chapter 12: factorising algebraic expressions

Factorising algebraic expressions

12

contents12:01 Factorising using common factors: A review12:02 Factorising by grouping in pairs12:03 Factorising using the difference of two squares

Challenge 12:03 The difference of two cubes12:04 Factorising quadratic trinomials

Fun spot 12:04 How much logic do you have?12:05 Factorising further quadratic trinomials

Challenge 12:05 Another method for factorising harder trinomials

12:06 Factorising: Miscellaneous typesFun spot 12:06 What did the caterpillar say when it saw the butterfly?

12:07 Simplifying algebraic fractions: Multiplication and division

12:08 Addition and subtraction of algebraic fractions Maths Terms, Diagnostic Test, Assignments

syllabus references (See pages x–xv for details.)

Number and AlgebraSelections from Algebraic Techniques [Stages 5.2, 5.3§]• Factorisealgebraicexpressionsbytakingoutacommonalgebraicfactor(ACMNA230)• Factorisemonicandnon-monicquadraticexpressions(ACMNA269)

Working Mathematically• Communicating • ProblemSolving • Reasoning • Understanding • Fluency

Can you crack thecode, Mr. x?

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352 Australian Signpost Mathematics New South Wales 9 Stages 5.1–5.3

InChapter3,youexpandedalgebraicexpressionsthatwerewrittenasproductsinfactorisedform.Asshowninthefollowingexamples,eachproductwaswrittenwithoutgroupingsymbols.

3a(5 − 2a) = 15a −6a2

(a − 2)(a + 7) = a2 + 5a − 14 (x + 5)2 = x2 +10x + 25 (m + 2)(m − 2) = m2 − 4

Nowyouwillreversethisprocessandfactorisealgebraicexpressions.

12:01 Factorising using common factors: A review

Expand: 1 3(x + 7) 2 a(a − 1) 3 4p(2p − 3q) 4 −5m(3 − 2n)

WritetheHCF(highestcommonfactor)of: 5 10x and 15x 6 8x2and6xy

Factorisebytakingoutacommonfactor: 7 4n + 8 8 w2 −6w 9 ab + bc 10 −6y −9

Thesimplestmethodoffactorisingis‘takingout’acommonfactor.ThismethodwasmetinChapter3.

Thisisanimportantskillbecauseitisthefirstmethodthatshouldbe consideredwhenfactorisinganyexpression.Itisalsosometimesusedinconjunctionwithothermethods.

Checkyourcompetencewiththisskillbycompletingtheshortexercisebelow.

Exercise 12:01

1 Factorisethefollowingbytakingoutthehighestcommonfactor.a 2x + 8 b 10m + 5 c 9a +6 d 4 +6ye 7w − 21 f 15n −20 g 12 − 18t h 16− 8zi x2 + 5x j am + an k 6p − p2 l ab − a

2 Factorisefully.a 2ax +6ay b 5m2 + 5mn c 9ab +6bc d 4yz +6xye 7tw − 14t2 f 15n −10n2 g 12p − 8pq h 6fg − 8gh i x2y+ xy2 j am2 + amn k 6p −6p2 l abc − bc

prep QUiZ 12:01

See Section 3:05 Page 64

ProductLink resources

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35312 Factorising algebraic expressions

3 Factorisebytakingoutanegativecommonfactor.a −2x − 8 b −5m + 5 c −9a + 12 d −4 − 8y e −7w2 − w f −mn − np g −12t2 + 8t h −16xy + 24yz

4 Factorise:a 3x + x2 − ax b ax + ay + az c 4m − 8n +6pd 5ab − 15ac +10ad e x2 − 7x + xy f 10ab − 5bc − 15b2

5 Rememberingthata(b + 2) + 4(b + 2) = (b + 2)(a +4),factorisethefollowing.a 2(a + x) + b(a + x) b x(3 + b) + 2(3 + b) c a(a + 3) − 2(a + 3)d 3m(m − 1) + 5(m − 1) e x(2x+5) − (2x + 5) f 7(7 − z) − z(7 − z)

12:02 Factorising by grouping in pairs

Factorisetheseexpressions. 1 3a + 18 2 5x + ax 3 pq − px 4 3ax −9bx 5 x2 − 2x 6 a3 + a2 7 9− 3a 8 −5m −10 9 9(a + 1) + x(a + 1) 10 x(x + y) − 1(x + y)

Forsomealgebraicexpressions,theremay notbeafactorcommonineveryterm. Considertheexpression3x + 3 + mx + m.

Thereisnofactorcommonineveryterm,but:•thefirsttwotermshaveacommonfactorof3•thelasttwotermshaveacommonfactorofm.

3x + 3 + mx + m = 3(x + 1) + m(x + 1)

Nowitcanbeseenthat(x +1)isacommonfactorforeachterm.

3(x + 1) + m(x + 1) = (x + 1)(3 + m)

Therefore:

3x + 3 + mx + m = (x + 1)(3 + m)

Theoriginalexpressionhasbeenfactorisedbygroupingthetermsinpairs.

a 2x + 2y + ax + ay = 2(x + y) + a(x + y) = (x + y)(2 + a)

b a2 + 3a + ax + 3x = a(a + 3) + x(a + 3) = (a + 3)(a + x)

prep QUiZ 12:02

WORKED EXAMPLE 1

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a ax − bx + am − bm = x(a − b) + m(a − b) = (a − b)(x + m)

b ab + b2 − a − b = b(a + b) − 1(a + b) = (a + b)(b − 1)

c 5x + 2y + xy +10= 5x +10+ 2y + xy = 5(x + 2) + y(2 + x) = (x + 2)(5 + y)

ab + ac + bd + cd = a(b + c) + d(b + c) = (b + c)(a + d)

Exercise 12:02

1 Completethefactorisationofeachexpressionbelow.a 2(a + b) + x(a + b) b a(x + 7) + p(x + 7) c m(x − y) + n(x − y)d x(m + n) − y(m + n) e a2(2 − x) + 7(2 − x) f q(q − 2) − 2(q − 2)g (x + y) + a(x + y) h x(1 − 3y) − 2(1 − 3y)

2 Factorisetheseexpressions.a pa + pb + qa + qb b 3a + 3b + ax + bx c mn + 3np + 5m + 15pd a2 + ab + ac + bc e 9x2 − 12x + 3xy − 4y f 12p2 −16p + 3pq − 4qg ab + 3c + 3a + bc h xy + y + 4x + 4 i a3 + a2 + a + 1j pq + 5r + 5p + qr k xy − x + y − 1 l 8a − 2 + 4ay − ym mn + m + n + 1 n x2 + my + xy + mx o x2 − xy + xw − ywp x2 + yz + xz + xy q 11a + 4c + 44 + ac r a3 − a2 + a − 1

3 Factorisethefollowing.a xy + xz − wy − wz b ab + bc − ad − cd c 5a + 15 − ab − 3bd 6x − 24 − xy + 4y e 11y + 22 − xy − 2x f ax2 − ax − x + 1

Can you find a formula that calculates the day of the week for any given date? How many different calendars are needed to describe any given year?

