Chapter2-FactorTheoremand

Inequalities

LessonPackage

MHF4U

Chapter2OutlineUnitGoal:Bytheendofthisunit,youwillbeabletofactorandsolvepolynomialsuptodegree4usingthefactortheorem,longdivision,andsyntheticdivision.Youwillalsolearnhowtosolvefactorablepolynomialinequalities.

Section Subject LearningGoals CurriculumExpectations

L1 LongDivision -dividepolynomialexpressionsusinglongdivision-understandtheremaindertheorem C3.1

L2 SyntheticDivision -dividepolynomialexpressionsusingsyntheticdivision 3.1

L3 FactorTheorem-beabletodeterminefactorsofpolynomialexpressionsbytestingvalues

C3.2

L4 SolvingPolynomialEquations

-solvepolynomialequationsuptodegree4byfactoring-makeconnectionsbetweensolutionsandx-interceptsofthegraph

C3.2,C3.3,C3.4,C3.7

L5 FamiliesofPolynomialFunctions

-determinetheequationofafamilyofpolynomialfunctionsgiventhex-intercepts

C1.8

L5 SolvingPolynomialInequalities

-Solvefactorablepolynomialinequalities C4.1,C4.2,C4.3

Assessments F/A/O MinistryCode P/O/C KTACNoteCompletion A P PracticeWorksheetCompletion F/A P

Quiz–FactorTheorem F P PreTestReview F/A P Test–FactorTheoremandInequalities O

C1.8C3.1,3.2,3.3,3.3,3.7

C4.1,4.2,4.3P K(21%),T(34%),A(10%),

C(34%)

L1–2.1–LongDivisionofPolynomialsandTheRemainderTheoremLessonMHF4UJensenInthissectionyouwillapplythemethodoflongdivisiontodivideapolynomialbyabinomial.Youwillalsolearntousetheremaindertheoremtodeterminetheremainderofadivisionwithoutdividing.Part1:DoYouRememberLongDivision(divide,multiply,subtract,repeat)?107 ÷ 4canbecompletedusinglongdivisionasfollows:Everydivisionstatementthatinvolvesnumberscanberewrittenusingmultiplicationandaddition.Wecanexpresstheresultsofourexampleintwodifferentways:107 = 4 26 + 3 OR

+,-.= 26 + /

.

Example1:Uselongdivisiontocalculate753 ÷ 22

4doesnotgointo1,sowestartbydetermininghowmanytimes4goesinto10.Itgoesin2times.Soweputa2inthequotientabovethe10.Thisisthedivisionstep.

Then,multiplythe2by4(thedivisor)andputtheproductbelowthe10inthedividend.Thisisthemultiplystep.

Now,subtract8fromthe10inthedividend.Thenbringdownthenextdigitinthedividendandputitbesidethedifferenceyoucalculated.Thisisthesubtractstep.

Youthenrepeatthesestepsuntiltherearenomoredigitsinthedividendtobringdown.

=26.75 =26+0.75=26.75

Part2:UsingLongDivisiontoDivideaPolynomialbyaBinomialThequotientof 3𝑥/ − 5𝑥3 − 7𝑥 − 1 ÷ 𝑥 − 3 canbefoundusinglongdivisionaswell…Theresultinquotientformis:3𝑥/ − 5𝑥3 − 7𝑥 − 1

𝑥 − 3 = 3𝑥3 + 4𝑥 + 5 +14𝑥 − 3

Theexpressionthatcanbeusedtocheckthedivisionis:3𝑥/ − 5𝑥3 − 7𝑥 − 1 = 𝑥 − 3 3𝑥3 + 4𝑥 + 5 + 14

Focusonlyonthefirsttermsofthedividendandthedivisor.Findthequotientoftheseterms.

Since3𝑥/ ÷ 𝑥 = 3𝑥3,thisbecomesthefirsttermofthequotient.Place3𝑥3abovethetermofthedividendwiththesamedegree.

Multiply3𝑥3bythedivisor,andwritetheanswerbelowthedividend.Makesuretolineup‘liketerms’.3𝑥3(𝑥 − 3) = 3𝑥/ − 9𝑥3.Subtractthisproductfromthedividendandthenbringdownthenextterminthedividend.

