Ch 06

6The area of a computer mouse pad is 500 cm2. List three possible dimensions (length and width) of the mouse pad, if it is rectangular in shape.

Simone is designing different sized mouse pads. As the dimensions can vary, she uses the algebraic expression x2 + 5x to represent the area, where x can be any positive value.

If the width of the mouse pad is x, what algebraic expression would represent the length of this mouse pad?

In this chapter you will learn how to find the product of factors which make up an algebraic expression using different techniques of factorising.

Factorising

MQ 9 Vic - 06 Page 179 Monday, September 17, 2001 9:19 AM

180

M a t h s Q u e s t 9 f o r V i c t o r i a

The highest common factor

Previously, the term

expanding

was defined as changing a compact form of anexpression to an expanded form. Now, we can define

factorising

as the reverse oper-ation, going from an expanded form to a more compact one. This compact form is aproduct of factors, or two or more factors multiplied together. First we need to knowhow to find all the factors of integers.

Factors

The factors of an integer are two or more integers which, when

multiplied

together,produce that integer.

3

×

2

=

6, so 3 and 2 are factors of 6.If an integer divides into a number without a remainder, it is a factor of the number.

16

÷

8

=

2, so 2 and 8 are factors of 16.The number itself and 1 are always factors of a given number.

7 and 1 are both factors of 7.

Find all the factors of each of the following integers.a 12 b −21 c −16

THINK WRITE

a Multiplications that make 12 are1 × 12, 2 × 6 and 3 × 4. All these numbers are factors.

a 1, 12; 2, 6; 3, 4

Because 12 is a positive number, it could be the result of multiplying 2 negative numbers. List the negative factor pairs.

−1, −12; −2, −6; −3, −4

List the factors of 12 in ascending order.

Factors of 12 are −12, −6, −4, −3, −2, −1, 1, 2, 3, 4, 6, 12.

b Multiplications that make −21 are 1 × −21 and 3 × −7.

b 1, −21; 3, −7

Since −21 is a negative integer, the factor pairs can change signs, so list the other pairs of factors.

−1, 21; −3, 7

List all the factors of −21 in ascending order.

Factors of −21 are −21, −7, −3, −1, 1, 3, 7, 21.

c Multiplications that make −16 are −16 × 1, −8 × 2 and −4 × 4.

c 1, −16; 2, −8; 4, −4

Since −16 is a negative integer, the factor pairs can change signs, so list the other pairs of factors. Note that −4 and 4 is the same as 4 and −4.

−1, 16; −2, 8

List all the factors of −16 in ascending order.

Factors of −16 are −16, −8, −4, −2, −1, 1, 2, 4, 8, 16.

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1WORKEDExample


C h a p t e r 6 F a c t o r i s i n g 181

Finding the highest common factorThe highest common factor or HCF of two or more numbers is the largest factor thatdivides into all of the given numbers without a remainder. This also applies to algebraicterms.The HCF of xyz and 2yz is yz because:

xyz = x × y × z2yz = 2 × y × z.

The HCF is yz (combining the common factors of each).For an algebraic term, the HCF is found by taking the HCF of the coefficients and

combining all common pronumerals.

Find the highest common factor (HCF) of each of the following.a 6 and 9 b 12, 16 and 56 c 4abc and 6bcd

THINK WRITE

a Find the factors of 6 and write them in ascending order. Consider positive factors only.

a 6: 1, 6; 2, 3Factors of 6 are 1, 2, 3, 6.

Find the factors of 9 and write them in ascending order.

9: 1, 9; 3, 3Factors of 9 are 1, 3, 9.

Write the common factors. Common factors are 1, 3.

Find the highest common factor (HCF). The HCF of 6 and 9 is 3.

b Find the factors of 12 and write them in ascending order.

b 12: 1, 12; 2, 6; 3, 4Factors of 12 are 1, 2, 3, 4, 6, 12.


16: 1, 16; 2, 8; 4, 4Factors of 16 are 1, 2, 4, 8, 16.


56: 1, 56; 2, 28; 4, 14; 7, 8Factors of 56 are 1, 2, 4, 7, 8, 14, 28, 56.

Write the common factors. Common factors are 1, 2, 4.

Find the highest common factor (HCF).

The HCF of 12, 16 and 56 is 4.

c Find the factors of 4. c 4: 1, 2, 4

Find the factors of 6. 6: 1, 2, 3, 6

Write the common factors of 4 and 6.

Common factors of 4 and 6 are 1, 2

Find the HCF of the coefficients. The HCF of 4 and 6 is 2.

List the pronumerals that are common to each term.

Common pronumerals are b and c.

Find the HCF of the algebraic terms by multiplying the HCF of the coefficients to all the common pronumerals.

The HCF of 4abc and 6bcd is 2bc.

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2WORKEDExample


182 M a t h s Q u e s t 9 f o r V i c t o r i a

Factorising expressions by finding the highest common factor An algebraic expression is made up of terms which are separated by either a + or a −sign. For example:

4xy + 3x − 4y is an algebraic expression which has 3 terms.4xy, 3x and 4y are all terms.

To factorise such an expression, find the HCF of the terms and ‘take it out’ bydividing it into the terms one by one, and then place the results inside the brackets.

If both terms are positive, the HCF will be positive, so include only positive factorsof the integer.

If both terms are negative or you want to have a negative HCF, include only negativefactors of the integers.

Factorise each of the following expressions by first finding the highest common factor (HCF).a 5x + 15y b −14xy − 7y c 15ab − 21bc + 18bf d 6x2y + 9xy2

THINK WRITEa Find the HCF of the coefficients, using

only positive integer factors. List the pronumerals common to each term.

a The HCF of 5 and 15 is 5.There are no common pronumerals.The HCF of 5x and 15y is 5.

Divide each term by the HCF. 5x ÷ 5 = x so 5x = 5 × x.15y ÷ 5 = 3y, so 15y = 5 × 3y.

Write the expression. 5x + 15y Place the HCF outside the brackets and the remaining terms inside the brackets.

= 5(x + 3y)

b Find the HCF of the coefficients, using negative integer factors. List the pronumerals common to each term.

b The HCF of −14 and −7 is −7.The common pronumeral is y.The HCF of −14xy and −7y is −7y.

Divide each term by the HCF. −14xy ÷ −7y = 2x, so −14xy = −7y × 2x.−7y ÷ −7y = 1, so −7y = −7y × 1.

Write the expression. −14xy − 7yPlace the HCF outside the brackets and the remaining terms inside the brackets.

= −7y(2x + 1)

c Find the HCF of the coefficients, using only positive integer factors. List the pronumerals common to each term.

c The HCF of 15, −21 and 18 is 3.Common pronumeral is b.The HCF of 15ab, −21bc and 18bf is 3b.

Divide each term by the HCF. 15ab ÷ 3b = 5a, so 15ab = 3b × 5a.−21bc ÷ 3b = −7c, so −21bc = 3b × −7c.18bf ÷ 3b = 6f, so 18bf = 3b × 6f.

Write the expression. 15ab − 21bc + 18bfPlace the HCF outside the brackets and the remaining terms inside the brackets.

= 3b(5a − 7c + 6f )

d Find the HCF of the coefficients, using only positive integer factors. List the pronumerals common to each term.

d The HCF of 6 and 9 is 3.Common pronumerals are x and y.The HCF of 6x2y and 9xy2 is 3xy.

Divide each term by the HCF. 6x2y ÷ 3xy = 2x, so 6x2y = 3xy × 2x.9xy2 ÷ 3xy = 3y, so 9xy2 = 3xy × 3y.

Write the expression. 6x2y + 9xy2

Place the HCF outside the brackets and the remaining terms inside the brackets.

= 3xy(2x + 3y)

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3WORKEDExample


C h a p t e r 6 F a c t o r i s i n g 183The key to HCF factoring is to find and divide each term by the HCF and place thatresult inside the brackets, leaving the HCF outside the brackets.