WORKED EXAMPLE 2

Note: Terms had to be rearranged to pair those with common factors.

Foundation worksheet 12:02grouping in pairsProductLink resources

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12:03 Factorising using the difference of two squares

Simplify: 1 16 2 49 3 121

If xispositive,simplify: 4 2x 5 9 2x 6 64 2x

Expandandsimplify. 7 (x − 2)(x + 2) 8 (x + 5)(x − 5) 9 (7 − a)(7 + a) 10 (3m + 2n)(3m − 2n)

Iftheexpressionwewanttofactoriseisthedifference oftwosquares,wecansimplyreversetheprocedure seeninSection3:07B.

a x2 −9= x2 − 32

= (x − 3)(x + 3)b 25a2 − b2 = (5a)2 − b2

= (5a − b)(5a + b)

c a4 −64= (a2)2 − 82

= (a2 − 8)(a2 + 8)c 36m2 −49n2 =(6m)2 − (7n)2

=(6m − 7n)(6m + 7n)

x2 − y2 = (x − y)(x + y)

Note: (x − y)(x + y) = (x + y)(x − y)

Exercise 12.03

1 Factoriseeachoftheseexpressions.a x2 − 4 b a2 −16 c m2 − 25 d p2 − 81e y2 −100 f x2 − 121 g 9− x2 h 1 − n2

i 49− y2 j a2 − b2 k x2 − a2 l y2 − a2

m9a2 − 4 n 16x2 − 1 o 25p2 −9 p 49− 4a2

q 25p2 − a2 r m2 − 81n2 s 100a2 −9b2 t 81x2 − 121y2

prep QUiZ 12.03

expand

(x − a)(x + a) x2 − a 2

factorise

WORKED EXAMPLE 1

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2 Factorisebyfirsttakingoutacommonfactor.a 2x2 − 32 b 3x2 −108c 4a2 −100 d 5y2 −20e 24a2 −6b2 f 3x2 − 27y2

g 8y2 − 128 h 80p2 − 5q2

i 4x2 −64 j 3x2 − 3k 72p2 − 2 l 2 − 18x2

m 8a2 − 18m2 n 125 −20a2

o 200x2 − 18y2 p 98m2 − 8n2

•Thelargecubehasavolumeofa3cubicunits Itismadeupoffoursmallerparts (acubeandthreerectangularprisms). Ouraimistofindanexpressionforthe differenceoftwocubes(a3 − b3).

1 Completethetablebelow.

2 Writeanexpressionforthevolumeofthelargecube(a3)intermsofthevolumesofthefoursmaller parts. a3 = V 1 + V 2 + V 3 + V 4

3 UseyouranswertoQuestion 2towriteanexpression for a3 − b3.

Volume 1 Volume 2 Volume 3 Volume 4

b × b × b

Expresseachvolumeasaproductofitsfactors.

Difference of two cubesa3 − b3 = (a − b)(a2 + ab + b2)

Applyingthistoalgebraicexpressions,wecouldfactoriseadifferenceoftwocubes. e.g. x3 − 8 = x3 − 23

= (x − 2)(x2 + 2x + 4)

exercisesFactorisetheseexpressionsusingthedifferenceoftwocubesrule. 1 m3 − n3 2 x3 − y3 3 a3 − 8 4 m3 − 27 5 x3 −1000 6 y3 − 125 7 64− n3 8 27 − k3 9 8m3 − 27 10 64x3 − 125y3 11 125x3 − 8y3 12 27m3 − 343n3

18x2 −50=2(9x2 − 25) = 2([3x]2 − 52) = 2(3x − 5)(3x + 5)

WORKED EXAMPLE 2

cHallenge 12:03 tHe DiFFerence oF tWo cUbes (extension)

a

a

a

b

b

b

1 2

3

4a − b

a − b

a − b

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12:04 Factorising quadratic trinomials

Expand: 1 (x + 2)(x + 3) 2 (a − 1)(a + 3) 3 (m − 7)(m − 2) 4 (x + 5)2 5 (a − 2)2

Findtwonumbersa and bwhere: 6 a + b = 5 and ab =6 7 a + b =9andab =20 8 a + b = −2 and ab = −15 9 a + b = 3 and ab = −4 10 a + b = 7 and ab = −18

•Anexpressionwiththreetermsiscalledatrinomial.•Expressionslikex2 + 3x − 4 are called quadratic trinomials.

Thehighestpowerofthevariableis2.•Factorisingisthereverseofexpanding.

(x + a)(x + b) = x2 + ax + bx + ab = x2 + (a + b)x + ab

Usingthisresult,tofactorisex2 + 5x +6welookfortwovaluesa and b,wherea + b = 5 and ab =6. Thesenumbersare2and3,so: x2 + 5x +6= (x + 2)(x + 3)

Factorise: 1 x2 + 7x +10 2 m2 −6m + 8 3 y2 + y − 12 4 x2 −9x −36 5 3y2 + 15y − 72

solutions

1 2 + 5 = 7 2 × 5 =10

∴ x2 + 7x +10 = (x + 2)(x + 5)

2 (−2) + (−4) = −6 (−2) × (−4) = 8

∴ m2 −6m + 8 = (m − 2)(m − 4)

3 (−3) + 4 = 1 (−3) × 4 = −12

∴ y2 + y − 12 = (y − 3)(y + 4)

4 3 + (−12) = −9 3 × (−12) = −36

∴ x2 −9x −36 = (x + 3)(x − 12)

5 (−3) + 8 = 5 (−3) × 8 = −24

3y2 + 15y − 72 = 3(y2 + 5y − 24) = 3(y − 3)(y + 8)

prep QUiZ 12:04

2 and 3 add to give5 and multiply togive 6.