Now,onceagain,findthequotientofthefirsttermsofthedivisorandthenewexpressionyouhaveinthedividend.Since4𝑥3 ÷ 𝑥 = 4𝑥,thisbecomesthenextterminthequotient.

Multiply4𝑥bythedivisor,andwritetheanswerbelowthelastlineinthedividend.Makesuretolineup‘liketerms’.4𝑥(𝑥 − 3) = 4𝑥3 − 12𝑥 .Subtractthisproductfromthedividendandthenbringdownthenextterm.

Now,findthequotientofthefirsttermsofthedivisorandthenewexpressioninthedividend.Since5𝑥 ÷ 𝑥 = 5,thisbecomesthenextterminthequotient.Multiply5(𝑥 − 3) = 5𝑥 − 15.Subtractthisproductfromthedividend.

Theprocessisstoppedoncethedegreeoftheremainderislessthanthedegreeofthedivisor.Thedivisorisdegree1andtheremainderisnowdegree0,sowestop.

Theresultofthedivisionof𝑃(𝑥)byabinomialoftheform𝑥 − 𝑏is:

𝑃(𝑥)𝑥 − 𝑏 = 𝑄(𝑥) +

𝑅𝑥 − 𝑏

Where𝑅istheremainder.Thestatementthatcanbeusedtocheckthedivisionis:

𝑃(𝑥) = (𝑥 − 𝑏)𝑄(𝑥) + 𝑅

Note:youcouldcheckthisanswerbyFOILingtheproductandcollectingliketerms.

Example2:Findthefollowingquotientsusinglongdivision.Expresstheresultinquotientform.Also,writethestatementthatcanbeusedtocheckthedivision(thencheckit!).a)𝑥3 + 5𝑥 + 7dividedby𝑥 + 2b)2𝑥/ − 3𝑥3 + 8𝑥 − 12dividedby𝑥 − 1c)4𝑥/ + 9𝑥 − 12dividedby2𝑥 + 1

Theresultinquotientformis:𝑥3 + 5𝑥 + 7

𝑥 + 2 = 𝑥 + 3 +1

𝑥 + 2Theexpressionthatcanbeusedtocheckthedivisionis:𝑥3 + 5𝑥 + 7 = (𝑥 + 2)(𝑥 + 3) + 1

Theresultinquotientformis:2𝑥/ − 3𝑥3 + 8𝑥 − 12

𝑥 − 1 = 2𝑥3 − 𝑥 + 7 +−5𝑥 − 1

Theexpressionthatcanbeusedtocheckthedivisionis:2𝑥/ − 3𝑥3 + 8𝑥 − 12 = (𝑥 − 1)(2𝑥3 − 𝑥 + 7) − 5

Theresultinquotientformis:4𝑥/ + 9𝑥 − 12

2𝑥 + 1 = 2𝑥3 − 𝑥 + 5 +−172𝑥 + 1

Theexpressionthatcanbeusedtocheckthedivisionis:4𝑥/ + 9𝑥 − 12 = (2𝑥 + 1)(2𝑥3 − 𝑥 + 5) − 17

Example3:Thevolume,incubiccm,ofarectangularboxisgivenby𝑉 𝑥 = 𝑥/ + 7𝑥3 + 14𝑥 + 8.Determineexpressionsforpossibledimensionsoftheboxiftheheightisgivenby𝑥 + 2.Expressionsforthepossibledimensionsoftheboxare𝑥 + 1, 𝑥 + 2,and𝑥 + 4.Part3:RemainderTheoremWhenapolynomialfunction𝑃(𝑥)isdividedby𝑥 − 𝑏,theremainderis𝑃(𝑏);andwhenitisdividedby𝑎𝑥 − 𝑏,theremainderis𝑃 @

A,where𝑎and𝑏areintegers,and𝑎 ≠ 0.

Example3:Applytheremaindertheorema)Usetheremaindertheoremtodeterminetheremainderwhen𝑃 𝑥 = 2𝑥/ + 𝑥3 − 3𝑥 − 6isdividedby𝑥 + 1Since𝑥 + 1is𝑥 − (−1),theremainderis𝑃(−1).𝑃 −1 = 2(−1)/ + (−1)3 − 3 −1 − 6= −2 + 1 + 3 − 6= −4Therefore,theremainderis−4

Dividingthevolumebytheheightwillgiveanexpressionfortheareaofthebaseofthebox.