Each factorisation can be checked by expanding your factored answer. For example,in the previous worked example:

a 5(x + 3y) = 5x + 15y b −7y(2x + 1) = –14xy − 7yc 3b(5a − 7c + 6f ) = 15ab − 21bc + 18bf d 3xy(2x + 3y) = 6x2y + 9xy2

The highest common factor

1 Find all the factors of each of the following integers.a 36 b 17 c 51 d −14 e −8 f 100 g −42 h 32 i −32 j −9 k −64 l −81 m 29 n −92 o 48 p −12

2 Find the highest common factor (HCF) of each of the following.a 4 and 12 b 6 and 15 c 10 and 25 d 24 and 32 e 12, 15 and 21 f 25, 50 and 200 g 17 and 23 h 6a and 12abi 14xy and 21xz j 60pq and 30q k 50cde and 70fgh l 6x2 and 15xm 6a and 9c n 5ab and 25 o 3x2y and 4x2z p 4k and 6

3What is 5m the highest common factor of?

4 Factorise each of the following expressions, by first finding the highest common factor(HCF).a 4x + 12y b 5m + 15n c 7a + 14bd 7m − 21n e −8a − 24b f 8x − 4yg −12p − 2q h 6p + 12pq + 18q i 32x + 8y + 16zj 16m − 4n + 24p k 72x − 8y + 64pq l 15x2 − 3ym 5p2 − 20q n 5x + 5 o 56q + 8p2

p 7p − 42x2y q 16p2 + 20q + 4 r 12 + 36a2b − 24b2

A 2m and 5m B 5m and m C 25mn and 15lmD 20m and 40m E 15m2n and 5n2

remember1. Factorising is the ‘opposite’ of expanding, going from an expanded form to a

more compact form.2. Factor pairs of a term are numbers and pronumerals which, when multiplied

together, produce the original term.3. The number itself and 1 are factors of every integer.4. The highest common factor (HCF) of given terms is the largest factor that

divides into all terms without a remainder.5. An expression is factorised by finding the HCF of each term, dividing it into

each term and placing the result inside the brackets, with the HCF outside the brackets.

remember

6AWWORKEDORKEDEExample

1

WWORKEDORKEDEExample

2

Mathcad

HCF

EXCEL Spreadsheet

HCF

GC program

HCF

mmultiple choiceultiple choice


3

Mathcad

Factorisingusing the

HCF

SkillSH

EET 6.1



5 Factorise each expression by taking out the highest common factor.a 9a + 21b b 4c + 18d2 c 12p2 + 20q2

d 35 − 14m2n e 25y2 − 15x f 16a2 + 20bg 42m2 + 12n h 63p2 + 81 − 27y i 121a2 − 55b + 110cj 10 − 22x2y3 + 14xy k 18a2bc − 27ab − 90c l 144p + 36q2 − 84pqm 63a2b2 − 49 + 56ab2 n 22 + 99p3q2 − 44p2r o 36 − 24ab2 + 18b2c

6 Factorise by taking out the highest negative common factor.a −x + 5 b −a + 7 c −b + 9d −2m − 6 e −6p − 12 f −4a − 8g −3n2 + 15m h −7x2y2 + 21 i −7y2 − 49zj −12p2 − 18q k −63m + 56 l −12m3 − 50x3 m −9a2b + 30 n −15p − 12q o −18x2 + 4y2 p −3ab + 18m − 21 q −10 − 25p2 − 45q r −90m2 + 27n + 54p3

7 Factorise by taking out the highest common factor.a a2 + 5a b m2 + 3m c x2 − 6xd 14q − q2 e 18m + 5m2 f 6p + 7p2 g 7n2 − 2n h a2 − ab + 5a i 7p − p2q + pqj xy + 9y − 3y2 k 5c + 3c2d − cd l 3ab + a2b + 4ab2

m 2x2y + xy + 5xy2 n 5p2q2 − 4pq + 3p2q o 6x2y2 − 5xy + x2y

8 Factorise each of the following expressions.a 5x2 + 15x b 10y2 + 2y c 12p2 + 4pd 24m2 − 6m e 32a2 − 4a f −2m2 + 8mg −5x2 + 25x h −7y2 + 14y i −3a2 + 9aj −12p2 − 2p k −15b2 − 5b l −26y2 − 13ym 4m − 18m2 n −6t + 36t2 o −8p − 24p2

EXCEL

Spreadsheet

Factorisingax2 + bx

MA

TH

SQUEST

C H A L L

EN

GE

MA

TH

SQUEST

C H A L L

EN

GE

1 This Noughts and Crosses game is to be finished. The squares labelled p,q, r, s and t are empty. What is the best play if:a O is to play next?b X is to play next?

2 The following is a famous problem studied by an Egyptian mathemat-ician called Hypatia.‘Find a number that is the sum of two squares such that its square isalso the sum of two squares.’There are a number of solutions to this problem. Can you find 3 poss-ible solutions?

O O X

X p q

r s t


C h a p t e r 6 F a c t o r i s i n g

185

Factorise the expressionsto find the puzzle’s code.

I just want yI just want you to rou to remember emember one thing! one thing!

7y (2 – x)

7x (2 – x) 2y (3 + 7x) –3b (a + b)

2b (b – a) 3a (2 + a)y (x + y) 2(x – 4) 5(x – 3)8(2 –x)11(2x – 3)a (a – b)

x (x + 7) y (1 – x) b (1 – b)2xy (2 + x)

5y (5 + y) 6(3 – 2x)

2(2x + 1)xy (x – 1)–x (3 + x)

4a (1 – b) 4x (2 – x)

4(5 – 2x)

–a (3 – b)3x (y – 4) –(x + 1) –(xy + 6) –2(y – 3)b 2(2 + a) b 2(a – b) x 2(1 –x)a (a + b)

= a 2 + ab

=

= 7x + x 2

=

= 2b 2 + ab 2

=

= ab 2 – b 3

=

= a 2 – ab

=

= –3ab – 3b 2

=

= 4xy + 2x 2y

=

= 14x – 7x 2

=

= 3xy – 12x

=

= 18 – 12x

=

= 20 – 8x

=

= 16 – 8x

=

= 6a + 3a 2

=

= –3x – x 2

=

= b – b 2

=

= x 2 – x 3

=

= –2y + 6

=

= –xy – 6

=

= 8x – 4x 2

=

= 25y – 5y 2

=

= 22x – 33

=

= xy + y 2

=

= x2y – xy

=

= y – xy

=

= 4a – 4ab

=

= 2b 2 – 2ab

=

= –3a + ab

=

= –x – 1

=

= 14y – 7xy

=

= 6y + 14xy

=

= 5x – 15

=

= 4x + 2

=

= 2x – 8

=

’’’’

MQ 9 Vic - 06 Page 185 Wednesday, September 19, 2001 11:21 AM


More factorising using the highest common factor

Remember, whenever factorising, look for a common factor first.

The binomial common factorSometimes the common factor may itself be in brackets.

Consider the following expression:5(x + y) + 6b(x + y).

While it may appear at first that no factorising can be done, consider the fact thatboth terms contain the bracketed expression (x + y) as a factor. This is called a binomialfactor because it contains two terms.

These answers can also be checked by performing an expansion.When factorising expressions that already have brackets in them, look to this method

of using the bracketed part as the common factor or, in some cases, as part of thecommon factor.

Factorising by grouping termsIf an algebraic expression has 4 terms and no common factor in all the terms, it may bepossible to group the terms in pairs and find a common factor in each pair.

Factorise each of the following expressions by taking out the binomial common factor.a 5(x + y) + 6b(x + y) b 2b(a − 3b) − (a − 3b)

THINK WRITE

a Identify the common factor. a Common factor is (x + y).Divide each term by the common factor.

