g) 6ab(x + 2)(x – 3)3.(j) 3xy2(1 + y)(1 – y)

(b) (7y + 12 )(y – 2)(e) (3p + 2)(p – 5)(h) (2x – 1)(x + 2)(k) (4x + 5)(4x + 5)(n) (4b + 3)(4b + 5)(q) (5x + 2)2

(b) – (2x + 19)(x + 1)

(b) (x + 6y)(x + 2y)(e) (x + y)(x – 4y)

(b) 2(x + 5y)(x – 5y)(e) 4(x + 1)(x – 2)(h) 2a2(x + 3)2

(k) 3a(2a + 5)(a – 2)

(c) (4x + 3)(4x + 3)(f) (5x + 6)(x – 1)(i) (2x – 1)(x – 5)(l) (4c – 9)(c – 1)(o) (3y – 1)(2y – 9)(r) (3d + 4)2

(c) – (4x + 13)(x – 3)

(c) (x – 2y)(x – 7y)(f) (x + 2y)(x – 6y)

(c) 11(x + 3y)(x – 3y)(f) 2a(x + 2)(x – 3)(i) ab(x – 5)(x – 1)(l) 2x2(2x – 1)(x + 3)

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Olympic College Topic 12 Factorisation Topic 12 Factorisation 1. How to find the greatest common factors of an algebraic expression. Definition:A factor