WORKED EXAMPLES

If x2 + 7x +10= (x + a)(x + b) thena + b = 7 and ab =10.

Step 1:Take out anycommon factor.

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Exercise 12:04

1 Factoriseeachofthesetrinomials.a x2 + 4x + 3 b x2 + 3x + 2 c x2 +6x + 5d x2 + 7x +6 e x2 +9x +20 f x2 +10x + 25g x2 + 12x +36 h x2 +10x + 21 i x2 +9x + 18j x2 + 14x +40 k x2 + 15x + 54 l x2 + 13x +36m x2 − 4x + 4 n x2 − 12x +36 o x2 − 7x + 12p x2 −9x +20 q x2 + 2x − 3 r x2 + x − 12s x2 + 4x − 12 t x2 + 7x −30 u x2 − x − 2v x2 −10x − 24 w x2 − 7x −30 x x2 − x −56

2 Factorise:a a2 +6a + 8 b m2 +9m + 18 c y2 + 13y + 42d p2 + 7p + 12 e x2 + 12x +20 f n2 + 17n + 42g s2 + 21s + 54 h a2 + 18a +56 i x2 − 3x − 4j a2 − 2a − 8 k p2 − 5p − 24 l y2 + y −6m x2 + 7x − 8 n q2 + 5q − 24 o m2 + 12m − 45p a2 + 18a −63 q y2 +6y − 55 r x2 − 2x + 1s k2 − 5k +6 t x2 − 13x +36 u a2 − 22a + 72v p2 + 22p +96 w q2 − 12q − 45 x m2 − 4m − 77

3 Factorisebyfirsttakingoutacommonfactor(SeeWorkedExample5).a 2x2 +6x + 4 b 3x2 −6x −9 c 5x2 −10x −40d 2x2 +16x + 32 e 3x2 −30x − 33 f 3x2 + 21x +36g 4a2 − 12a −40 h 2n2 + 8n +6 i 5x2 −30x +40j 3x2 − 21x +36 k 3a2 − 15a −108 l 5x2 + 15x −350

Seeifyoucansolvethethreeproblemsbelow.

1 Whatisthenextletterinthissequence? O,T,T,F,F,S,S,...?

2 Amanpassingapilotinanairportexclaimed,‘Iamthatpilot’sfather!’Butthepilotwasnottheman’sson.Howcanthisbe?

3 Twoguardsareguardingtwosacks.Oneguardalwaystellsthetruth,buttheotherguardalwayslies,butyoudonotknowwhichguardiswhich.Oneofthesacksisfullofgold;theotherisfullofpeanuts.Youarepermittedtotakeoneofthesacksbutyouarenotsurewhichonecontainsthegold. Youarealsoallowedtoaskoneoftheguardsjustonequestion. Whatquestionshouldyouasktoensureyougetthesackofgold?

Foundation worksheet 12:04Factorising trinomialsProductLink resources

FUn spot 12:04 HoW MUcH logic Do YoU HaVe?

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12:05 Factorising further quadratic trinomials

•Inallquadratictrinomialsfactorisedsofar,thecoefficient of x2has been1.Wewillnowconsidercaseswherethecoefficient of x2 is not 1.

To expand (5x − 1)(x +3)wecanuseacrossdiagram.

∴ (5x − 1)(x + 3) = 5x2 + 14x − 3•Onemethodusedtofactorisetrinomialslike5x2 + 14x − 3 is called

thecrossmethod.

cross methodTo factorise 5x2 + 14x −3,weneedtoreversetheexpandingprocess. Weneedtochoosetwofactorsof5x2andtwofactorsof−3towrite onthecross.

•If(5x − 3) and (x +1)arethefactorsof5x2 + 14x −3,thentheproductsofnumbersontheendsofeacharmwillhaveasumof+14x.

•Whenweaddthecross-productshere,weget:(5x) + (−3x) = 2x Thisdoesnotgivethecorrectmiddletermof14x, so (5x − 3) and (x + 1) are not factors.

•Varythetermsonthecross.

Try:5xx and

+1−3

Cross-product= (−15x) + (x) = −14x

Try:5xx and

−1+3

Cross-product= (15x) + (−x) = 14x ∴Thismustbethecorrectcombination. ∴ 5x2 + 14x − 3 = (5x − 1)(x + 3)

Examinetheexamplesonthenextpage.Makesureyouunderstandthemethod.

3x2 + 7x + 9

coefficient of x2

Remember5x2 −3

(5x − 1)(x + 3)

−x+15x

= 5x2 + 15x − x − 3= 5x2 + 14x − 3

5x

x

−1

+3

5x2istheproductofthetwoleftterms. −3istheproductofthetworightterms. 14xisthesumoftheproductsalongthecross. i.e. 15x + (−x)

Keeptrying.

5x

5x

5x

x

x

x

−1

+1

+1

+3

−3

−3

Try:5xx

and −3+1

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Findthefactorsof: 1 3x2 −19x +6 2 4x2 − x − 3 3 2x2 + 25x + 12

solutions 1

2

3 Inpractice,wewouldnotdrawaseparatecross foreachnewsetoffactors.Wesimplycrossout thefactorsthatdon’tworkandtryanewset.

alternative method: splitting the middle termTo factorise 5x2 + 14x −3wefirstmultiplythetwo‘outer’numberstogetherandnoteitabovethelastnumber.

i.e. 5x2 + 14x − 3

5 × −3 = −15

Wethenseektwonumbersaswedidformonictrinomials;thosethatmultiplytoequal−15 and addtoequal+14.Theonlypossiblenumbersare+15 and −1.