Factortheareaofthebasetogetpossibledimensionsforthelengthandwidthofthebox.

b)VerifyyouranswerusinglongdivisionExample4:Usetheremaindertheoremtodeterminetheremainderwhen𝑃 𝑥 = 2𝑥/ + 𝑥3 − 3𝑥 − 6isdividedby2𝑥 − 3Theremainderis𝑃 /

3

𝑃32

= 232

/

+32

3

− 332− 6

= 2 3-

C+ D

.− 3 /

3− 6

= E.

C+ D

.− D

3− 6

= 3-

.+ D

.− +C

.− 3.

.

= − /

3

Example5:Determinethevalueof𝑘suchthatwhen3𝑥. + 𝑘𝑥/ − 7𝑥 − 10isdividedby𝑥 − 2,theremainderis8.Theremainderis𝑃(2).Solvefor𝑘when𝑃(2)issettoequal8.𝑃 2 = 3 2 . + 𝑘 2 / − 7 2 − 108 = 3 2 . + 𝑘 2 / − 7 2 − 108 = 3 16 + 8𝑘 − 14 − 108 = 48 + 8𝑘 − 248 = 24 + 8𝑘−16 = 8𝑘−2 = 𝑘Therefore,thevalueof𝑘is−2.

Therefore,theremainderis−/3

L2–2.1–SyntheticDivisionLessonMHF4UJensenInthissectionyouwilllearnhowtousesyntheticdivisionasanalternatemethodtodividingapolynomialbyabinomial.Syntheticdivisionisanefficientwaytodivideapolynomialbyabinomialoftheform𝑥 − 𝑏.

Part1:Syntheticdivisionwhenthebinomialisoftheform𝒙 − 𝒃Divide3𝑥' − 5𝑥) − 7𝑥 − 1by𝑥 − 3.Inthisquestion,𝑏 = 3.Don’tforgetthattheanswercanbewrittenintwoways…3𝑥' − 5𝑥) − 7𝑥 − 1

𝑥 − 3 = 𝟑𝒙𝟐 + 𝟒𝒙 + 𝟓 +𝟏𝟒𝒙 − 𝟑

OR3𝑥' − 5𝑥) − 7𝑥 − 1 = 𝐱 − 𝟑 𝟑𝒙𝟐 + 𝟒𝒙 + 𝟓 + 𝟏𝟒

Listthecoefficientsofthedividendinthefirstrow.Totheleft,writethe𝑏value(thezeroofthedivisor).Placea+signabovethehorizontallinetorepresentadditionanda×signbelowthehorizontallinetoindicatemultiplicationofthedivisorandthetermsofthequotient.

Bringthefirsttermdown,thisisthecoefficientofthefirsttermofthequotient.Multiplyitbythe𝑏valueandwritethisproductbelowthesecondtermofthedividend.

Nowaddthetermstogether.

Multiplythissumbythe𝑏valueandwritetheproductbelowthethirdtermofthedividend.Repeatthisprocessuntilyouhaveacompletedthechart.

Thelastnumberbelowthechartistheremainder.Thefirstnumbersarethecoefficientsofthequotient,startingwithdegreethatisonelessthanthedividend.

IMPORTANT:WhenusingPolynomialORSyntheticdivision…

• Termsmustbearrangedindescendingorderofdegree,inboththedivisorandthedividend.• Zeromustbeusedasthecoefficientofanymissingpowersofthevariableinboththedivisor

andthedividend.

Example1:Usesyntheticdivisiontodivide.Thenwritethemultiplicationstatementthatcouldbeusedtocheckthedivision.a)(𝑥7 − 2𝑥' + 13𝑥 − 6)÷(𝑥 + 2)

b)(2𝑥' − 5𝑥) + 8𝑥 + 4) ÷ (𝑥 − 3)

= 𝑥7 − 2𝑥' + 0𝑥) + 13𝑥 − 6

Note:sincetheremainderiszero,boththequotientanddivisorarefactorsofthedividend.

