quadratic factorisation - haese mathematics · quadratic factorisation chapter 8 ... factorise by...

Quadraticfactorisation

8Chapter

Contents: A Factorisation by removal ofcommon factors

B Difference of two squaresfactorisation

C Perfect square factorisation

D Factorising expressions with fourterms

E Quadratic trinomial factorisation

F Miscellaneous factorisationG Factorisation of

ax2 + bx+ c; a 6= 1

IB MYP_4magentacyan yellow black

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Y:\HAESE\IB_MYP4\IB_MYP4_08\177IB_MYP4_08.CDR Thursday, 13 March 2008 12:17:03 PM PETERDELL

178 QUADRATIC FACTORISATION (Chapter 8)

A quadratic expression in x is an expression of the form ax2 + bx+ c where x is the

variable, and a, b and c are constants with a 6= 0:

ax2

the x2 term

+ bx

the x term

+ c

the constant term

For example: x2 + 5x + 6, 4x2 ¡ 9 and 9x2 + 6x + 1 are quadratic expressions.

Factorisation is the process of writing an expression as a product of factors.

For example:

Since x2 + 5x + 6 = (x + 2)(x + 3), we say that

(x + 2) and (x + 3) are factors of x2 + 5x + 6.

You should remember the following expansion rules from Chapter 3:

(x+ p)(x+ q) = x2 + (p + q)x+ pq sum and product expansion

(x+ a)2 = x2 + 2ax+ a2 perfect square expansion

(x+ a)(x¡ a) = x2 ¡ a2

These statements are called identities because they are true for all values of the variable x.

Notice that the RHS of each identity is a quadratic expression which has been formed by

expanding the LHS.

The LHS of the identities above can be obtained by factorising the RHS.

Some quadratic expressions can be factorised by removing the Highest Common Factor (HCF)

of the terms in the expression. In fact, we should always look to remove the HCF before

proceeding with any other factorisation.

FACTORISATION BY REMOVAL OFCOMMON FACTORS

A

(x+ 2) (x+ 3) = x2 + x+ 65

expansion

factorisation

In we studied the expansion of algebraic factors, many of which resulted inquadratic expressions. In this chapter we will consider , which is the reverseprocess of expansion. We will find later that factorisation is critical in the solution ofproblems that convert to quadratic equations.

Chapter 3

factorisation

difference of two squares expansion


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QUADRATIC FACTORISATION (Chapter 8) 179

Factorise by removing a common factor:

a 2x2 + 3x b ¡2x2 ¡ 6x

a 2x2 + 3x has HCF x

) 2x2 + 3x = x(2x + 3)

b ¡2x2 ¡ 6x has HCF ¡2x

) ¡2x2 ¡ 6x = ¡2x(x + 3)

Check your

factorisations by

expansion!

Notice the use of the

square brackets.

Fully factorise by removing a common factor:

a (x¡ 5)2 ¡ 2(x¡ 5) b (x + 2)2 + 2x + 4

a (x¡ 5)2 ¡ 2(x¡ 5)

= (x¡ 5)(x¡ 5) ¡ 2(x¡ 5) fHCF = (x¡ 5)g= (x¡ 5)[(x¡ 5) ¡ 2]

= (x¡ 5)(x¡ 7) fsimplifyinggb (x + 2)2 + 2x + 4

= (x + 2)(x + 2) + 2(x + 2) fHCF = (x + 2)g= (x + 2)[(x + 2) + 2]

= (x + 2)(x + 4)

EXERCISE 8A

1 Fully factorise by first removing a common factor:

a 3x2 + 5x b 2x2 ¡ 7x c 3x2 + 6x

d 4x2 ¡ 8x e ¡2x2 + 9x f ¡3x2 ¡ 15x

g ¡4x + 8x2 h ¡5x¡ 10x2 i 12x¡ 4x2

j x3 + x2 + x k 2x3 + 11x2 + 4x l ab + ac + ad

m ax2 + 2ax n ab2 + a2b o ax3 + ax2

2 Fully factorise by removing a common factor:

a (x + 2)2 ¡ 5(x + 2) b (x¡ 1)2 ¡ 3(x¡ 1) c (x + 1)2 + 2(x + 1)