15 × −1 = −15 and 15 + −1 = 14

Wenow‘split’themiddleterm,+14xusingthesenumbers,i.e.+15x, −1xandrewritethetrinomialwithfourterms.

i.e. 5x2 + 14x − 3 = 5x2 + 15x − 1x − 3

Thisexpressionwenowfactorisebygroupinginpairs:

5x2 + 15x − 1x − 3 = 5x(x + 3) − 1(x + 3) = (x + 3)(5x − 1) ∴ 5x2 + 14x − 3 = (x + 3)(5x − 1)

WORKED EXAMPLES

Thiscross-productgivesthecorrectmiddle term of −19x. 3x2 −19x +6= (3x − 1)(x −6) Note:Thefactorsof+6hadtobebothnegativetogiveanegativemiddle term.

3x3x3x

xxx

−1

−2

−2

−6−3

−3

(−9x) (−11x) (−19x)

4x2x2x

x2x2x

+3

+1

+3

−1−1

−3

(−4x) (+4x)

4x

x

−3

+1

(+x) (−x)

∴ 4x2 − x − 3 = (4x + 3)(x − 1)

2x 3

4

4 2 6 12 1

3 6 2 1 12x

∴ 2x2 + 25x + 12 = (2x + 1)(x + 12)

To factorise a quadratic trinomial, ax2 + bx + c, when a (the coefficient of x2) is not 1, use the cross method.

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Factorise: 1 2x2 − 5x − 12 2 12a2 −29a + 15

solutions 1 2x2 − 5x − 12 = 2x2 − 5x − 12 (−8 × 3 = −24 and −8 + 3 = −5)

= 2x2 − 8x + 3x − 12 = 2x(x − 4) + 3(x − 4) = (x − 4)(2x + 3)

2 12a2 −29a + 15 = 12a2 −29a + 15 (−9× −20=180and−9+ −20= −29) = 12a2 −9a −20a + 15 = 3a(4a − 3) − 5(4a − 3) = (4a − 3)(3a − 5)

Exercise 12:05

1 a Whichdiagramwillgivethefactorsof2x2 + 13x +6? i ii iii iv

b Whichdiagramwillgivethefactorsof9x2 −9x −4? i ii iii iv

c Whichdiagramwillgivethefactorsof5x2 −19x +12? i ii iii iv

d Whichdiagramwillgivethefactorsof12x2 + 7x −10? i ii iii iv

WORKED EXAMPLES

−24

With this method, sometimes the product can be a big number.

180


2x +3

x +2

2x +2

x +3

2x +1

x +6

+6

x +1

2x

9x −4

+1x

9x −1

+4x

3x

3x

−2

+2

3x

3x

−4

+1

5x −3

−4x

5x −4

−3x

5x −2

−6x

5x −6

−2x

6x +10

−12x

12x −10

+1x

3x +5

−24x

4x +5

−23x

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2 Factorisetheseexpressions.a 2x2 + 7x + 3 b 3x2 + 8x + 4 c 2x2 + 7x +6d 2x2 + 11x + 5 e 3x2 + 5x + 2 f 2x2 + 11x + 15g 4x2 + 13x + 3 h 5x2 + 17x +6 i 2x2 + 13x + 15j 2x2 − 5x + 2 k 3x2 − 11x +6 l 5x2 − 17x +6m 4x2 − 11x +6 n 10x2 − 21x +9 o 5x2 − 22x + 21p 2x2 + x −10 q 3x2 + 4x − 15 r 4x2 + 11x − 3s 2x2 − x −6 t 2x2 − 5x − 3 u 3x2 − x −30v 6x2 − 5x − 21 w 2x2 − 5x − 12 x 4x2 − x − 18

3 Findthefactorsofthefollowing.a 12x2 + 7x + 1 b 6a2 + 5a + 1 c 6p2 + 7p + 2d 10y2 −9y + 2 e 12x2 − 7x + 1 f 9a2 − 21a +10g 8m2 + 18m − 5 h 6n2 − 7n − 3 i 21q2 −20q + 4j 20x2 − x − 1 k 8m2 − 2m − 15 l 18y2 − 3y −10m6a2 + 5a −6 n 15k2 +26k + 8 o 8x2 + 18x +9p 4 − 3a − a2 q 2 + m −10m2 r 6+ 7x − 3x2

s 6− 7x − 3x2 t 15 − x − 28x2 u 2 +9n − 35n2

v 3x2 +10xy + 8y2 w 2x2 − 5xy + 2y2 x 5m2 − 2mn − 7n2

4 Factorisebyfirsttakingoutthecommonfactor.a 6x2 +10x − 4 b 6a2 − 2a − 4 c 6a2 +9a − 27d 8x2 + 12x −36 e 6x2 + 28x +16 f 12p2 + 12p −9g 30q2 + 55q − 35 h 10m2 −46m + 24 i 50a2 + 15a − 5j 4 −6x −10x2 k 36− 3t − 3t2 l 9+ 24x + 12x2

5 Completeeachinasmanywaysaspossiblebywritingpositivewholenumbersintheboxesand inserting operation signs.a (x … )(x … ) = x2 … x … 15b (x … )(x … ) = x2 … x − 12c (x … )(x … ) = x2 … 5x + d (5x … )(x … ) = 5x2 … x … 2

This is an exercise you can sink your teeth into!

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12:06 Factorising: Miscellaneous types

1 4x2 −36

= 4(x2 −9) common factor = 4(x − 3)(x + 3) differenceoftwosquares

2 15x2y −20xy +10xy2

= 5xy(3x − 4 + 2y) common factor

3 8x2 −40x + 32 = 8(x2 − 5x + 4) common factor = 8(x − 4)(x − 1) quadratictrinomial

4 12 − a −6a2 = (3 + 2a)(4 − 3a) quadratictrinomial

5 ap − aq − 3p + 3q

= a(p − q) − 3(p − q) groupinginpairs = (p − q)(a − 3)

Exercise 12:06

1 Factoriseeachoftheseexpressions.a x2 −6x + 5 b x2 −9 c xy + 2y +9x + 18d a2 −9a e a2 −6a +9 f 4x2 − 1g 12x2 − x − 35 h a2 − 13a +40 i 5a2b −10ab3

j p2 − q2 k pq − 3p +10q −30 l 7x2 + 11x −6m a2 + 3a − ab n 16− 25a2 o 1 − 2a − 24a2

p 4m + 4n − am − an q 5ay −10y + 15xy r 15x2 − x − 28s x2y2 − 1 t x2 − x −56 u 2mn + 3np + 4m +6p v 100a2 −49x2 w 2 − 5x − 3x2 x k2 + 2k − 48