Part2:Syntheticdivisionwhenthebinomialisoftheform𝒂𝒙 − 𝒃Divide6𝑥' + 5𝑥) − 16𝑥 − 15by2𝑥 + 32𝑥 + 3 = 2 𝑥 + '

)à𝑏 = − '

)

Checkanswerusinglongdivision

Tousesyntheticdivision,thedivisormustbeintheform𝑥 − 𝑏.Re-writethedivisorbyfactoringoutthecoefficientofthe𝑥.

Wecannowdivide6𝑥' + 5𝑥) − 16𝑥 − 15by@𝑥 + ')Ausing

syntheticdivisionaslongasyouremembertodividethequotientby2after.

Note:Syntheticdivisioncanonlybeusedwithalineardivisor.Itismostusefulwithadivisoroftheform𝑥 − 𝑏.Ifthedivisoris𝑎𝑥 − 𝑏,itcanbeusedbutlongdivisionmaybeeasier.

Example2:Findeachquotientbychoosinganappropriatestrategy.a)Divide𝑥' − 4𝑥) + 2𝑥 + 3by𝑥 − 3b)Divide12𝑥7 − 56𝑥' + 59𝑥) + 9𝑥 − 18by2𝑥 + 1

c)Divide𝑥7 − 2𝑥' + 5𝑥 + 3by𝑥) + 2𝑥 + 1d)Divide𝑥7 − 𝑥' − 𝑥) + 2𝑥 + 1by𝑥) + 2

L3–2.2–FactorTheoremLessonMHF4UJensenInthissection,youwilllearnhowtodeterminethefactorsofapolynomialfunctionofdegree3orgreater.Part1:RemainderTheoremRefreshera)Usetheremaindertheoremtodeterminetheremainderwhen𝑓 𝑥 = 𝑥$ + 4𝑥' + 𝑥 − 6isdividedby𝑥 + 2𝑓 −2 = (−2)$ + 4(−2)' + −2 − 6𝑓 −2 = −8 + 16 − 2 − 6𝑓 −2 = 0Theremainderwhendividedby𝑥 + 2is0.Thismeansthat𝑥 + 2isafactorofthedividend.b)Verifyyouranswertoparta)bycompletingthedivisionusinglongdivisionorsyntheticdivision.FactorTheorem:𝑥 − 𝑏isafactorofapolynomial𝑃(𝑥)ifandonlyif𝑃 𝑏 = 0.Similarly,𝑎𝑥 − 𝑏isafactorof𝑃(𝑥)ifandonlyif𝑃 3

4= 0.

RemainderTheorem:Whenapolynomialfunction𝑃(𝑥)isdividedby𝑥 − 𝑏,theremainderis𝑃(𝑏);andwhenitisdividedby𝑎𝑥 − 𝑏,theremainderis𝑃 53

46,

where𝑎and𝑏areintegers,and𝑎 ≠ 0.

Note:Ichosesyntheticsinceitisalineardivisoroftheform𝑥 − 𝑏.

Example1:Determineif𝑥 − 3and𝑥 + 2arefactorsof𝑃 𝑥 = 𝑥$ − 𝑥' − 14𝑥 + 24𝑃 3 = (3)$ − 3 ' − 14 3 + 24𝑃 3 = 27 − 9 − 42 + 24𝑃 3 = 0Sincetheremainderis0,𝑥 − 3dividesevenlyinto𝑃(𝑥);thatmeans𝑥 − 3isafactorof𝑃(𝑥).𝑃 −2 = (−2)$ − −2 ' − 14 −2 + 24𝑃 −2 = −8 − 4 + 28 + 24𝑃 −2 = 40Sincetheremainderisnot0,𝑥 + 2doesnotdivideevenlyinto𝑃(𝑥);thatmeans𝑥 + 2isnotafactorof𝑃(𝑥).Part2:HowtodetermineafactorofaPolynomialWithLeadingCoefficient1Youcouldguessandcheckvaluesof𝑏thatmake𝑃 𝑏 = 0untilyoufindonethatworks…OryoucanusetheIntegralZeroTheoremtohelp.IntegralZeroTheoremIf𝑥 − 𝑏isafactorofapolynomialfunction𝑃(𝑥)withleadingcoefficient1andremainingcoefficientsthatareintegers,then𝒃isafactoroftheconstanttermof𝑃(𝑥).Note:Onceoneofthefactorsofapolynomialisfound,divisionisusedtodeterminetheotherfactors.