5(x + y) ÷ (x + y) = 5 6b(x + y) ÷ (x + y) = 6b

Write the expression. 5(x + y) + 6b(x + y)Factorise, by taking out the binomial common factor and placing the remaining terms inside brackets.

= (x + y)(5 + 6b)

b Identify the common factor. b Common factor is (a − 3b).Divide each term by the common factor.

2b(a − 3b) ÷ (a − 3b) = 2b (a − 3b) ÷ (a − 3b) = 1

Write the expression. 2b(a − 3b) − (a − 3b)Factorise, by taking out the binomial common factor and placing the remaining terms inside brackets.

= (a − 3b)(2b − 1)

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4WORKEDExample



The answers found in worked example 5 can each be checked by expanding.

There are only 3 possible pair groupings to consider with this technique:1st and 2nd terms + 3rd and 4th terms; 1st and 4th terms + 2nd and 3rd terms; 1st and3rd terms + 2nd and 4th terms.

So in the worst case you would have to try all 3.

Factorise each of the following expressions by grouping the terms in pairs.a 5a + 10b + ac + 2bc b x − 3y + ax − 3ay c 5p + 6q + 15pq + 2

THINK WRITE

a Write the expression. a 5a + 10b + ac + 2bc

Look for a common factor of all 4 terms. (There isn’t one.) If necessary, rewrite the expression so that the terms with common factors are next to each other.

Take out a common factor from each group.

= 5(a + 2b) + c(a + 2b)

Factorise by taking out a binomial common factor.

= (a + 2b)(5 + c)

b Write the expression. b x − 3y + ax − 3ay


Take out a common factor from each pair of terms.

= 1(x − 3y) + a(x − 3y)


= (x − 3y)(1 + a)

c Write the expression. c 5p + 6q + 15pq + 2


= 5p + 15pq + 6q + 2

Take out a common factor from each pair of terms.

= 5p(1 + 3q) + 2(3q + 1)= 5p(1 + 3q) + 2(1 + 3q)


= (1 + 3q)(5p + 2)

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5WORKEDExample



More factorising using the highest common factor

1 Factorise each of the following expressions by taking out the binomial common factor.a 2(a + b) + 3c(a + b) b 4(m + n) + p(m + n)c 7x(2m + 1) − y(2m + 1) d 4a(3b + 2) − b(3b + 2)e z(x + 2y) − 3(x + 2y) f 12p(6 − q) − 5(6 − q)g 3p2(x − y) + 2q(x − y) h 4a2(b − 3) + 3b(b − 3)i p2(q + 2p) − 5(q + 2p) j 6(5m + 1) + n2(5m + 1)

2 Factorise each of the following expressions by grouping the terms in pairs.a xy + 2x + 2y + 4 b ab + 3a + 3b + 9c xy − 4y + 3x − 12 d 2xy + x + 6y + 3e 3ab + a + 12b + 4 f ab − 2a + 5b − 10g m − 2n + am − 2an h 5 + 3p + 15a + 9api 15mn − 5n − 6m + 2 j 10pq − q − 20p + 2k 6x − 2 − 3xy + y l 16p − 4 − 12pq + 3qm 10xy + 5x − 4y − 2 n 6ab + 9b − 4a − 6o 5ab − 10ac − 3b + 6c p 4x + 12y − xz − 3yzq 5pr + 10qr − 3p − 6q r ac − 5bc − 2a + 10b

1 List all the factors of −36.

2 Find the highest common factor of 45 and 21.

3 Find the highest common factor of 8ab and 12ad.

4 Factorise 27xy − 45y.

5 Factorise 25y2 − 85xy + 35x2y2.

6 Factorise −27abc + 36ac − 18a2bc.

7 Factorise 9(x − 2y) + b(x − 2y).

8 Factorise 7x + 14 − 2xy − 4y.

9 Factorise 3abc + 6ac + b + 2.

10 Factorise 12xy − 8 + 3x2y − 2x.

remember1. When factorising any number of terms, look for a factor which is common to

all terms.2. A binomial factor has 2 terms.3. The HCF of an algebraic expression may be a binomial factor, which is in brackets.4. When factorising expressions with 4 terms which have no factor common to all 4:

(a) group the terms in pairs with a common factor(b) factorise each pair(c) factorise the expression by taking out a binomial common factor.

5. (x + y) = 1(x + y)

remember

6BWWORKEDORKEDEExample

4

Mathca

d

Factorising using grouping


5

WorkS

HEET 6.1

1



Factorising using the difference of two squares rule

In chapter 5, we saw expansions of the form (x + 2)(x − 2) = x2 − 4. The result of thisexpansion is called a difference of two squares because one square is subtracted fromanother square.

To factorise a difference of two squares, we use the rule or formula in reverse.a2 − b2 = (a + b)(a − b)

We recall that a and b are any terms, and that the expression x2 − 4 can be written asx2 − 22 , so that it can be factorised as (x + 2)(x − 2) or (x − 2)(x + 2) as the order of thetwo factors is unimportant.

Factorise each of the following expressions using the difference of two squares rule.a x2 − 9 b 25x2 − 49 c 64m2 − 25n2 d (c + 7)2 − 16 e 3x2 − 48

THINK WRITE

a Write the given expression. a x2 − 9Rewrite, showing the two squares. = x2 − 32 Factorise using the formulaa2 − b2 = (a + b)(a − b) where a = x and b = 3.

= (x + 3)(x − 3)

b Write the given expression. b 25x2 − 49Rewrite, showing the two squares. = (5x)2 − 72 Factorise, using the formula a2 − b2 = (a + b)(a − b) where a = 5x and b = 7.

= (5x + 7)(5x − 7)

c Write the given expression. c 64m2 − 25n2

Rewrite, showing the two squares. = (8m)2 − (5n)2 Factorise, using the formula a2 − b2 = (a + b)(a − b) where a = 8m and b = 5n.

= (8m + 5n)(8m − 5n)

d Write the expression. d (c + 7)2 − 16Rewrite, showing the two squares. = (c + 7)2 − 42

Factorise, using the formula a2 − b2 = (a + b)(a − b) where a = c + 7 and b = 4.

= (c + 7 + 4)(c + 7 − 4)

Simplify. = (c + 11)(c + 3)e Write the expression. e 3x2 − 48

Look for a common factor and factorise, taking out the common factor.

= 3(x2 − 16)

Rewrite the two terms in the bracket showing the two squares.

= 3(x2 − 42)

Factorise, using the formula a2 − b2 = (a + b)(a − b) where a = x and b = 4.

= 3(x + 4)(x − 4)

123

123

123

123

412

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6WORKEDExample



Factorising using the difference of two squares rule

1 Factorise each of the following expressions using the difference of two squares rule.a x2 − 25 b x2 − 81 c a2 − 16 d c2 − 64e y2 − 144 f 16 − x2 g 25 − p2 h 121 − a2

i 36 − y2 j 4b2 − 25 k 9a2 − 16 l 25d2 − 1

2 Factorise each of the following expressions.a x2 − y2 b a2 − b2 c p2 − q2 d c2 − d2

e 9x2 − y2 f 16p2 − q2 g 25m2 − n2 h 81x2 − y2

i p2 − 36q2 j m2 − 4n2 k a2 − 49b2 l y2 − 100z2

m 36m2 − 25n2 n 16q2 − 9p2 o 4m2 − 49n2

3 Factorise each of the following expressions.a (x + 9)2 − 16 b (p + 8)2 − 25 c (p − 2)2 − q2 d (c − 6)2 − d2

e (x + 7)2 − y2 f (p + 5)2 − q2 g (r − 9)2 − s2 h (a − 1)2 − b2

i (x + 1)2 − y2 j (a + 3)2 − b2 k (a − 3)2 − 1 l (b − 1)2 − 36

4 Factorise each of the following expressions, using the difference of two squares rule,after removing any common factors.a 2m2 − 32 b 5y2 − 45 c 6p2 − 24 d 4m2 − 100e 288 − 2x2 f 80 − 5a2 g 36x2 − 9y2 h 9y2 − 81z2

i 4m2 − 36n2 j 100x2 − 25y2 k 144p2 − 4q2 l 32y2 − 2x2

m 3m2 − 27n2 n 4(x + 2)2 − 36 o 3(m + 1)2 − 12 p 7(y − 3)2 − 28q 2(y − 4)2 − 50 r 3(b + 5)2 − 48