d (x¡ 2)2 + 3x¡ 6 e x + 3 + (x + 3)2 f (x + 4)2 + 8 + 2x

g (x¡ 3)2 ¡ x + 3 h (x + 4)2 ¡ 2x¡ 8 i (x¡ 4)2 ¡ 5x + 20

j 3x + 6 + (x + 2)2 k (x + 1)3 + (x + 1)2 l (a + b)3 + a + b

m 2(x + 1)2 + x + 1 n 3(x¡ 2)2 ¡ (x¡ 2) o 4(a + b)2 ¡ 2a¡ 2b

Example 2 Self Tutor



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Y:\HAESE\IB_MYP4\IB_MYP4_08\179IB_MYP4_08.CDR Monday, 17 March 2008 3:29:38 PM PETERDELL


We know the expansion of (a + b)(a¡ b) is a2 ¡ b2.

Thus, the factorisation of a2 ¡ b2 is (a + b)(a¡ b):

The difference between

a2 and b2 is a2 ¡ b2

which is the differenceof two squares.

a2 ¡ b2 = (a+ b)(a¡ b)

Note: The sum of two squares does not factorise into two real linear factors.

Use the rule a2 ¡ b2 = (a + b)(a¡ b) to factorise fully:

a 9 ¡ x2 b 4x2 ¡ 25

a 9 ¡ x2

= 32 ¡ x2 fdifference of squaresg= (3 + x)(3 ¡ x)

b 4x2 ¡ 25

= (2x)2 ¡ 52 fdifference of squaresg= (2x + 5)(2x¡ 5)

Fully factorise: a 2x2 ¡ 8 b ¡3x2 + 48

a 2x2 ¡ 8

= 2(x2 ¡ 4)

= 2(x2 ¡ 22)

= 2(x + 2)(x¡ 2)

fHCF is 2gfdifference of squaresg

b ¡3x2 + 48

= ¡3(x2 ¡ 16)

= ¡3(x2 ¡ 42)

= ¡3(x + 4)(x¡ 4)

fHCF is ¡ 3gfdifference of squaresg

DIFFERENCE OF TWO SQUARESFACTORISATION

B



Always look to remove

a common factor first.


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Y:\HAESE\IB_MYP4\IB_MYP4_08\180IB_MYP4_08.CDR Wednesday, 5 March 2008 2:50:37 PM PETERDELL


Even though 7 is not a perfect square, we can still factorise x2 ¡ 7 by writing 7 = (p

7)2:

So, x2 ¡ 7 = x2 ¡ (p

7)2 = (x +p

7)(x¡p7):

We say that x +p

7 and x¡p7 are the linear factors of x2 ¡ 7.

Factorise into linear factors: a x2 ¡ 11 b (x + 3)2 ¡ 5

a x2 ¡ 11

= x2 ¡ (p

11)2

= (x +p

11)(x¡p

11)

b (x + 3)2 ¡ 5

= (x + 3)2 ¡ (p

5)2

= [(x + 3) +p

5][(x + 3) ¡p

5]

= [x + 3 +p

5][x + 3 ¡p

5]

Factorise using the difference between two squares:

a (3x + 2)2 ¡ 9 b (x + 2)2 ¡ (x¡ 1)2

a (3x + 2)2 ¡ 9

= (3x + 2)2 ¡ 32

= [(3x + 2) + 3][(3x + 2) ¡ 3]

= [3x + 5][3x¡ 1]

b (x + 2)2 ¡ (x¡ 1)2

= [(x + 2) + (x¡ 1)][(x + 2) ¡ (x¡ 1)]

= [x + 2 + x¡ 1][x + 2 ¡ x + 1]

= [2x + 1][3]

= 3(2x + 1)