2 Factorisecompletely.a 2 − 8x2 b 5x2 −10x − 5xy +10y c 2a2 − 22a + 48d 3m2 − 18m + 27 e x4 − 1 f p3 − 4p2 − p + 4g 4x2 −36 h a3 − a i 3a2 −39a +120j 9−9p2 k 3k2 + 3k − 18 l 24a2 − 42a +9m ax2 + axy + 3ax + 3ay n (x + y)2 + 3(x + y) o 5xy2 −20xz2

p 6ax2 + 5ax −6a q x2 − y2 + 5x − 5y r 3x2 − 12x + 12s 63x2 − 28y2 t a4 −16 u (a − 2)2 − 4v 1 + p + p2 + p3 w 8t2 − 28t −60 x 8 − 8x −6x2

1 Take out any common factors.2 If there are two terms, is it a difference of two

squares? a2 − b2.3 If there are three terms, is it a quadratic trinomial?

ax2 + bx + c.4 lf there are four terms, can it be factorised by

grouping in pairs?

When factorising anyalgebraic expressions,

remember thischecklist . . .

WORKED EXAMPLES

6 −6a a −3a 2a

a −6a 2a −3a2

3

4


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Answereachquestionandwritetheletterforthat questionintheboxabovethecorrectanswer.

For the number plane on the right, find: E theequationofthex-axis E thedistanceBC E themidpointofAB E theequationofAB F thegradientof DF H theintersectionofDF and EF I thedistanceofFfromtheorigin G theequationofthey-axis E they-interceptofDF E thedistanceAB H theequationofDF G thegradientofAB I theintersectionofAB and BC.

Simplify: I 10x2 + x2 L 10x2 − x2 L 10x2 × x2

M 10x2 ÷ x2 N 14 of 8x4 N 5%of40x

N (2x2)3 N +2 2x x

Aplayingcardischosenatrandomfromastandardpack. Find the probability that it is: O theAceofspades P aheart R a King S apicturecard T aredcardgreaterthan3butlessthan9.

Expand and simplify: O (x + 1)(x + 7) O (a − b)(a + b) S (a + b)2 T (a − b)2

On this number plane, find the length of: T OC U OA U AC V BC W OB Y AB O Whatistheareaof∆ABC?

FUn spot 12:06 WHat DiD tHe caterpillar saY WHen it saW tHe bUtterFlY?

y

x

5

4

3

2

1

−1

−2

A

D F

E

C

B

−1 10 2 3 4 5 6 7 8−2

y

x

A

CB

40

30−90 0

40 90 5

120 50310

11x2

x2 + 8

x +

7

y =

x +

1

a2 − 2

ab +

b2

a2 + 2

ab +

b2

y =

3

8x6

2x x

2400

uni

ts2

( ,

1 ) 0 30

(8, 3

)

(−2,

−1) 1 3 13

1 21 2 1 52

1 45 261 13

y =

0

x =

0

2x4

10x4

9x2

a2 − b

2

√73

√970

0

√50

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12:07 Simplifying algebraic fractions: Multiplication and division

Simplifythefollowing:

1 510

2aa

2 12

8

2

2xy

x y 3 ×2

36xx

4 ÷23

49

x x

Factorise: 5 6x2 +9x 6 x2 + 7x + 12 7 x2 −49 8 3x2 +6x + 3 9 3x + 3y + ax + ay 10 2x2 +9x − 5

Justasnumericalfractionscanbesimplifiedbycancellingcommonfactorsinthenumeratorandthedenominator,soalgebraicfractionscanoftenbesimplifiedinthesamewayafterfirstfactorisingwherepossible.

a 2 2

3 3

2 ( 1)

3 ( 1)

2

3

1

1

xx

x

x−−

=−−

=

b + +

−=

+ +− +

= +−

7 12

9

( 4)( 3)

( 3)( 3)

43

2

2

1

1

x x

x

x x

x x

xx

c 6 6

3 3

6(1 )(1 )

(3 )(1 )

6(1 )

3

2 1

1

aa x ax

a a

x a

ax

−+ + +

=− ++ +

= −+

d −−

=−

− +

=+

3 9

3 27

3 ( 3)

3 ( 3) ( 3)

3

2

2

1

1

x x

x

x x

x x

xx

Algebraicfractionsshouldalsobefactorisedbeforecompletingamultiplicationoradivisionbecausethecancellingofcommonfactorsoftensimplifiestheseprocesses.

prep QUiZ 12:07

WORKED EXAMPLE 1Simplifying looks simple.

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a ++

× + = ++

×+

= +

5 151

2 25

5( 3)( 1)

2 ( 1)

5

2( 3)

1

1

1

1

xx

x xx

x

x

b −+ +

× +− −

=− ++ +

×+

− +

=+

9

5 6

3 6

2 3

( 3) ( 3)

( 2) ( 3)

3 ( 2)

( 3)( 1)

31

2

2 2

1 1

1 1

1

1

x

x x

x

x x

x x

x x

x

x x

x

c −−

÷ −−

=−

−×

−

−

=

6 143 9

3 75 15

2 (2 7)

3 ( 3)

5 ( 3)

(3 7)

103

or 313

1

1

1

1xx

xx

x

x

x

x

d −−

÷ − −+ +

=− +

− −×

+ +− +

= + +− +

16

25

2 8

10 25

( 4)( 4)

( 5)( 5)

( 5)( 5)

( 4)( 2)

( 4)( 5)( 5)( 2)

2

2

2

2

1

1

1

1

a

a

a a

a a

a a

a a

a a

a a

a aa a

Tosimplifyalgebraicfractions,factorisebothnumeratoranddenominatorwherepossible,andthencancel.