5a The expression x2 − 121 factorises to:

b What does the expression 36m2 − 1 factorise to?

c What does the expression 16a2 − 25b2 factorise to?

d The expression 5c2 − 20 factorises to:

A (x − 11)2 B (x − 11)(x + 12) C (x + 11)(x − 11)D (x − 11)(x − 11) E (x + 11)(x − 12)

A (6m + 1)(6m − 1) B 36(m + 1)(m − 1) C (6m − 1)2

D 6(6m − 1)(6m − 1) E (6m + 1)(6m + 1)

A 16(a + 5b)(a − 5b) B 4(4a − 5b)2 C (16a + 25b)(16a − 25b)D (4a + 5b)(4a − 5b) E (4a − 5b)(4a − 5b)

A 5(c + 2)(c − 2) B 5(c + 4)(c − 4) C (5c + 4)(5c − 4)D (5c + 20)(5c − 20) E (5c + 10)(5c − 10)

rememberWhen factorising an algebraic expression with 2 terms follow these steps.1. Look for a common factor first. If there is one, factorise by taking it out.2. Rewrite the expression showing the two squares and identifying the a and b

parts of the expression.3. Factorise, using the rule a2 − b2 = (a + b)(a − b).

remember

6C

SkillSH

EET 6.2 WWORKEDORKEDEExample

6a, b

EXCEL

Spreadsheet

Difference of two squares rule


6c

Mathca

d

Difference of two squares rule


6d


6e



C h a p t e r 6 F a c t o r i s i n g 191e The expression (x + 4)2 − 9 factorises to:

6 A circular pool with a radius of r metres is surrounded by a circular path 1 m wide.

a Find the surface area of the pool.b Find, in terms of r, the distance from the centre of the pool to the outer edge of the

path.c Find the area of the circle that includes the path and the pool. (Don’t expand the

expression.)d Write an expression for the area of the path.e Simplify this expression.f If the pool had a radius of 5 m, what would be the area of the path, to the nearest

square metre?g If the pool had a radius of 7 m, what would be the area of the path, to the nearest

square metre?

A 9(x + 4)(x − 4) B (x + 13)(x − 5) C (x + 7)(x + 1)D (x − 1)(x + 5) E (x − 7)(x + 11)

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1 Work out the values of the following without using a calculator or longmultiplication.a 92 – 72

b 852 – 152 c 7652 – 2352

2 The difference between the squares of two consecutive odd numbers is48. What are the numbers?

3 The difference between the squares of two consecutive even numbers is92. What are the numbers?



What has area got to do with factorising?

Copy the diagram shown at right on to paper and use scissors to cut around the outline of the large square.1 What is the area of the large square in

terms of the pronumeral a?2 What is the area of the small square in

terms of the pronumeral b?

Cut out the small square from the large square.3 Write an expression for the area of the

shape left over. 4 Find the lengths, in terms of a and b, of the

two sides not yet labelled on your shape. Include these on your shape.

Now cut your shape into two pieces as shown in the diagram at right. (Cut along the dotted line.) Rearrange the two pieces into a rectangle. 5 Write an expression for the length of the

rectangle.6 Write an expression for the width of the

rectangle.7 Use your answers from parts 5 and 6 to

write an expression for the area of the rectangle.

8 Relate this expression for the area of this shape to the one found earlier in part 3.

9 Where have you seen this formula before?

a

ab

b

a

ab

b

a

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Quadratic trinomialsAn expression with 3 terms is called a trinomial, tri meaning three.

A quadratic trinomial is an algebraic expression that has 3 terms and the highestpower is a squared term.

The expressions y2 + 3y − 4 and 5x2 − 2x + 4 are both quadratic trinomials.The expression x3 + 3x2 − 4 is not a quadratic trinomial, because x2 is not the highest

power.Therefore, we could write a quadratic trinomial as:

constant × (pronumeral)2 + constant × pronumeral + constant.Note: The + signs could be replaced by − signs or the constants could be positive ornegative.

The general caseUsing the above definition, a general quadratic trinomial can be written, replacing thewords ‘constant’ with a, b and c respectively and perhaps x as the pronumeral.

ax2 + bx + cNote: We can allow for minus signs in the trinomial by giving a, b and c negativevalues.There are only two cases, each of which will require a different technique:a = 1 The coefficient of the squared term is 1.a ≠ 1 The coefficient of the squared term is not equal to 1.

Factorising quadratic trinomialsFactorising is the reverse of expanding. Consider the expansion below:

(x + 3)(x + 5) = x(x + 5) + 3(x + 5) = x2 + 5x + 3x + 15 = x2 + 8x + 15

Now, if we were to factorise x2 + 8x + 15, the answer must be (x + 3)(x + 5), sincefactorising is the reverse of expanding. Looking for a relationship between the quad-ratic and its factors, we can see that 3 + 5 = 8 and 3 × 5 = 15.

Now consider the more general expansion:(x + p)(x + q) = x(x + q) + p(x + q)

= x2 + qx + px + pq = x2 + (p + q)x + pq

If p and q are the constant terms in the factors, we can see that the coefficient of thex term is the sum of p and q, while the constant term is the product of p and q. In otherwords, we are looking for two numbers p and q which multiply to give c and add togive b.

Factorising ax2 + bx + c when a = 1The following method works for every possible trinomial (with a = 1) that can befactorised.Step 1 Put the trinomial into the correct order or standard form x2 + bx + c.Step 2 Find all the factor pairs of c (the constant term).Step 3 Identify the factor pair whose sum equals b; that is, p + q = b.Step 4 The factorisation for x2 + bx + c = (x + p) (x + q).



Let us perform these 4 steps on the trinomial y2 + 5y + 4.Step 1 The quadratic is already in standard form, with b = 5 and c = 4.Step 2 Find the factor pairs of 4 (the c).

1 and 42 and 2−1 and −4−2 and −2

Step 3 Find the sum of each factor pair.1 + 4 = 5 (equals b)2 + 2 = 4−1 + −4 = −5−2 + −2 = −4

Step 4 The first pair’s sum equals 5 (the b), so p = 1, q = 4 and the factorisation is(y + 1)(y + 4).

Factorise each of the following quadratic trinomials.a y2 + 6y + 8 b y2 − 5y + 4 c y2 − 3y − 10 d y2 − 4 + 3y

THINK WRITE

a Make sure that the expression is in the correct order.

a y2 + 6y + 8

Find the factor pairs of c (+8). 8: 1 and 8, 2 and 4, −1 and −8, −2 and −4Find the sum of each factor pair and identify the pair with a sum of b (+6).