EXERCISE 8B

1 Use the rule a2 ¡ b2 = (a + b)(a¡ b) to fully factorise:

a x2 ¡ 4 b 4 ¡ x2 c x2 ¡ 81 d 25 ¡ x2

e 4x2 ¡ 1 f 9x2 ¡ 16 g 4x2 ¡ 9 h 36 ¡ 49x2

2 Fully factorise:

a 3x2 ¡ 27 b ¡2x2 + 8 c 3x2 ¡ 75

d ¡5x2 + 5 e 8x2 ¡ 18 f ¡27x2 + 75

3 If possible, factorise into linear factors:

a x2 ¡ 3 b x2 + 4 c x2 ¡ 15

d 3x2 ¡ 15 e (x + 1)2 ¡ 6 f (x + 2)2 + 6

g (x¡ 2)2 ¡ 7 h (x + 3)2 ¡ 17 i (x¡ 4)2 + 9



We notice that x2¡9 is the difference of two squares and therefore we can factorise it using

a2 ¡ b2 = (a + b)(a¡ b).


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4

a (x + 1)2 ¡ 4 b (2x + 1)2 ¡ 9 c (1 ¡ x)2 ¡ 16

d (x + 3)2 ¡ 4x2 e 4x2 ¡ (x + 2)2 f 9x2 ¡ (3 ¡ x)2

g (2x + 1)2 ¡ (x¡ 2)2 h (3x¡ 1)2 ¡ (x + 1)2 i 4x2 ¡ (2x + 3)2

We know the expansion of (x + a)2 is x2 + 2ax + a2,

so the factorisation of x2 + 2ax + a2 is (x + a)2. (x + a)2 and

(x ¡ a)2 are perfect

squares!x2 + 2ax+ a2 = (x+ a)2

Notice that (x¡ a)2

= (x + (¡a))2

= x2 + 2(¡a)x + (¡a)2

= x2 ¡ 2ax + a2

So, x2 ¡ 2ax+ a2 = (x¡ a)2

Use perfect square rules to fully factorise:

a x2 + 10x + 25 b x2 ¡ 14x + 49

a x2 + 10x + 25

= x2 + 2 £ x£ 5 + 52

= (x + 5)2

b x2 ¡ 14x + 49

= x2 ¡ 2 £ x£ 7 + 72

= (x¡ 7)2

Fully factorise:

a 9x2 ¡ 6x + 1 b ¡8x2 ¡ 24x¡ 18

a 9x2 ¡ 6x + 1

= (3x)2 ¡ 2 £ 3x£ 1 + 12

= (3x¡ 1)2

b ¡8x2 ¡ 24x¡ 18

= ¡2(4x2 + 12x + 9) fHCF = ¡2g= ¡2([2x]2 + 2 £ 2x£ 3 + 32)

= ¡2(2x + 3)2

PERFECT SQUARE FACTORISATIONC



Factorise using the difference of two squares:


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EXERCISE 8C

1 Use perfect square rules to fully factorise:

a x2 + 6x + 9 b x2 + 8x + 16 c x2 ¡ 6x + 9

d x2 ¡ 8x + 16 e x2 + 2x + 1 f x2 ¡ 10x + 25

g y2 + 18y + 81 h m2 ¡ 20m + 100 i t2 + 12t + 36

2 Fully factorise:

a 9x2 + 6x + 1 b 4x2 ¡ 4x + 1 c 9x2 + 12x + 4

d 25x2 ¡ 10x + 1 e 16x2 + 24x + 9 f 25x2 ¡ 20x + 4

g ¡x2 + 2x¡ 1 h ¡2x2 ¡ 8x¡ 8 i ¡3x2 ¡ 30x¡ 75

Sometimes we can factorise an expression containing four terms by grouping them in two

pairs.

For example, ax2 + 2x + 2 + ax can be rewritten as

ax2 + ax| {z } + 2x + 2| {z }= ax(x + 1)+ 2(x + 1) ffactorising each pairg= (x + 1)(ax + 2) f(x + 1) is a common factorg

Fully factorise: a ax + by + bx + ay b 2x2 ¡ 15 + 3x¡ 10x

a ax + by + bx + ay

= ax + ay| {z } + bx + by| {z }= a(x + y)+ b(x + y)

= (x + y)(a + b)

fputting terms containing a togethergffactorising each pairgf(x + y) is a common factorg

b 2x2 ¡ 15 + 3x¡ 10x

= 2x2 ¡ 10x| {z } + 3x¡ 15| {z }= 2x(x¡ 5) + 3(x¡ 5)