Example2:Factor𝑥$ + 2𝑥' − 5𝑥 − 6fully.Let𝑃 𝑥 = 𝑥$ + 2𝑥' − 5𝑥 − 6Findavalueof𝑏suchthat𝑃 𝑏 = 0.Basedonthefactortheorem,if𝑃 𝑏 = 0,thenweknowthat𝑥 − 𝑏isafactor.Wecanthendivide𝑃(𝑥)bythatfactor.Theintegralzerotheoremtellsustotestfactorsof−𝟔.Test±𝟏,±𝟐,±𝟑, 𝒂𝒏𝒅 ± 𝟔.Onceonefactorisfound,youcanstoptestingandusethatfactortodivide𝑃(𝑥).𝑃 1 = (1)$ + 2(1)' − 5(1) − 6𝑃 1 = 1 + 2 − 5 − 6𝑃 1 = −8Since𝑃(1) ≠ 0,weknowthat𝑥 − 1isNOTafactorof𝑃(𝑥).𝑃 2 = (2)$ + 2(2)' − 5(2) − 6𝑃 2 = 8 + 8 − 10 − 6𝑃 2 = 0Since𝑃 2 = 0,weknowthat𝑥 − 2isafactorof𝑃(𝑥).YoucannowuseeitherlongdivisionorsyntheticdivisiontofindtheotherfactorsMethod1:Longdivision Method2:SyntheticDivision

Example3:Factor𝑥G + 3𝑥$ − 7𝑥' − 27𝑥 − 18completely.

Let𝑃 𝑥 = 𝑥G + 3𝑥$ − 7𝑥' − 27𝑥 − 18

Findavalueof𝑏suchthat𝑃 𝑏 = 0.Basedonthefactortheorem,if𝑃 𝑏 = 0,thenweknowthat𝑥 − 𝑏isafactor.Wecanthendivide𝑃(𝑥)bythatfactor.Theintegralzerotheoremtellsustotestfactorsof−𝟏𝟖.Test±𝟏,±𝟐,±𝟑,±𝟔,±𝟗𝒂𝒏𝒅 ± 𝟏𝟖.Onceonefactorisfound,youcanstoptestingandusethatfactortodivide𝑃(𝑥).

Since𝑷 −𝟏 = 𝟎,thistellusthat𝒙 + 𝟏isafactor.Usedivisiontodeterminetheotherfactor.Wecannowfurtherdivide𝑥$ + 2𝑥' − 9𝑥 − 18usingdivisionagainorbyfactoringbygrouping.

Method1:Division Testfactorsof−18𝑓 −2 = (−2)$ + 2(−2)' − 9(−2) − 18

𝑓 −2 = 0

𝑥 + 2isafactor

𝑃(1) = (1)G + 3(1)$ − 7(1)' − 27(1) − 18𝑃(1) = −48𝑥 − 1isNOTafactorof𝑃(𝑥).

𝑃(−1) = (−1)G + 3(−1)$ − 7(−1)' − 27(−1) − 18𝑃(−1) = 0𝑥 + 1ISafactorof𝑃(𝑥).

Method2:FactoringbyGrouping

Therefore,𝑥G + 3𝑥$ − 7𝑥' − 27𝑥 − 18 = (𝑥 + 1) 𝑥$ + 2𝑥' − 9𝑥 − 18 = (𝑥 + 1)(𝑥 + 2)(𝑥' − 9) = (𝑥 + 1)(𝑥 + 2)(𝑥 − 3)(𝑥 + 3)Example4:TryFactoringbyGroupingAgain𝑥G − 6𝑥$ + 2𝑥' − 12𝑥= 𝒙𝟒 − 𝟔𝒙𝟑 + 𝟐𝒙𝟐 − 𝟏𝟐𝒙 = 𝒙𝟑 𝒙 − 𝟔 + 𝟐𝒙 𝒙 − 𝟔 = 𝒙 − 𝟔 𝒙𝟑 + 𝟐𝒙 = (𝒙 − 𝟔)(𝒙) 𝒙𝟐 + 𝟐

Note:Factoringbygroupingdoesnotalwayswork…butwhenitdoes,itsavesyoutime!