1 + 8 = 9 2 + 4 = 6 (equals b)−1 + −8 = −9−2 + −4 = −6

Write the expression and its factorised form.

y2 + 6y + 8 = (y + 2)(y + 4)

b Make sure that the expression is in the correct order.

b y2 − 5y + 4

Find the factor pairs of c (+4). 4: 1 and 4, 2 and 2, −1 and −4, −2 and −2Identify the pair with a sum of b (−5). −1 + −4 = −5 Write the expression and its factorised form.

y2 − 5y + 4 = (y − 1)(y − 4)

c Make sure that the expression is in the correct order.

c y2 − 3y − 10

Find the factor pairs of c (−10). 10: 1 and −10, 2 and −5, −1 and 10, −2 and 5Identify the pair with a sum of b (−3). 2 − 5 = −3Write the expression and its factorised form.

y2 − 3y − 10 = (y + 2)(y − 5)

d Make sure that the expression is in the correct order.

d y2 + 3y − 4

Find the factor pairs of c (−4). −4: 1 and −4, 2 and −2, −1 and 4Identify the pair with a sum of b (+3). −1 + 4 = 3Write the expression and its factorised form.

y2 + 3y − 4 = (y − 1)(y + 4)

1

23

4

1

234

1

234

1

234

7WORKEDExample


C h a p t e r 6 F a c t o r i s i n g 195Again, each of the factorisations can be checked by expanding the pair of brackets.Note: In the four parts of the previous worked example, we have covered all cases,namely b either positive or negative and c either positive or negative, without resortingto any different method.

Quadratic trinomials

Factorise each of the following quadratic trinomials.1 x2 + 4x + 3 2 x2 + 12x + 11 3 a2 + 6a + 5

4 b2 − 8b + 7 5 x2 − 5x + 6 6 y2 − 7y + 12

7 c2 + c − 20 8 x2 + 4x − 21 9 x2 + 9x − 10

10 p2 − 3p − 28 11 q2 − 6q − 27 12 x2 − 6x − 16

13 y2 + 10y + 9 14 x2 − 12x + 32 15 c2 − 5c + 36

16 m2 − 5m − 14 17 x2 − 6x − 55 18 m2 − 14m + 24

19 x2 + 12x + 35 20 x2 − 17x + 72 21 x2 + 10x + 25

22 p2 + 4p + 4 23 a2 + 8a + 16 24 y2 − 6y + 9

25 x2 − 16x + 64 26 x2 − 10x + 25 27 m2 − 7m + 6

28 x2 − 12x + 27 29 k2 − 9k − 22 30 x2 + 11x − 12

31 x2 − 13x + 42 32 a2 + a − 6 33 a2 − 3a − 4

34 x2 − 3x − 10 35 x2 + 2x − 8 36 x2 − 5x − 6

37 x2 + 4x − 5 38 b2 − 2b − 35 39 c2 − 15c − 16

40 x2 − 16x − 36 41 x2 + x − 30 42 d2 + 5d − 14

43 f 2 − 3f − 40 44 m2 + 5m − 6 45 q2 + 7q − 18

46 x2 + 5x − 24 47 x2 − x − 2 48 y2 + 3y − 28

49 a2 + 7a − 18 50 x2 + x − 12

remember1. A quadratic trinomial consists of 3 terms, the highest power being a squared

term.

2. The general form of a quadratic trinomial is ax2 + bx + c.

3. There are 2 main cases: either a = 1 or a ≠ 1.

4. Always look for a common factor first.

5. To factorise a quadratic trinomial when a = 1, (x2 + bx + c) use the following steps:(a) put the trinomial into standard form x2 + bx + c(b) find all the factor pairs of c(c) identify the factor pair; call one factor p and the other q, whose sum equals

b; that is, p + q = b(d) the factorisation for x2 + bx + c = (x + p) (x + q).

remember

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Factorisingx2 + bx + c

Mathcad

Factorisingx2 + bx + c



1 Find the HCF of 48a and 54.

2 Factorise 12x2 − 32x3y.

3 Factorise 8(x − 2y) − 3q(x − 2y).

4 Factorise x2 − 81.

5 Factorise (x + 6)2 − y2.

6 Factorise 49x2 − 100y2.

7 Factorise y2 − 7y − 8.

8 Factorise y2 + y − 12.

9 Factorise c2 + 8c + 15.

10 Factorise x2 − 10x + 16.

Mouse pad dimensionsAt the start of the chapter, we considered the dimensions (length and width) of a rectangular mouse pad. 1 If the area of a rectangular mouse pad

is 500 cm2 and the width is 20 cm, what is the length? Explain how you are able to work this out.

2 If the area of the mouse pad is (x2 + 5x) cm2 and the width is x cm, what is the length? Explain how factorising the expression helps you find the length.

3 For a particular mouse pad, x is 22 cm. a Find the width and length of this

mouse pad.b Multiply the length and width

together to find the area. c Substitute x = 22 into the area

expression x2 + 5x to calculate the area. Verify that this gives the same result as that obtained in part b.

4 Simone also designs rectangular floor mats for under desk chairs. She uses the algebraic expression x2 + 2x − 15. a If the length of the mat is (x + 5) cm, find an expression for the width.b If the length of the mat is 70 cm, what is the width? c If the width of the mat is 1 m, what is the length?

2



More quadratic trinomialsRemoving a common factorSometimes a quadratic trinomial will have a common factor in each term. Afterremoving the common factor, the remaining factor is a case where a = 1.

Factorising ax2 + bx + c when a ≠ 1When no common factor can be removed, as in the previous worked example, a modi-fied method for factorisation can be used.

The factors of ac must be considered instead of just the factors of c. We need to find2 numbers which multiply to give ac and add to give b. These 2 numbers will be usedto rewrite the middle term of the expression, before using the method of factorising bygrouping terms.

Remember to check that the terms are in the correct order before you start.

Factorise each of the following quadratic trinomials by first taking out a common factor.a 2x2 − 16x − 18 b −x2 − 5x + 6THINK WRITEa Write the expression and take out the

common factor.a 2x2 − 16x − 18 = 2(x2 − 8x − 9)

List the factor pairs of the last term. −9: 1 and −9, 3 and −3, −1 and 9Find the pair with a sum equal to the coefficient of the middle term.

1 + −9 = −8

Write the expression and its factorised form with the common factor outside the brackets.

2x2 − 16x − 18 = 2(x + 1)(x − 9)

b Write the expression and take out the common factor, which, in this case, is −1.

b −x2 − 5x + 6 = −(x2 + 5x − 6)

List the factor pairs of the last term. −6: 1 and −6, 2 and −3, −1 and 6, −2 and 3Find the pair with a sum of the coefficient of the middle term.

−1 + 6 = 5

Write the expression and its factorised form with the common factor outside the brackets.

−x2 − 5x + 6 = −(x − 1)(x + 6)

1

23

4

1

23

4

8WORKEDExample

Factorise each of the following quadratic trinomials.a 3x2 + 5x + 2 b 2x2 − 11x − 6 c 4x2 − 19x + 12THINK WRITEa Write the expression. a 3x2 + 5x + 2

List the factor pairs of ac (6). 6: 1 and 6, 2 and 3, −1 and −6, −2 and −3Identify which pair has a sum of b (5). 2 + 3 = 5 Rewrite the expression by breaking the middle term into 2 terms using the factor pair from step .

3x2 + 5x + 2 = 3x2 + 3x + 2x + 2

Factorise by grouping terms. = 3x(x + 1) + 2(x + 1) = (x + 1)(3x + 2)

1234

35

9WORKEDExample

Continued over page



With practice you will be able to reduce the number of factor pairs to consider, insteadof looking at all of them. For example, in part c of worked example 10, we consideredonly negative factor pairs as we were looking for a negative sum. Similarly, in part awe need only have considered the positive factor pairs.Note: This method can also be used for expressions where a = 1.

THINK WRITE

b Write the expression. b 2x2 − 11x − 6List the factor pairs of ac (−12). −12: 1 and −12, −1 and 12, 2 and −6, −2 and

6, 3 and −4, −3 and 4Identify which pair has a sum of b (−11).

1 + −12 = −11

Rewrite the expression by breaking the middle term into 2 terms using the factor pair from step .

2x2 − 11x − 6 = 2x2 − 12x + x − 6

Factorise by grouping terms. = 2x(x − 6) + 1(x − 6) = (x − 6)(2x + 1)

c Write the expression. c 4x2 − 19x + 12List the factor pairs of ac (48). Consider only the negative factors as we are looking for a pair with a negative sum.