= (x¡ 5)(2x + 3)

fsplitting into two pairsgffactorising each pairgf(x¡ 5) is a common factorg

FACTORISING EXPRESSIONS WITHFOUR TERMS

D



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EXERCISE 8D

1 Fully factorise:

a bx + cx + by + cy b 2px + 3q + 2qx + 3p c 6ax + 3bx + 2b + 4a

d am¡ bn¡ an + bm e 3dr + r ¡ 3ds¡ s f 2ac¡ 5a + 2bc¡ 5b

g x2 + 5x + 7x + 35 h x2 ¡ 2x¡ 6x + 12 i x2 + 3x + 9 + 3x

j x2 + 8 + 8x + x k 2x2 + 3x + 3 + 2x l 3x2 + x¡ 3x¡ 1

m 2x2 + 3x + 10x + 15 n 6x2 ¡ 3x¡ 2 + 4x o 4x2 + x + 8x + 2

p 6x2 + 3x + 10x + 5 q 6x2 ¡ 4x + 9x¡ 6 r 4x2 + x¡ 8x¡ 2

s 3x2 + 4x + 33x + 44 t 18x2 + 3x¡ 12x¡ 2 u 10x2 + 4x¡ 35x¡ 14

2 Fully factorise:

a x2 ¡ 2x + 1 ¡ a2 b x2 ¡ a2 + x + a c b2 ¡ x2 ¡ 4x¡ 4

d c2 ¡ x2 + 6x¡ 9 e x2 ¡ y2 + y ¡ x f a2 + 2ab + a2 ¡ 4b2

g x2 + 4x + 4 ¡m2 h x2 + 2ax + a2 ¡ b2 i x2 ¡ y2 ¡ 3x¡ 3y

A quadratic trinomial is an expression of the form ax2 + bx+ c where x is a variable

and a, b, c are constants, a 6= 0.

For example: x2 + 7x + 6 and 3x2 ¡ 13x¡ 10 are both quadratic trinomials.

Consider the expansion of the product (x + 1)(x + 6):

(x + 1)(x + 6) = x2 + 6x + x + 1 £ 6 fusing FOILg= x2 + [6 + 1]x + [1 £ 6]

= x2 + [sum of 1 and 6]x + [product of 1 and 6]

= x2 + 7x + 6

More generally, (x + p)(x + q) = x2 + qx + px + pq

= x2 + (p + q)x + pq

and so x2 + (p + q)x+ pq = (x+ p)(x+ q)

the coefficient of x is

the sum of p and qthe constant term is the

product of p and q

So, if we are asked to factorise x2 + 7x + 6, we need to look for two numbers with a

product of 6 and a sum of 7. These numbers are 1 and 6, and so x2+7x+6 = (x+1)(x+6):

We call this the sum and product method.

QUADRATIC TRINOMIALEFACTORISATION


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Use the sum and product method to fully factorise:

a x2 + 5x + 4 b x2 ¡ x¡ 12

a x2 + 5x + 4 has p + q = 5 and pq = 4:

) p and q are 1 and 4:

) x2 + 5x + 4 = (x + 1)(x + 4)

b x2 ¡ x¡ 12 has p + q = ¡1 and pq = ¡12:

) p and q are ¡4 and 3:

) x2 ¡ x¡ 12 = (x¡ 4)(x + 3)

Fully factorise by first removing a common factor:

a 3x2 ¡ 9x + 6 b ¡2x2 + 2x + 12

a 3x2 ¡ 9x + 6

= 3(x2 ¡ 3x + 2)

= 3(x¡ 2)(x¡ 1)

fremoving 3 as a common factorgfsum = ¡3 and product = 2

) the numbers are ¡2 and ¡1gb ¡2x2 + 2x + 12

= ¡2(x2 ¡ x¡ 6)