Exercise 12:07

1 Factoriseandsimplify.

a +5 105

x b

+4

2 6x c

−12

3 9x

d −−

2 105

xx

e ++

73 21xx

f −−

5 58 8aa

g +

+3 96 18

aa

h −−

7 283 12mm

i +−

2

2x x

x x

j −

−42

2xx

k +−1

12a

a l

−+

4 94 6

2yy

m −−

4

3

2

2a a

a a n

−−

2 22 2

2xx

o −−

363 18

2xx

p − −+3 4

1

2a aa

q − +−6 9

3

2x xx

r −

+ +4

3 2

2

2x

x x

WORKED EXAMPLE 2

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s + ++ +

3 2

5 6

2

2x x

x x t

+ −− +

5 24

7 12

2

2m m

m m u

+ +−

7 12

9

2

2t t

t

v −

+ + +3 3

2 2

2a x

a a ax x w

− −−

2 1

4 1

2

2x x

x x

−+ −

18 8

6 2

2

2a

a a

2 Simplifythefollowing.

a + ×

+2 4

36

2x x

x b

−+

× +5 152 8

410

yy

y

c −−

× −−

2 43 9

5 157 14

xx

xx

d ++

× ++

5 103

6 184 8

nn

nn

e + ×

+7 28

216

6 24y

y f

++

× +−

1 210 30

6 181 2

aa

aa

g ++

× ++2 8

4 63 3

2y yy

yy

h − × +

−3 2 5

9 27

2

2

2x x

x

x xx

i +−

× −+

3

9

312

x

x

xx

j +−

× −+

3 15

25

493 212

2x

x

xx

k + +−

× − −−

5 6

4

2

1

2

2

2

2a a

a

a a

a l

+ ++ +

× + ++ +

3 2

5 6

7 12

5 4

2

2

2

2y y

y y

y y

y y

m + ++ +

× + ++ +

6 5

5 4

7 12

12 35

2

2

2

2x x

x x

x x

x x n

−− +

× − +−

1

6 5

10 25

25

2

2

2

2m

m m

m m

m

o −

+ −× −

+ −4

3 4

16

2 8

2

2

2

2a

a a

a

a a p

+ +−

× + −+

2 4 2

1

3 44 4

2

2

2x x

x

x xx

q + +− −

× + −+ +

3 5 2

2

6

3 11 6

2

2

2

2x x

x x

x x

x x r

+ +−

× −+ −

5 16 3

25 1

5

2 5 3

2

2

2

2a a

x

a a

a a

s − + −− +

× −+ +2

10 105 5 5

2 2

2 2x y x y

x xy y

x yx y

t + −

+ + +× + − −

+ +( )2 2

2

2a b c

a ab ac bc

a ab ac bca b c

3 Simplify:

a + ÷ +3 62

24

a a b

+ ÷ +25

7 410

xx

xx

c −+

÷ −+

5 101

3 63 3

mm

mm

d +−

÷ +−

6 92 8

2 33 12

mm

mm

e +

÷ ++

35 15 3

2xx

x xx

f −+

÷ −+

24 164 6

3 28 12

yy

yy

g −+

÷ −+

5 204 6

5 20

2 32mm

m

m m h

+−

÷ +25 153 3

5 33

kk

kk

i −+

÷ +92 4

32

2nn

n j

+−

÷ −−

77

49

7

2

2yy

y

y y

k + +−

÷ −− −

5 4

16

9

12

2

2

2

2a a

a

a

a a l

+ ++ +

÷ + ++ +

6 9

8 15

5 6

7 10

2

2

2

2x a

x x

x x

x x

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m −

− +÷ − −

− −4

7 10

6

3 10

2

2

2

2x

x x

x x

x x n

+ +− −

÷ + −+ −

7 10

2 8

2 15

12

2

2

2

2p p

p p

p p

p p

o −−

÷ + +− +

49

9

14 49

6 9

2

2

2

2n

n

n n

n n p

− −+ +

÷ − ++ −

2 8 42

6 9

9 14

6

2

2

2

2x x

x x

x x

x x

q −

− −÷ +

−3 48

3 4

42

2

2

3x

x x

x x

x x r

− −−

÷ + −+ −

2 1

1

6 1

3 2 1

2

2

2

2a a

a

a a

a a

s + + −+ +

÷ + −+2

12 2

2 2

2 2x y x y

x xy y

x yx y

t − +

+ − −÷ − −

− − +( )2 2

2 2p q r

p pq pr qr

p q r

p pq pr qr

12:08 Addition and subtraction of algebraic fractions

Simplify:

1 +12

35

2 +34

38

3 −910

35

4 −715

320

5 +52

7x x

6 +2 12a a

7 +23

32a a

8 −1 14x x

9 +2

2ax

ax

10 −52

43

mn

mn

Theprepquizaboveshouldhaveremindedyouthatthelowestcommondenominatorneedstobefoundwhenaddingorsubtractingfractions.Ifthedenominatorsinvolvetwoormoreterms,factorisingfirstmayhelpinfindingthelowestcommondenominator. Forexample:

−+

+ +=

− ++

+ +

= + + −− + +

= + + −− + +

= −− + +

2

9

5

5 6

2( 3)( 3)

5( 3)( 2)

2( 2) 5( 3)( 3)( 3)( 2)

2 4 5 15( 3)( 3)( 2)

7 11( 3)( 3)( 2)

2 2x x x x x x x

x xx x x

x xx x x

xx x x

prep QUiZ 12:08

LCDstands for

lowestcommon

denominator.

Here, LCD = (x − 3)(x + 3)(x + 2).Note that the factors of each denominator are present without repeating any factor common to both. Each numerator is then multiplied by each factor not present in its original denominator.

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Whenaddingorsubtractingfractions:• factorisethedenominatorofeachfraction• findthelowestcommondenominator• rewriteeachfractionwiththiscommondenominatorandsimplify.

1 2

2

1

3

2( 3) 1( 2)

( 3)( 3)

2 6 2

( 2)( 3)

3 8

( 2)( 3)

x xx xx x

x xx x

xx x

++

+= + + +

− +

= + + ++ +

= ++ +

2 3

2 1

4

3 1

3(3 1) 4(2 1)

(2 1)(3 1)

9 3 8 4

(2 1)(3 1)

7

(2 1)(3 1)

x xx xx x

x xx x

xx x

+−

−= − − +

+ −

= − − −+ −

= −+ −

3 1

5 6

2

31

( 2)( 3)

2

( 3)

1 2( 2)

( 2)( 3)

1 2 4

( 2)( 3)

2 5

( 2)( 3)

2x x x

x x x

xx x

xx x

xx x

+ ++

+

=+ +

++

= + ++ +

= + ++ +

= ++ +

4 4 3

14

( 1)

2

( 1)( 1)

4( 1) 3

( 1)( 1)

4 4 3

( 1)( 1)

4

( 1)( 1)

2 2x x x

x x x x

x xx x x

x xx x x

xx x x

+−

−

=+

+− +

= − −+ −

= − −+ −

= −+ −

5 3

2 1

1

23

( 1)( 1)

1

( 1)( 2)

( 3)( 2) ( 1)( 1)

( 1)( 1)( 2)

6 ( 1)

( 1)( 1)( 2)

5

( 1) ( 2)

2 2

2 2

2

x

x x

x

x xx

x xx

x x

x x x xx x x

x x xx x x

x

x x

++ +

− −− −

= ++ +

− −+ −

= + − − − ++ + −

= + − − −+ + −

= −+ −

WORKED EXAMPLES

No factorising was needed in these first two examples.