48: −4 and −12, −6 and −8, −2 and −24, −3 and −16

Identify which pair has a sum of b (−19).

−3 + −16 = −19

Rewrite the expression by breaking the middle term into 2 terms using the factor pair from step .

4x2 − 19x + 12 = 4x2 − 16x − 3x + 12

Factorise by grouping terms. = 4x(x − 4) − 3(x − 4) = (x − 4)(4x − 3)

1

2

3

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3

5

1

2

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3

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remember1. Always look for a common factor before factorising.

2. To factorise any quadratic trinomial of the form ax2 + bx + c, follow these steps.

Step 1 List the factor pairs of ac.

Step 2 Identify which pair has a sum of b.

Step 3 Rewrite the expression by breaking the middle term into two terms using the factor pair from Step 2.

Step 4 Factorise by grouping the terms.

3. You can always check your answer by expanding.

remember



More quadratic trinomials

1 Factorise each of the following quadratic trinomials by first taking out a commonfactor.a 2x2 + 10x + 12 b 3x2 + 15x + 12 c 7x2 + 42x + 56d 6x2 + 54x + 120 e 2x2 − 12x + 16 f 3x2 − 12x + 9g 5x2 − 45x + 70 h 6x2 − 42x + 72 i 4x2 − 4x − 24j 3x2 + 9x − 30 k 2x2 + 8x − 42 l 7x2 + 35x − 252m 3x2 − 9x − 54 n 8x2 + 40x − 48 o 5x2 + 20x − 60

2 Factorise each of the following quadratic trinomials.a 2x2 + 7x + 3 b 2x2 + 7x + 6 c 3x2 + 7x + 2d 3x2 + 10x + 3 e 5x2 + 11x + 2 f 6x2 + 13x + 6g 2x2 − 7x + 3 h 3x2 − x − 2 i 5x2 + 3x − 2j 7x2 − 17x + 6 k 10x2 − 11x − 6 l 2x2 + 5x − 12m 7x2 − 26x + 15 n 11x2 − 28x + 12 o 2x2 + x − 36p 4x2 + 8x − 5 q 2x2 + 5x − 18 r 7x2 − 3x − 4

3a The expression 3x2 + 21x + 36 factorises to:

b What does the expression 5x2 + 20x − 60 factorise to?

c The expression 2x2 − 16x + 14 factorises to:

d What does the expression 3x2 − 13x + 4 factorise to?

e What does the expression 12x2 + 58x − 10 factorise to?

A 3(x + 6)(x + 2) B 3(x + 4)(x + 3) C (3x + 9)(x + 4)D (3x + 2)(x + 18) E (3x + 2)(x + 9)

A (5x − 12)(x + 5) B (5x + 15)(x − 4) C 5(x + 6)(x − 2)D 5(x − 6)(x + 2) E 5(x − 6)(x − 2)

A 2(x − 1)(x − 7) B 2(x + 1)(x + 7) C (2x − 7)(x − 2)D (2x − 2)(x − 7) E (2x − 2)(2x − 7)

A (3x − 2)2 B (3x − 2)(x + 2) C (3x − 1)(x − 4)D (3x − 4)(x − 1) E (3x − 4)(x + 1)

A (12x − 2)(x − 5) B 2(6x + 1)(x − 5) C 2(6x + 5)(x − 1)D 2(6x − 1)(x + 5) E 2(2x − 5)(3x + 1)

6EWWORKEDORKEDEExample

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EXCEL Spreadsheet

Factorisingax2 + bx + c


9

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Factorisingax2 + bx + c


GAMEtime

Factorising— 001

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1 In numbering the pages of a book, it was determined that 348 digits areneeded to completely number the book. If the numbering begins at thefirst page (that is, the first page is numbered 1), how many pages are inthe book?

2 Volumes 15 and 16 of a series of books stand side by side, in order, on abookshelf. Volume 15, without its cover, is 3.2 cm thick and volume 16,without its cover, is 2.8 cm thick. The cover of each book is 3 mm thick. Ifan insect begins at the first page of volume 15 and eats its way in a straightline to the last page of volume 16, how far will the insect have travelled?


Isaac NeIsaac Newton demonstrated this in 1wton demonstrated this in 1665.665.

Factorise the trinomials onthe left of the page. Join

the dot next to each with the dots next to its factors,

using straight lines. Thetwo lines will pass through

a letter and numbergiving the puzzle’s code.

= 2x 2 + 5x – 3= (x + 1)

(x – 1)

(x – 2)

(x + 3)

(x – 4)

(x + 4)

(x – 5)

(2x + 1)

(2x – 1)

(2x + 3)

(2x – 3)

(2x – 5)

(3x + 1)

(3x – 1)

(3x + 2)

(3x – 2)

= 2x 2 – 3x – 2=

= 2x 2 – 9x + 4=

= 2x 2 + 5x – 12=

= 2x 2 – 7x + 6=

= 2x 2 – 11x + 12=

= 3x 2 – 7x + 2=

= 2x2 – x – 3=

= 3x 2 + 2x – 1=

= 3x 2 + x – 2=

= 3x 2 – 13x – 10=

= 3x 2 – 11x – 4=

= 3x 2 – 8x + 4=

= 3x 2 + 4x + 1=

= 3x 2 – 10x – 8=

= 4x 2 – 4x – 15=

= 3x 2 + 14x + 8=

= 2x 2 + 3x – 9=

= 2x 2 – x – 1=

1

11 3 15 15 5 17 5 18 4 19 14 14 12 17 86

12 13 14 15 10 9 3 16 4 12 17 5 14 15

2 3 4 5 6 3 7 2 4 3 8 9 10 11 5

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Mixed factorising practiceYou may like to review the techniques demonstrated in the previous worked examples,or go over the solutions in your workbook before attempting this set of mixedproblems.

Mixed factorising practice

Factorise each of the following.

1 2m + 8n 2 −m + 6 3 ab − 3a + 2b − 6

4 p2 − 16 5 m2 + 3m + 2 6 2m2 − 18

7 3a − 9b 8 x2 + 6x + 5 9 −m2 − 3m

10 8m2 + 4 − 16q 11 x2 + 4x − 21 12 2ab − 6b − ac + 3c

13 2m2 − 5m − 3 14 18a2 − 3a 15 x2 − 9x + 18

16 (x + 2)2 − 1 17 5m2 + 33m − 14 18 m2 − 9m + 20

19 x2 − x − 30 20 1 − x2 21 2m2 + 3m − 2

22 4a2 + 4a + 1 23 3x2 − x − 2 24 7 − 14a2

25 8p + 9p2 26 2mn + 2m + 3n + 3 27 2x2 + 2x − 12

28 p2 − 36q2 29 −5x2y − x 30 m2 + 8m + 16

31 10p2 − 25q 32 5x2 − 30x + 40 33 (x + 3)2 − 25

34 2x2 + 6x − 20 35 3a2 − 48b2 36 −18m − 12n2

37 2ac + bc − 6a − 3b 38 (a − 7)2 − 36 39 2m2 − 9m + 10

40 110a2b + 121b − 10a2 − 11

remember1. Always look for a common factor to take out if possible.

2. For an expression with 2 terms: try the difference of two squares formula a2 − b2 = (a + b)(a − b).

3. For an expression with 3 terms: if it is a quadratic trinomial of the form ax2 + bx + c, rewrite the expression by breaking the middle term into 2 terms using a factor pair which multiplies to give ac and adds to give b. Factorise by grouping terms.

4. For an expression with 4 terms: try factorising by grouping terms.

remember

6FMathcad

Mixedfactorising

GAMEtime

Factorising— 002



Simplifying algebraic fractionsIn this section we look at using factorisation techniques to simplify more complexalgebraic fractions. After factorising a numerator or denominator, terms may becancelled, resulting in a simplified fraction, or even no fraction at all.