= ¡2(x¡ 3)(x + 2)

fremoving ¡2 as a common factorgfsum = ¡1 and product = ¡6

) the numbers are ¡3 and 2g

EXERCISE 8E

1 Use the x2 + (p + q)x + pq = (x + p)(x + q) factorisation to fully factorise:

a x2 + 3x + 2 b x2 + 5x + 6 c x2 ¡ x¡ 6

d x2 + 3x¡ 10 e x2 + 4x¡ 21 f x2 + 8x + 16

g x2 ¡ 14x + 49 h x2 + 3x¡ 28 i x2 + 7x + 10

j x2 ¡ 11x + 24 k x2 + 15x + 44 l x2 + x¡ 42

m x2 ¡ x¡ 56 n x2 ¡ 18x + 81 o x2 ¡ 4x¡ 32

2 Fully factorise by first removing a common factor:

a 2x2 ¡ 6x¡ 8 b 3x2 + 9x¡ 12 c 5x2 + 10x¡ 15

d 4x2 + 4x¡ 80 e 2x2 ¡ 4x¡ 30 f 3x2 + 12x¡ 63

g ¡2x2 + 2x + 40 h ¡3x2 + 12x¡ 12 i ¡7x2 ¡ 21x + 28

j ¡x2 ¡ 3x¡ 2 k ¡x2 + 5x¡ 6 l ¡x2 + 9x¡ 18

m 5x2 + 15x¡ 50 n ¡2x2 ¡ 8x + 42 o 4x¡ x2 + 32


Example 10 Self TutorThe sum of thenumbers is the

coefficient of Theproduct of the numbers

is the constant term.

x:


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Use the following steps in order to factorise quadratic expressions:

Step 1: Look carefully at the quadratic expression to be factorised.

Step 2: If there is a common factor, take it out.

Step 3: Look for a perfect square factorisation: x2 + 2ax + a2 = (x + a)2

or x2 ¡ 2ax + a2 = (x¡ a)2

Step 4: Look for the difference of two squares: x2 ¡ a2 = (x + a)(x¡ a)

Step 5: Look for the sum and product type: x2 + (p + q)x + pq

= (x + p)(x + q)

EXERCISE 8F

1 Where possible, fully factorise the following expressions:

a 3x2 + 9x b 4x2 ¡ 1 c 5x2 ¡ 15

d 3x¡ 5x2 e x2 + 3x¡ 40 f 2x2 ¡ 32

g x2 + 9 h x2 + 10x + 25 i x2 ¡ x¡ 6

j x2 ¡ 16x + 39 k x2 ¡ 7x¡ 60 l x2 ¡ 2x¡ 8

m x2 + 11x + 30 n x2 + 6x¡ 16 o x2 ¡ 5x¡ 24

p 3x2 + 6x¡ 72 q 4x2 ¡ 8x¡ 60 r 3x2 ¡ 42x + 99

s ¡x2 + 9x¡ 14 t ¡x2 ¡ 13x¡ 36 u ¡2x2 ¡ 14x + 36

In the previous section we revised techniques for factorising quadratic expressions in the form

ax2 + bx + c where:

² a = 1For example: x2 + 5x + 6

= (x + 3)(x + 2)

² a was a common factor

For example: 2x2 + 10x + 12

= 2(x2 + 5x + 6)

= 2(x + 3)(x + 2)

² we had a perfect square or difference of two squares type

For example: 4x2 ¡ 9 = (2x)2 ¡ 32

= (2x + 3)(2x¡ 3)

Factorising a quadratic expression such as 3x2 + 11x+ 6 appears to be more complicated

because it does not fall into any of these categories.

MISCELLANEOUS FACTORISATIONF

G FACTORISATION OF ax bx c; a2 + + =16


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We need to develop a method for factorising this type of quadratic expression.

Two methods for factorising ax2 + bx + c where a 6= 1 are commonly used:

² trial and error ² ‘splitting’ the x-term

FACTORISATION BY TRIAL AND ERROR

Consider the quadratic

Since 3 is a prime number,

To fill the gaps we need two numbers with a product of 4 and so the sum of the inner and

outer terms is 13x:

As the product is 4 we will try 2 and 2, 4 and 1, and 1 and 4.