Factorisefirst.

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Exercise 12:08

1 Simplifyeachofthefollowing.(Note:Nofactorisingisneeded.)

a +

+−

11

11x x

b +

++

15

13a a

c −

−+

17

11y y

d +

++

23

35x x

e +

−−

51

32m m

f +

−+

610

32t t

g −

+−

12 1

31x x

h +

−+

93 2

72 5x x

i −

++

85 1

73 1x x

j ++

32

57x x

k +

−92 5

53x x

l −+

12

32 1a a

m +

++3 1

xx

xx

n +

−−2 12

4 11a

aa

a o

++

+ ++

12

21

xx

xx

2 Simplify.(Note:Thedenominatorsarealreadyfactorised.)

a + +

++

1( 1)( 2)

12x x x

b +

++

1( 2)

12x x x

c +

−+

13

1( 3)x x x

d −

−− +

15

1( 5)( 2)x x x

e + +

++

3( 2)( 3)

42x x x

f +

−+ +

54

3( 1)( 4)x x x

g + +

++ +

1( 1)( 2)

1( 2)( 3)x x x x

h − +

++ +

2( 3)( 3)

4( 3)( 1)x x x x

i + −

++ +

3( 7)( 1)

5( 7)( 1)x x x x

j + +

−+ −

9( 9)( 3)

7( 3)( 1)x x x x

k + +

++ +

1(2 1)( 5)

3( 5)( 2)x x x x

l − +

−−

5(2 1)(3 2)

6(2 1)x x x x

m −

+ ++ +

+ −1

( 3)( 1)1

( 3)( 1)x

x xx

x x n

++

− −+

2( 3)

1( 2)

xx x

xx x

3 Simplify,byfirstfactorisingeachdenominatorwherepossible.

a +

++

1 112x x x

b +

−+

13 9

13x x

c +

++

22 3

34 6x x

d −

+−

5

1

312x x

e −

+−

1

9

12 62x x

f +

−−

1 1

12 2x x x

g + +

+−

1

2 1

1

12 2x x x h

+ ++

+ +1

7 12

1

8 162 2x x x x

i + +

++ +

2

6 8

4

5 62 2x x x x j

+ ++

+ +2

7 12

4

5 42 2x x x x

k − −

−− −

3

2

4

2 32 2x x x x l

− −−

− −3

6

2

2 32 2x x x x

Foundation worksheet 12:08addition and subtraction of algebraic fractions

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m − −

−− −

5

3 4

3

22 2x x x x n

+ −−

+ +3

2 7 4

4

3 14 82 2x x x x

o −

+− −

2

49

4

4 212 2x x x p

+ −−

−4

2 1

1

12 2x x x

q +

+ ++ −

−1

5 6

1

92 2x

x x

x

x r

+−

− +−

3

16

2

42 2x

x

x

x x

s −

+ ++ +

2

5 20

1

4 42 2x

x

x

x x t

+− −

+ −−

5 2

2 5 3

3 1

4 12 2x

x x

x

x

binomial• analgebraicexpressionconsistingoftwo

terms e.g. 2x + 4, 3x − 2y

coefficient• thenumberthatmultipliesapronumeral

inanalgebraicexpression e.g. In 3x − 5y: -thecoefficientofx is 3 - thecoefficientofy is −5

expand• toremovegroupingsymbolsby

multiplyingeachterminsidegroupingsymbolsbythetermortermsoutside

factorise• towriteanexpressionasaproductofits

factors• thereverseofexpanding

product• theresultofmultiplyingtermsor

expressionstogetherquadratic trinomial• expressionssuchasx2 + 4x + 3• thehighestpowerofthevariableis2

trinomial• analgebraicexpressionconsistingofthree

terms

This spiral or helix is a mathematical shape.Discover how it can be drawn.Investigate its links to the golden rectangle.

MatHs terMs 12

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Eachpartofthistesthassimilaritemsthattestaparticularskill.Errorsmadewillindicateareasofweakness.Eachweaknessshouldbetreatedbygoingbacktothesectionlisted.

1 Factorisebytakingoutacommonfactor.a 3x − 12 b ax + ay c −2x −6 d ax + bx − cx

12:01

2 Factorisebygroupinginpairs.a ax + bx + 2a + 2b b 6m +6n + am + anc xy − x + y − 1 d ab + 4c + 4a + bc

12:02

3 Factorisethesedifferencesoftwosquares.a x2 − 25 b a2 − x2 c 4 − m2 d 9x2 − 1

12:03

4 Factorisethesetrinomials.a x2 + 7x + 12 b x2 − 5x +6 c x2 − 3x −10 d x2 + x −20

12:04

5 Factorise:a 2x2 + 11x + 5 b 3x2 − 11x +6 c 4x2 − x − 18 d 6x2 + 5x + 1

12:05

6 Simplify,byfirstfactorisingwherepossible.

a +6 126

x b

−−

12 1814 21

aa

c ++

55

2x xax a

d + −

−3 10

4

2

2x x

x

12:07

7 Simplify:

a + ×

+3 6

48

2x x

x b

+ +−

× −+ +

5 6

9

1

3 2

2

2

2

2a a

a

a

a a

c −+

÷ −+

3 63

5 103 9

mm

mm

d − −− −

÷ − +−

3 10

6

7 10

4

2

2

2

2x x

x x

x x

x

12:07

8 Simplify:

a +

+−

23

11x x

b +

−+ +

1( 2)