Simplifying after taking out a common factorLook for common factors in the numerators and denominators before cancelling.

Simplifying after factorising quadratic trinomialsIf there is a quadratic trinomial in either the numerator, the denominator or both, thenfactorise before cancelling.

Simplify each of the following fractions by factorising the numerator and denominatorand cancelling as approriate.

a b c d

THINK WRITE

a Write the fraction. a

Factorise both numerator and denominator. =

Cancel any common factors (3). =

b Write the fraction. b


Cancel any common factors (5). =

c Write the fraction. c


Cancel any common factors [the binominal factor (2x + 3)]. =

d Write the fraction. d


Cancel any common factors (x). =

3x 9+15

--------------- 25–10x 20+--------------------- 10x 15+

6x 9+--------------------- x2 4x+

x2 5x–------------------

13x 9+

15---------------

23 x 3+( )

15--------------------

3x 3+

5------------

125–

10x 20+---------------------

225–

10 x 2+( )-----------------------

35–

2 x 2+( )--------------------

110x 15+6x 9+

---------------------

25 2x 3+( )3 2x 3+( )-----------------------

353---

1x2 4x+x2 5x–-----------------

2x x 4+( )x x 5–( )--------------------

3x 4+x 5–------------

10WORKEDExample



Simplify each of the following algebraic fractions by first factorising the numerator and denominator.

a b c

THINK WRITE

a Write the numerator, making sure that the expression is in the correct order.

a x2 + 3x − 4

Find the factor pair of c (−4) which adds to b (3).

Factor sum of −4: −1 + 4 = 3

Write the expression and its factorised form.

x2 + 3x − 4 = (x − 1)(x + 4)

Write the original fraction.

Replace the numerator with the factorised form.

=

Cancel common factors and simplify if appropriate.

= x + 4

b Write the numerator, making sure that the expression is in the correct order.

b x2 − 7x − 8

Find the factor pair of c (−8) which adds to b (−7).

Factor sum of −8: 1 − 8 = −7

Write the expression in its factorised form, using the appropriate factor pair.

x2 − 7x − 8 = (x + 1)(x − 8)

Write the denominator, making sure that the expression is in the correct order.

x2 + 3x + 2

Find the factor pair of c (2) that adds to b (3).

Factor sum of 2: 1 + 2 = 3

Factorise using the appropriate factor pair.

x2 + 3x + 2 = (x + 1)(x + 2)


Replace the numerator and denominator with the factorised form.

=

Cancel any common factors and simplify if appropriate.

=

x2 3x 4–+x 1–

--------------------------- x2 7x– 8–x2 3x 2+ +--------------------------- x2 6x– 5+

2x2 16x– 30+-------------------------------------

1

2

3

4x2 3x 4–+

x 1–--------------------------

5x 1–( ) x 4+( )

x 1–----------------------------------

6

1

2

3

4

5

6

7x

2 7x– 8–

x2 3x 2+ +

---------------------------

8 x 1+( ) x 8–( )x 1+( ) x 2+( )

----------------------------------

9 x 8–x 2+------------

11WORKEDExample

Continued over page



Note: We cannot simplify a fraction like any further. It may be tempting to cancel

x in the numerator or denominator, but x is not a factor of x − 8 or x − 2 and hence wecannot cancel further. (The same applies to trying to cancel 8 and 2 by dividing by 2.)We can cancel only terms or expressions that are factors; that is, they are multiplied toanother term or expression.

Factorising will also help when we have to multiply or divide algebraic fractions.

Note: If 2 fractions are divided, you must change the division sign to a multiplicationsign and tip the second fraction (times and tip).

THINK WRITE

c Write the numerator, making sure that the expression is in the correct order.

c x2 − 6x + 5

Find the factor pair of c (5) which adds to b (−6).

Factor sum of 5: −1 + −5 = −6

Factorise using the appropriate factor pair. x2 − 6x + 5 = (x − 1)(x − 5)Write the denominator, making sure that the expression is in the correct order and take out the common factor of 2.

2x2 − 16x + 30 = 2(x2 − 8x + 15)

Find the factor pair of c (15) that adds to b (−8).

Factor sum of 15: −3 + −5 = −8

Factorise using the appropriate factor pair. 2x2 − 16x + 30 = 2(x − 3)(x − 5)


Replace the numerator and denominator with the factorised form.

=

Cancel any common factors and simplify if appropriate.

=

1

2

3

4

5

6

7x2 6x– 5+

2x2 16x– 30+------------------------------------

8x 1–( ) x 5–( )

2 x 3–( ) x 5–( )-------------------------------------

9x 1–

2 x 3–( )--------------------

x 8–x 2+------------

Simplify , by factorising the numerator and denominator and cancelling as

appropriate.THINK WRITE

Write the expression.

Factorise each numerator and denominator as appropriate and cancel common factors; one in any numerator and one in any denominator. The factors that can be cancelled are 3 and (x + 2).

=

Multiply numerators and multiply denominators and simplify. = 2(x − 1)

2x 4+3

--------------- 3x 3–x 2+

---------------×

12x 4+

3--------------- 3x 3–

x 2+---------------×

22 x 2+( )

3-------------------- 3 x 1–( )

x 2+--------------------×

3

12WORKEDExample



Simplifying algebraic fractions

1 Simplify each of the following by factorising the numerator and denominator andcancelling as appropriate.

a b c d

e f g h

i j k l

m n o p

2 Simplify each of the following fractions by factorising numerator and denominator andcancelling as appropriate.

a b c d

e f g h

i j k l

m n o p

3 Simplify each of the following algebraic fractions by first factorising the numerator andthe denominator.

a b c d

e f g h

i j k l

m n o p

rememberWhen simplifying algebraic fractions:1. factorise the numerator and the denominator2. cancel factors where appropriate3. if 2 fractions are multiplied, factorise where possible then cancel any factors,

one from the numerator and one from the denominator4. if 2 fractions are divided, remember to times and tip before factorising and

cancelling.

remember

6G

SkillSH

EET 6.4WWORKEDORKEDEExample

10a, b, c2a 2+

2--------------- 3a 6+

9--------------- 4a 4–

4--------------- 7 p 21–

7------------------

4x 8+8

--------------- 3x 12–6

------------------ 9x 54+3

------------------ 12x 96–4

--------------------- Mathcad

Simplifyingalgebraicfractions

12x 24–18

--------------------- 9x 45–15

------------------ 22x 10+------------------ 6–

6x 24–------------------

2x 2+3x 3+--------------- 3x 6–

4x 8–--------------- 5x 15–

6x 18–------------------ 9x 18+

3x 6+------------------


10dx2 3x+

x----------------- p2 5 p–

p------------------ y2 4y–

y----------------- a2 10a+

a---------------------

7x5x x2–----------------- 5b

5b b2–----------------- 9y

y2 9y–----------------- 4m

m2 4m–--------------------

7xx 2x2–----------------- 3m

2m2 m–-------------------- a 2a2–

a2 3a+------------------ 2b2 b+

5b 8b2–---------------------

3x2 x+9x x2–----------------- 7b 2b2–

b2 5b+--------------------- 9m2 18m+

12m2 3m–--------------------------- 32a 16a2–

24a 16a2+---------------------------


11x2 5x 6+ +

x 3+--------------------------- x2 7x 12+ +

x 4+------------------------------ x2 9x– 20+

x 5–----------------------------- b2 8b– 7+

b 1–---------------------------

a2 18a– 81+a 9–

--------------------------------- a2 22a– 121+a 11–

------------------------------------ x2 49–x 7–

----------------- m2 64–m 8–

------------------

a2 7a– 12+a2 16–

------------------------------ p2 4 p– 5–p2 25–

--------------------------- x2 6x 9+ +x2 2x 3–+--------------------------- m2 2m– 1+

m2 5m 6–+-----------------------------

y2 4y– 12–y2 36–

----------------------------- x2 4x– 4+x2 4–

-------------------------- x2 3x 40–+x2 6x 16–+----------------------------- x2 3x 18–+

x2 6x– 9+-----------------------------



4 Simplify each of the following by factorising the numerator and denominator and can-celling as appropriate.

a b c

d e f

g h

i j

k l

Equal or not equal?Algebraic terms have a different value depending on the value assigned to the pronumeral. Some algebraic terms however, will be equal regardless of the value of the pronumeral.