𝑓(𝑥) = 𝑥$ + 2𝑥' − 9𝑥 − 18Groupthefirst2termsandthelast2termsandseparatewithanadditionsign.

𝑓(𝑥) = (𝑥$ + 2𝑥') + (−9𝑥 − 18)Commonfactorwithineachgroup

𝑓(𝑥) = 𝑥'(𝑥 + 2) − 9(𝑥 + 2)Factoroutthecommonbinomial

𝑓(𝑥) = (𝑥 + 2)(𝑥' − 9)

Part3:HowtodetermineafactorofaPolynomialWithLeadingCoefficientNOT1Theintegralzerotheoremcanbeextendedtoincludepolynomialswithleadingcoefficientsthatarenot1.Thisextensionisknownastherationalzerotheorem.RationalZeroTheorem:Suppose𝑃 𝑥 isapolynomialfunctionwithintegercoefficientsand𝑥 = 3

4isazeroof𝑃(𝑥),where𝑎and

𝑏areintegersand𝑎 ≠ 0.Then,

• 𝑏isafactoroftheconstanttermof𝑃(𝑥)• 𝑎isafactoroftheleadingcoefficientof𝑃(𝑥)• (𝑎𝑥 − 𝑏)isafactorof𝑃(𝑥)

Example5:Factor𝑃 𝑥 = 3𝑥$ + 2𝑥' − 7𝑥 + 2Wemuststartbyfindingavalueof3

4where𝑃 3

4= 0.

𝑏mustbeafactoroftheconstantterm.Possiblevaluesfor𝑏are:±𝟏,±𝟐𝑎mustbeafactoroftheleadingcoefficient.Possiblevaluesof𝑎are:±𝟏,±𝟑Therefore,possiblevaluesfor3

4are:±𝟏,± 𝟏

𝟑, ±𝟐,± 𝟐

𝟑

Testvaluesof3

4for𝑥in𝑃(𝑥)tofindazero.

𝑃 1 = 3 1 $ + 2 1 ' − 7 1 + 2 = 0Since𝑷 𝟏 = 𝟎,𝒙 − 𝟏isafactorof𝑃(𝑥).Usedivisiontofindtheotherfactors.

Example6:Factor𝑃 𝑥 = 2𝑥$ + 𝑥' − 7𝑥 − 6Possiblevaluesfor𝑏are:±𝟏,±𝟐,±𝟑,±𝟔

Possiblevaluesof𝑎are:±𝟏,±𝟐

Therefore,possiblevaluesfor34are:±𝟏,± 𝟏

𝟐, ±𝟐,±𝟑,± 𝟑

𝟐, ±𝟔

𝑓 −1 = 2 −1 $ + −1 ' − 7 −1 − 6 = 0Therefore,𝑥 + 1isafactorof𝑃(𝑥)Part4:ApplicationQuestionExample7:When𝑓 𝑥 = 2𝑥$ − 𝑚𝑥' + 𝑛𝑥 − 2isdividedby𝑥 + 1,theremainderis−12and𝑥 − 2isafactor.Determinethevaluesof𝑚and𝑛.

Hint:Usetheinformationgiventocreate2equationsandthenusesubstitutionoreliminationtosolve.

L4–2.3–SolvingPolynomialEquationsLessonMHF4UJensenInthissection,youwilllearnmethodsofsolvingpolynomialequationsofdegreehigherthantwobyfactoring(usingthefactortheorem).Youwillalsoidentifytheconnectionbetweentherootsofpolynomialequations,thex-interceptsofthegraphofapolynomialfunction,andthezerosofthefunction.Part1:Investigation

a)Usetechnologytographthefunction𝑓 𝑥 = 𝑥$ − 13𝑥( + 36b)Determinethex-interceptsfromthegraphThe𝑥-interceptsare:(-3,0),(-2,0),(2,0),and(3,0)c)Factor𝑓(𝑥).Then,usethefactorstodeterminethezeros(roots)of𝑓(𝑥).d)Howarethe𝑥-interceptsfromthegraphrelatedtotheroots(zeros)oftheequation?ThezerosoftheequationAREthe𝑥-interceptsofthegraphofthefunction.