1( 2)( 1)x x x x

c −

+−

5

9

32 62x x

d + +

− ++ −7 12

2

2 32 2x

x x

x

x x

12:08

Diagnostic test 12 Factorising algebraic expressions

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assignMent 12A Chapter review 1 Factorisethefollowingexpressions.

a a2 +9a +20b 2p − 4qc m2 − 4m − 45d 5x3 +10x2 + x + 2e 4x2 − 1f x2y − xyg 6a2 − 13a + 5h x2 + x −30i 3a2 − 4a − 15j xy + xz + py + pzk 2x2 + x − 1l x3 − 3x2 + 2x −6m −5ab −10a2b2

n x2 − y2 + 2x − 2yo 2 − 3x −9x2

2 Factoriseeachexpressionfully.a 2y2 − 18b 3r2 +9r − 84c 4x3 +6x + 4x2 +6d 2 − 18x2

e a3 + a2 − 72af 33 +36a + 3a2

g (x − y)2 + x − yh (x − 2)2 − 4

3 Simplifyeachofthefollowing.

a + −

−9 36

9

2

2x x

x

b −

+ −20 5

2 5 3

2

2x

x x

c +

++

32

23x x

d −

−−12

2x

xx

x

e − × +

+ +1

5 2 1

2 2

2x

xx x

x x

f −−

÷ − +− −

1

4

4 3

62

2

2x

x

x x

x x

g ++

− ++

12

21

xx

xx

h −

+−

23 1

1

(3 1)2x x

i +

−+

43 2

32 3x x

j + −

−× + +

−5 14

5 20

4 4

49

2

2

2

2x x

x

x x

x

8

7

6

5

4

3

2

1

A B C D E F G H

Chess is played on an 8 by 8 square grid. Each square is named using a letter and a number. The Knight pictured is standing on the square A1.

A Knight can move 3 squares from its starting position to its finishing position. The squares must form an ‘L’ shape in any direction. Some possible moves are shown below.

Which squares can the Knight move to from square A1?

If the Knight was standing on the square C1 what squares could it move to?

Give a sequence of squares showing how the Knight could move from A1 to B1 to C1 to … H1.

Finish Start

Start

Start Start

Finish

Finish

Finish

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assignMent 12B Working mathematically 1 Whatisthesmallestpossiblenumberof

childreninafamilyifeachchildhas:a atleastonebrotherandatleast

one sisterb atleasttwobrothersandatleast

twosisters?

2 Ifyouturnaright- handedgloveinsideoutisitstillaright-handedgloveorhasitbecomealeft-handedglove?

3 Iftheexteriorangles x°, y° and z° of a triangleareintheratio4:5:6,whatistheratiooftheinterior angles a°, b° and c°?

4 Theaverageoffivenumbersis11.Asixthnumberisaddedandthenewaverageis12.Whatisthesixthnumber?

5 Thissectorgraph showsthemethod of travelling to workforall persons.a Whatpercentage

oftheworkforce caughtatrain towork?

b Whatpercentageoftheworkforcewasdriventowork?

c Whatisthesizeofthesectoranglefor‘other’meansoftransport?Donotusea protractor.

d Whatpercentageoftheworkforceusedacartogettowork?

6 Usethisgraphtoanswerthefollowingquestions.a Whohasthegreaterchanceofhavingheartdisease:a60-year-oldwomanora

60-year-oldman?b Whohasthegreaterchanceofhavingcancer:a50-year-oldwomanora50-year-oldman?c Whichofthethreediseasesrevealsthegreatestgenderdifferenceforthe20-to-50-year-

oldrange?d Wouldthenumberof80-year-oldmensufferingfromheartdiseasebegreaterorlessthan

thenumberof80-year-oldwomensufferingfromheartdisease?Giveareasonforyouranswer.

MALE FEMALE

Health risks

15

12

9

6

3

0

0 10 20 30 40 50 60 70 80 Age

0 10 20 30 40 50 60 70 80

Source: Australian Institute of Health and Welfare

Per cent

Heart disease

Heartdisease

Cancer

CancerDiabetes

Diabetes

b c z

ax

y

Bus40°

Train36°

Walked36°

Car(passengers)45°

Car(driver)180°

Other

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assignMent 12C Cumulative review 1 Solvetheinequalities.

a 2a − 7 >10 b −3x > 12 c − ≥35

12m 7:06

2 Underafringebenefitstaxagreementanemployeecanreducetheirtaxableincomebyputting30%oftheirsalaryintoanexpenseaccount.Iftheirsalaryis$55200perannum,find:a howmuchcanbepaidintotheexpenseaccountb thetaxsavingifeverydollarthatgoesintotheexpenseaccountreducesthetax

payableby32⋅5 cents.

2:01

3 a Findthelengthoftheintervaljoiningthepoint(1,−1)tothepoint(−3, 2).b Findthex-andy-interceptsoftheline3x + 2y =9.c Whatistheslopeofthelinethathasanx-interceptof−2 and a y-intercept

of −4?

9:01,9:04,9:03

4 Calculatethevolumeofthefollowingprism. (Measurementsareincentimetres.)

5:05

5 Jimbuys2000sharesintheWeDiGiTminingcompanyatacostof$2.10pershare.After3monthshesells500sharesat$2.50pershare.Hethenwaitsforthemarkettoimproveandthensellsanother750sharesfor$3.20pershare.FindJim’sprofitandthepercentagereturnontheoriginalcostofthose1250shares.

8:05

6 Hannahearns$34.75perhour.Find:a hergrossweeklywageifsheworksa38hourweekb hernetweeklywageifshehasthefollowingdeductions:PAYG$287,

Superannuation$223,Unionmembership$5.20,Healthinsurance$69.75,CreditUniondeduction(savings)$200.

8:03

7 Findtheequationofalineinthenumberplanewhich:a hasagradientof3anday-interceptat(0,4)b hasagradientof−2andpassesthroughthepoint(3,−1)cpassesthroughthetwopoints(−2, 5) and (4, −7).

9:05,9:06,9:07

8 Solvethesepairsofsimultaneousequations.a 3x + 4y =6 and 5x − 3y =39b y = 7 − 3x and 5x − 2y + 25 =0

10:02

5·5

2·2

4·8

1·8 5·4

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chapter 12: factorising algebraic expressions

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