(3x + 2)(x + 2) = 3x2 + 6x + 2x + 4 fails

(3x + 4)(x + 1) = 3x2 + 3x + 4x + 4 fails

(3x + 1)(x + 4) = 3x2 + 12x + x + 4 is successful

So, 3x2 + 13x + 4 = (3x + 1)(x + 4)

We could set these trials out in table form:

For the general case ax2 + bx + c where a and c are not prime, there can be many

possibilities.

For example, consider 8x2 + 22x + 15:

By using trial and error, the possible factorisations are:

We could set these trials out in table form:

or

As you can see, this process can be very tedious and time consuming.

3x 2 4 1

x 2 1 4

8x 7x 13x

This entry is 3x£ 2 + x£ 2

(8x + 5)(x + 3) £ (4x + 5)(2x + 3) X this is correct

(8x + 3)(x + 5) £ (4x + 3)(2x + 5) £(8x + 1)(x + 15) £ (4x + 15)(2x + 1) £(8x + 15)(x + 1) £ (4x + 1)(2x + 15) £

8x 5 3 1 15

x 3 5 15 1

29x 43x 121x 23x

4x 5 3 1 15

2x 3 5 15 1

22x 26x 62x 34x

3x2 + 13x + 4.

3x2 + 13x + 4 = (3x )(x )

outers

inners


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FACTORISATION BY ‘SPLITTING’ THE x-TERM

Using the FOIL rule, we see that (2x + 3)(4x + 5)

= 8x2 + 10x + 12x + 15

= 8x2 + 22x + 15

We will now reverse the process to factorise the quadratic expression 8x2 + 22x + 15.

Notice that: 8x2 + 22x + 15

= 8x2 + 10x + 12x + 15 f‘splitting’ the middle termg= (8x2 + 10x) + (12x + 15) fgrouping in pairsg= 2x(4x + 5) + 3(4x + 5) ffactorising each pair separatelyg= (4x + 5)(2x + 3) fcompleting the factorisationg

But how do we correctly ‘split’ the middle term? How do we determine that 22x must be

written as + 10x + 12x?

When looking at we notice that 8 £ 15 = 120 and 10 £ 12 = 120

and also 10 + 12 = 22:

So, for 8x2 + 22x + 15, we need two numbers whose sum is 22 and whose product is

8 £ 15 = 120: These numbers are 10 and 12.

Likewise, for 6x2 + 19x + 15 we would need two numbers with sum 19 and product

6 £ 15 = 90.

These numbers are 10 and 9, so 6x2 + 19x + 15

= 6x2 + 10x + 9x + 15

= (6x2 + 10x) + (9x + 15)

= 2x(3x + 5) + 3(3x + 5)

= (3x + 5)(2x + 3)

The following procedure is recommended for factorising ax2 + bx + c by ‘splitting’ the

x-term:

Step 1: Find ac and then the factors of ac which add to b.

Step 2: If these factors are p and q, replace bx by px + qx.

Step 3: Complete the factorisation.

Show how to split the middle term of the following so that factorisation can occur:

a 3x2 + 7x + 2 b 10x2 ¡ 23x¡ 5

a In 3x2 + 7x + 2, ac = 3 £ 2 = 6 and b = 7.

We need two numbers with a product of 6 and a sum of 7. These are 1 and 6.

So, the split is 7x = x + 6x.


8x2+10x+12x+15


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b In 10x2 ¡ 23x¡ 5, ac = 10 £¡5 = ¡50 and b = ¡23.

We need two numbers with a product of ¡50 and a sum of ¡23. These are ¡25and 2.

So, the split is ¡23x = ¡25x + 2x.

Factorise by ‘splitting’ the x-term:

a 6x2 + 19x + 10 b 3x2 ¡ x¡ 10

a 6x2+19x+10 has ac = 60 and b = 19. We need two numbers with a product

of 60 and a sum of 19.

Searching amongst the factors of 60, only 4 and 15 have a sum of 19.

) 6x2 + 19x + 10

= 6x2 + 4x + 15x + 10 fsplitting the x-termg= 2x(3x + 2) + 5(3x + 2) ffactorising in pairsg= (3x + 2)(2x + 5) ftaking out the common factorg

b 3x2¡x¡10 has ac = ¡30 and b = ¡1. We need two numbers with a product

of ¡30 and a sum of ¡1.

