[PDF] Worksheet 2.6 Factorizing Algebraic Expressions

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Section1FindingFactors

numberofdierentways:

48=316=412=224

number23hasnofactors).Forexample: x where(x2)3isinfullyfactoredform.

3xy+9xy2+6x2y=3xy(1)+3xy(3y)+3xy(2x)

=3xy(1+3y+2x) todosomefactoring.

Example1

9a2b+3a2+5b+5b2a=3a2(3b+1)+5b(1+ba)

Example2

10x2+5x+2xy+y=5x(2x+1)+y(2x+1)LetT=2x+1

=5xT+yT =T(5x+y) =(2x+1)(5x+y)

Example3

x

2+2xy+5x3+10x2y=x(x+2y)+5x2(x+2y)

=(x+5x2)(x+2y) =x(1+5x)(x+2y)

Exercises:

(a)6x+24 (b)8x24x (c)6xy+10x2y (d)m43m2 (e)6x2+8x+12yx (f)8m212m+10m15 (g)x2+5x+2x+10 (h)m24m+3m12 (i)2t24t+t2 (j)6y215y+4y10

Section2Somestandardfactorizations

Recallthedistributivelawsofsection1.10.

Example1

(x+3)(x3)=x(x3)+3(x3) =x23x+3x9 =x29 =x232

Example2

(x+9)(x9)=x(x9)+9(x9) =x29x+9x81 =x281 =x292 Page2 example1,wehave A

2B2=x29

=(x+3)(x3) A

2B2=x281

=(x+9)(x9) A

2B2=(A+B)(AB)

(x+5)2=(x+5)(x+5) =x(x+5)+5(x+5) =x2+5x+5x+25 =x2+10x+25 =x2+2(5x)+52

Theperfectsquareiswrittenas:

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

Similarly,

(xa)2=(xa)(xa) =x(xa)a(xa) =x2axax+a2 =x22ax+a2

Forexample,

(x7)2=(x7)(x7) =x(x7)7(x7) =x27x7x+72 =x214x+49 Page3

Exercises:

(a)(x+2)(x2) (b)(y+5)(y5) (c)(y6)(y+6) (d)(x+7)(x7) (e)(2x+1)(2x1) (f)(3m+4)(3m4) (g)(3y+5)(3y5) (h)(2t+7)(2t7)

2.Factorizethefollowing:

(a)x216 (b)y249 (c)x225 (d)4x225(e)16y2 (f)m236 (g)4m249 (h)9m216

3.Expandthefollowingandcollectliketerms:

(a)(x+5)(x+5) (b)(x+9)(x+9) (c)(y2)(y2) (d)(m3)(m3)(e)(2m+5)(2m+5) (f)(t+10)(t+10) (g)(y+8)2 (h)(t+6)2

4.Factorizethefollowing:

(a)y26y+9 (b)x210x+25 (c)x2+8x+16 (d)x2+20x+100(e)m2+16m+64 (f)t230t+225 (g)m212m+36 (h)t2+18t+81 Page4

Section3IntroductiontoQuadratics

ax

2+bx+c

makeaboutthequadraticax2+bx+c:

1.aisthecoecientofthesquaredtermanda6=0.

2.bisthecoecientofxandcanbeanynumber.

anynumber.

Quadraticsmayfactorintotwolinearfactors:

ax

2+bx+c=a(x+k)(x+l)

Exercises:

(a)x23x+4 (b)4x2+6x1(c)x36x+2 (d) 1 x2+2x+1(e)x24 (f)6x2 Page5

Section4FactorizingQuadratics

(x+2)(x+4)=x2+4x+2x+8 =x2+6x+8 (x+5)(x3)=x23x+5x15 =x2+2x15 sowewrite x

2+7x+12=(x+3)(x+4)

Example1

:Factorizex2+9x+14. thenumbersthatdothisare2and7.Therefore x

2+9x+14=(x+2)(x+7)

andisshowntoequalx2+9x+14.