Are the expressions and equal for all values of x?

1 From our knowledge of numbers, we know that any number divided by itself will be equal to 1. The same will apply to algebraic terms. If two algebraic terms or expressions are equal, the result of division of one by another will be 1. Perform

the division by first fully factorising the expression.

2 Are the two expressions equal?

3 Check your answer by substituting the value x = 2 into each expression.

4 What happens when you attempt to substitute x = 1 into each expression?

What’s the problem?Each expression below has been factorised, but incorrectly! Identify what the problem is with each and then factorise the expression correctly. (If it cannot be factorised, write ‘Cannot be factorised’ as your answer.)

1 x2 + 7x – 8 = (x – 8)(x + 1)

2 x2 + 16 = (x + 4)(x + 4)

3 3x2 – 6x + 3 = 3x(x – 6)

4 x2 + 6x + 5 = (x + 2)(x + 3)

SkillSH

EET 6.5 WWORKEDORKEDEExample

1220x 40+

2x 8+--------------------- x 4+

x 2+------------× 9a 27+

3a 3–------------------ a 1–

a 3+------------× m 5+

m 2–------------- 8m 16–

2m 10+--------------------×

SkillSH

EET 6.6 q 3–q 1+------------ 15q 15+

5q 15–---------------------× 3x 6–

x 6+--------------- x 6+

4x 8–---------------× 3a 6+

2a 6–--------------- a 3+

8a 24–------------------÷

10x 5–4x 28+------------------ 20x 10–

6x 1+---------------------÷ m2 8m 15+ +

m2 7m 10+ +--------------------------------- m2 6m 8+ +

m2 10m 21+ +------------------------------------×

WorkS

HEET 6.3x2 6x 8+ +x2 5x 6+ +--------------------------- x2 8x 15+ +

x2 7x 12+ +------------------------------× x2 x 6–+

x2 5x 14–+----------------------------- x2 9x 14+ +

x2 x– 12–------------------------------×

y2 y 20–+y2 7y 10+ +------------------------------ y2 y– 6–

y2 3y– 4–--------------------------× x2 7x– 6+

x2 x 2–+-------------------------- x2 2x– 8–

x2 x– 12–--------------------------×

x 5+x2 x+-------------- x2 4x 5–+

x3 x–--------------------------

x 5+x2 x+-------------- x2 4x 5–+

x3 x–--------------------------÷



Copy the sentences below. Fill in the gaps by choosing the correct word, words or expression from the word list that follows.

1 The opposite operation to expanding is .

2 To factorise is to find the of two or more factors.

3 The number 3 has pairs of different factors.

4 The highest common factor of two or more numbers is the largest factorwhich divides into all of the given numbers without a .

5 Before factorising any expression, look for a factor.

6 To factorise 6abc + 9bcd − 12cdf you need to find the .

7 The expression x2 − 64 is an example of .

8 The difference of two squares rule is a2 − b2 = .

9 When factorising an expression with 2 terms, look for a common then try using the difference of two squares rule.

10 When factorising an expression with 4 terms, look for a common factorthen use factor .

11 To simplify an algebraic fraction, factorise the numerator and then cancel where appropriate.

12 To simplify 2 algebraic fractions multiplied together, factorise and denominators then cancel.

13 To simplify 2 algebraic fractions when dividing, change the ÷ to × and the second fraction upside down.

summary

W O R D L I S Ttwocommonproducttip

highest commonfactor

numeratorsremainder

difference of twosquares

denominatorfactorising

factor(a + b)(a − b)grouping



1 Factorise each of the following by finding common factors.a 6x + 12 b 6x + 12xy c 6x2 + 12x3yd 8x + 4p + 2pq e 8a2 − 4b f 22xy + 33ax − 55xpg 16x2 − 24xy h 50ab + 25bc − 15b2c i −2x − 4j −6xy + 3x k b2 − 3b + 4bc l 6p2q + 9p3q2 − 15pq2

2 Simplify each of the following using common factor techniques.a 5(x + y) − 4a(x + y) b 6x(b − 4c) + 5(b − 4c) c 7a(b + 5c) − 6c(b + 5c)d 15x(d − 2e) + 4x(d − 2e) e 15x(d + 2e) + 25xy(d + 2e) f 6x2(5y − 2) − (5y − 2)g 2x + 2y + ax + ay h 4x − ax + 8y − 2ay i 6xy + 4x − 6y − 4j x2 + ax − 2x − 2a k pq − r + p − rq l px + qy − py − qx

3 Factorise each of the following using the difference of two squares rule.a x2 − 64 b 81 − y2 c 121 − z2

d a2 − 144 e 49b2 − 1 f c2 − d2

g 4f 2 − 9g2 h 81h2 − 4k2 i l − 100m2

j (n + 1)2 − m2 k (p − 1)2 − 25 l (r − 1)2 − 4s2

4 Factorise each of the following quadratic trinomials.a c2 + 5c + 4 b x2 − 6x − 7 c x2 − 4x + 4d p2 + 10p − 24 e q2 + q − 42 f x2 − 2x − 24g y2 − 10y + 24 h x2 + 3x + 2 i c2 − 11c − 26j m2 − 7m + 10 k x2 + 6x − 27 l m2 + 24m + 44

5 Factorise each of the following.a 2a2 + 16a + 24 b 3b2 − 24b + 36 c 4c2 − 16c − 48 d 2d2 + 2d − 12 e 5e2 − 605 f 6x2 + 24x − 126g 2x2 + 3x + 1 h 3x2 − x − 2 i 6x2 + 13x + 6j 2x2 − 7x + 5 k 8x2 + 2x − 3 l 6x2 − 17x − 28

6 Factorise each of the following.a 3a + 9b b x2 − 16 c 11p + 33 + pq + 3qd m2 + 2m − 3 e ab + 2b + 9a + 18 f 2t2 − 8g 16e2 − 25 h h2 − 7h + 12 i x(x − 2) + 3(x − 2)j b2 + 7bc + 5b k (x − 2)2 − 16 l 2a2 + 3a − 5

7 Simplify the following fractions.

a b c d e f

g h i j k l

8 Simplify each of the following.

a × b ÷

c × d ÷

6A

CHAPTERreview

6B

6C

6D

6E

6F

6G2x 4–

2--------------- 14 p 7–

21------------------ 5

10y 25–--------------------- 16–

4d 12+------------------ 4x 6+

8x 12+------------------ 9q 12+

6q 8+------------------

x2 5x+2x

----------------- 8y2y2 4y+--------------------- p3 p2–

p4 p2+------------------ x2 3x–

4x x2–----------------- x2 8x– 15+

x 5–----------------------------- m2 m 6–+

m2 2m 8–+-----------------------------

6G4x 12–2x 2+------------------ x 1+

x 3–------------ 10x 20+

3x 3–--------------------- 5x 10+

6x 6+------------------

x2 7x 12+ +

x2 4x 3+ +

------------------------------ x2 6x– – 7

x2 2x 8–+

---------------------------- x2 25–

x2 4x 5–+

-------------------------- x2 6x– 5+

x2 5x 6–+

--------------------------testtest

CHAPTERyyourselfourself

testyyourselfourself

6


Technology

Ch 06