Searching amongst the factors of ¡30, only 5 and ¡6 have a sum of ¡1.

) 3x2 ¡ x¡ 10

= 3x2 + 5x¡ 6x¡ 10 fsplitting the x-termg= x(3x + 5) ¡ 2(3x + 5) ffactorising in pairsg= (3x + 5)(x¡ 2) ftaking out the common factorg

EXERCISE 8G

1 Fully factorise:

a 2x2 + 5x + 3 b 2x2 + 7x + 5 c 7x2 + 9x + 2

d 3x2 + 7x + 4 e 3x2 + 13x + 4 f 3x2 + 8x + 4

g 8x2 + 14x + 3 h 21x2 + 17x + 2 i 6x2 + 5x + 1

j 6x2 + 19x + 3 k 10x2 + 17x + 3 l 14x2 + 37x + 5

2 Fully factorise:

a 2x2 ¡ 9x¡ 5 b 3x2 + 5x¡ 2 c 3x2 ¡ 5x¡ 2

d 2x2 + 3x¡ 2 e 2x2 + 3x¡ 5 f 5x2 ¡ 14x¡ 3

g 5x2 ¡ 8x + 3 h 11x2 ¡ 9x¡ 2 i 3x2 ¡ 7x¡ 6

j 2x2 ¡ 3x¡ 9 k 3x2 ¡ 17x + 10 l 5x2 ¡ 13x¡ 6


Remember tocheck your

factorisationsby expansion!


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INVESTIGATION ANOTHER FACTORISATION TECHNIQUE


m 3x2 + 10x¡ 8 n 2x2 + 17x¡ 9 o 2x2 + 9x¡ 18

p 2x2 + 11x¡ 21 q 15x2 + x¡ 2 r 21x2 ¡ 62x¡ 3

s 9x2 ¡ 12x + 4 t 12x2 + 17x¡ 40 u 16x2 + 34x¡ 15

Fully factorise: ¡5x2 ¡ 7x + 6

We remove ¡1 as a common factor first.

¡5x2 ¡ 7x + 6

= ¡1[5x2 + 7x¡ 6]

= ¡[5x2 + 10x¡ 3x¡ 6]

= ¡[5x(x + 2) ¡ 3(x + 2)]

= ¡[(x + 2)(5x¡ 3)]

= ¡(x + 2)(5x¡ 3)

Here, ac = ¡30 and b = 7. We need two

numbers with a product of ¡30 and a sum of 7.

These are 10 and ¡3.

3 Fully factorise by first removing ¡1 as a common factor:

a ¡3x2 ¡ x + 14 b ¡5x2 + 11x¡ 2 c ¡4x2 ¡ 9x + 9

d ¡9x2 + 12x¡ 4 e ¡8x2 ¡ 14x¡ 3 f ¡12x2 + 16x + 3

What to do:

1 By expanding the brackets, show that

(ax + p)(ax + q)

a= ax2 + [p + q]x +

hpqa

i:

2 If ax2 + bx + c =(ax + p)(ax + q)

a, show that p + q = b and pq = ac.

3 Using 2 on 8x2 + 22x + 15, we have

8x2 + 22x + 15 =(8x + p)(8x + q)

8where

(p + q = 22

pq = 8 £ 15 = 120:

So, p = 12 and q = 10 (or vice versa)

) 8x2 + 22x + 15 =(8x + 12)(8x + 10)

8

=4(2x + 3)2(4x + 5)

8

= (2x + 3)(4x + 5)

a Use the method shown to factorise:

i 3x2 + 14x + 8 ii 12x2 + 17x + 6 iii 15x2 + 14x¡ 8

b Check your answers to a using expansion.