Example2

:Factorizex2+7x18. x

2+7x18=(x2)(x+9)

showntoequalx2+7x18. Page6

Exercises:

1.Factorizethefollowingquadratics:

(a)x2+4x+3 (b)x2+15x+44 (c)x2+11x26 (d)x2+7x30 (e)x2+10x+24(f)x214x+24 (g)x27x+10 (h)x25x24 (i)x2+2x15 (j)x22x15

1.Usingthe`ACE'method,orby

2.Usingthequadraticformula

thenyouwillbeabletofollowthistechnique. Page7

ExampleFactorize6x2x12

1:Multiplytherstterm6x2by

thelastterm(12)

2:Findfactorsof72x2which

addtox.

3:Returntotheoriginalex-

pressionandreplacexwith

9x+8x.

4:Factorize(6x29x)and(8x

12).

5:Onecommonfactoris(2x

3).Theotherfactor,(3x+4),

isfoundbydividingeachterm by(2x3).72x2 (9x)(8x)=72x2

9x+8x=x

6x2x12

=6x29x+8x12 =3x(2x3)+4(2x3) =(2x3)(3x+4)

6:Verifythefactorizationbyex-

pansion(3x+4)(2x3) =3x(2x3)+4(2x3) =6x29x+8x12 =6x2x12

Example3:Factorize4x2+21x+5.

1.Multiplyrstandlastterms:4x25=20x2

20xandx.

4x2+21x+5=4x2+20x+x+5

4x2+20x+x+5=4x(x+5)+(x+5)

Page8

5.Factorizefurther:

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

Exercises:

(a)2x2+11x+12 (b)3x2+16x+5 (c)6x2+17x+12 (d)2x2+9x+10 (e)12x2+11x+2(f)2x25x3 (g)3x210x8 (h)3x211x20 (i)5x2+17x+6 (j)10x2+19x+6

Section5Thequadraticformula

solutionstoax2+bx+c=0aregivenby x=bp b24ac 2a

Ifwelettherootsbekandl,say,then

k=b+p b24ac 2a l=bp b24ac 2a Then ax

2+bx+c=a(xk)(xl)

Example1

Page9 x=5p

524(1)(3)

2(1) =5 2p 13 2 sothatthetworootsare k 1=5+p 13

2andk2=5p

13 2 Then x

2+5x+3=(x5+p

13

2)(x5p

13 2)

Example2

:Factorize2x2x5. x=bp b24ac 2a 1p (1)242(5) 22
1p 41
4

Sothetwofactorsof2x2x5are

(x1+p 41

4)and(x1p

41
4) andsothefactorizationis

2x2x5=2

x1+p 41
4! x1p 41
4!

Page10

Exercises:

(a)3x2+2x4 (b)x2+3x+1 (c)2x2+8x+3 (d)3x2+5x+1 (e)3x2+6x+2(f)5x2+7x2 (g)3x2+5x4 (h)2x2+4x+1 (i)5x2+2x2 (j)2x2+x7

Section6Usesoffactorization

fractions.

Example1

x 29
x3=(x3)(x+3)(x3) x3 x3(x+3) =x+3

Example2

x x2+4x+4+xx+2=x(x+2)2+xx+2 x (x+2)2+xx+2x+2x+2 x (x+2)2+x2+2x(x+2)2 x2+x+2x (x+2)2 x(x+3) (x+2)2

Page11

Example3:Solve(x+3)2=x+5.

x

2+6x+9=x+5

x

2+5x+4=0

(x+4)(x+1)=0 following: a>0a<0 >0Therewillbe2distinctso- lutions,sothecurvecrosses thex-axistwice. 6 O- 6 ?W =0Therewillonlybeonesolu- tion,sothecurvewillonly touchthex-axis.Thisis calledadoubleroot. 6 66
6 <0Thecurvedoesnottouch thex-axis.Wewilldealwith thiscaseindetaillater. 6 66
6

Exercises:

(a) x2+3x x+3 (b) 6x28 2x (c) x2+3x+2 3x+6 (d) x27x18 x26x27 (e) x216 2x+8 (f)3x29x 18x

Page12

(g)x225x23x10 (h) 2x232 x2+6x+8 (i) x39x2 3x27 (j) 2x2x6 x2+x6 (a) 3 x+2+5xx+3quotesdbs_dbs17.pdfusesText_23

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