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REVIEW SET 8A

REVIEW SET 8B


1 Fully factorise:

a 3x2 + b 15x¡ 3x2 c (x + 3)2 ¡ 4(x + 3)

d 4x2 ¡ 9 e 6x2 ¡ 24y2 f x2 ¡ 13

g (x + 1)2 ¡ 6 h x2 + 8x + 16 i x2 ¡ 10x + 25

2 Fully factorise:

a 2x2 + 8x + 6 b 5x2 ¡ 10x + 5

c ax + 2a + 2b + bx d 2cx¡ 2dx + d¡ c

e 3x2 + 2x + 8 + 12x f 6x2 + 9x¡ 2x¡ 3

g x2 ¡ 8x + 24 ¡ 3x h x2 + 4x + 4 ¡ a2

3 Fully factorise:

a 2x2 + 17x + 8 b 2x2 + 15x¡ 8 c 2x2 ¡ 17x + 8

d 6x2 ¡ 11x¡ 10 e 12x2 + 5x¡ 2 f 12x2 ¡ 8x¡ 15

1 Fully factorise:

a 4x2 ¡ 8x b 16x¡ 8x2 c (2x¡ 1)2 + 2x¡ 1

d 9 ¡ 25x2 e 18 ¡ 2a2 f x2 ¡ 23

g (x + 2)2 ¡ 3 h x2 ¡ 12x + 36 i 2x2 + 8x + 8

2 Fully factorise:

a 3x2 ¡ 6x¡ 9 b 7x2 + 28x + 28 c mx + nx¡my ¡ ny

d 3a2 + ab¡ 2b2 ¡ 6ab e 3x + 2x2 + 8x + 12 f 6x + 4x2 ¡ 2x¡ 3

3 Fully factorise:

a 3x2 ¡ 17x¡ 6 b 3x2 ¡ 19x + 6 c 3x2 + 17x¡ 6

d 12x2 + 7x + 1 e 12x2 ¡ 23x¡ 2 f 9x2 + 12x + 4

LINKSclick here

THE GOLDEN RATIOAreas of interaction:Human ingenuity

12x


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Y:\HAESE\IB_MYP4\IB_MYP4_08\191IB_MYP4_08.CDR Monday, 17 March 2008 3:30:36 PM PETERDELL

HISTORICAL NOTE SRINIVASA RAMANUJAN 1887 - 1920


Ramanujan was born in India in 1887.

His parents were poor, but were able to

send him to school. He was fascinated

by mathematics.

Early attempts to study at University failed because

he was required to study other subjects as well as

mathematics, and mathematics was the only subject

at which he excelled. Also, he was extremely poor.

He taught himself from books and worked at home

on mathematical research. He was always meticulous

in the recording of his work and his results, but only

rarely did he work on proofs of his theories.

Fortunately, he was able to obtain a position at the Madras Port Trust Office, a job that

paid a small wage and left him with enough time to continue his research. He was

able to take away used wrapping paper on which to write his mathematics. Eventually,

Ramanujan obtained a grant from Madras University which enabled him to have access

to, and time to use, the library and research facilities, and he was able to undertake his

studies and research in a logical way. As a result, Ramanujan was awarded a scholarship

to Cambridge University in England. Some of his work revealed amazing discoveries,

but it also revealed a lack of background knowledge and Ramanujan spent most of his

time improving his basic knowledge and establishing proofs for some of his discoveries.

The English climate and food did not agree with Ramanujan, but he continued working on

mathematics and he published 32 important papers between 1914 and 1921 even though

he was ill with tuberculosis.

In 1918 Ramanujan was made a Fellow of the Royal Society and was awarded Fellowship

of Trinity College. He was too ill to accept the position of professor of mathematics at

Madras University. He returned to India and died in 1920.

The famous English mathematician Godfrey Hardy wrote of Ramanujan:

“One gift he has which no-one can deny: profound and invincible originality. He would

probably have been a greater mathematician if he had been caught and tamed a little in

his youth; he would have discovered more that was new, and that, no doubt, of greater

importance. On the other hand he would have been less of a Ramanujan, and more of a

European professor, and the loss might have been greater than the gain.”


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Y:\HAESE\IB_MYP4\IB_MYP4_08\192IB_MYP4_08.cdr Wednesday, 9 April 2008 3:57:57 PM PETERDELL

quadratic factorisation - haese mathematics · quadratic factorisation chapter 8 ... factorise by...

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