Remember:Thezerosofthefunctionarethevaluesof𝑥thatmake𝑓(𝑥) = 0.Ifthepolynomialequationisfactorable,thenthevaluesofthezeros(roots)canbedeterminedalgebraicallybysolvingeachlinearorquadraticfactor.

Example1:Statethesolutionstothefollowingpolynomialsthatarealreadyinfactoredforma)𝑥 2𝑥 + 3 𝑥 − 5 = 0 b)(2𝑥( − 3)(3𝑥( + 1)

Example2:Solveeachpolynomialequationbyfactoringa)𝑥0 − 𝑥( − 2𝑥 = 0 b)3𝑥0 + 𝑥( − 12𝑥 − 4 = 0

Methodsoffactoring:

• Longdivisionandsyntheticdivision• Factorbygrouping• Differenceofsquares𝑎( − 𝑏( = (𝑎 − 𝑏)(𝑎 + 𝑏)• CommonFactoring• Trinomialfactoring(sumandproduct)• Sumanddifferenceofcubes𝑎0 + 𝑏0 = (𝑎 + 𝑏)(𝑎( − 𝑎𝑏 + 𝑏()

𝑎0 − 𝑏0 = (𝑎 − 𝑏)(𝑎( + 𝑎𝑏 + 𝑏()

Solution(s):(0,0),(2,0),and(-1,0)

Solution(s):4−5

0, 07,(1,0),and(-1,0)

Example3:a)Usethefactortheoremtosolve𝑓 𝑥 = 2𝑥0 + 3𝑥( − 11𝑥 − 6Possiblevaluesof𝑏are:±1,±2,±3,±6Possiblevaluesforare:±1,±2Possiblevaluesfor9

:= ±1,± 5

(, ±2,±3,± 0

(, ±6

𝑓 2 = 2(2)0 + 3(2)( − 11 2 − 6 = 0,∴ 𝑥 − 2isafactorb)Whatdoyouranswerstoparta)represent?Thevaluesof2,− 5

(,and−3aretherootsoftheequation2𝑥0 + 3𝑥( − 11𝑥 − 6 = 0whichmeanstheyare

the𝑥-interceptsofthegraphofthefunction𝑓 𝑥 = 2𝑥0 + 3𝑥( − 11𝑥 − 6.

Solution(s):(2,0),(-3,0),and4− 5

(, 07

Example4:Findthezerosofthepolynomialfunction𝑓 𝑥 = 𝑥$ − 2𝑥0 − 7𝑥( + 8𝑥 + 12Possiblevaluesfor𝑏are±1,±2,±3,±4,±6𝑓 −1 = (−1)$ − 2 −1 0 − 7 −1 ( + 8 −1 + 12 = 0,∴ 𝑥 + 1isafactor

Solution(s):(-1,0),(3,0),(2,0),and(-2,0)

Example5:a)Findtherootsofthepolynomialfunction𝑓 𝑥 = 𝑥0 + 𝑥 − 3𝑥( − 3Startbyrearrangingindescendingorderofdegree:𝑓 𝑥 = 𝑥0 − 3𝑥( + 𝑥 − 3

b)Usetechnologytolookatthegraphofthefunction𝑓(𝑥).Commentonhow𝑥-intercept(s)ofthegrapharerelatedtotheREALandNON-REALrootsoftheequation.The𝑥-interceptsofthegraphofapolynomialfunctioncorrespondtoonlytheREALrootsoftherelatedpolynomialequation.Thereareno𝑥-interceptsonthegraphthatcorrespondtotheNON-REALrootsoftheequation.

Note:Sincethesquarerootofanegativenumberisnotarealnumber,theonlyREALrootis𝑥 = 3.𝑥 = ±√−1isconsideredaNON-REALroot.

Solution(s):(3,0)

Example6:Findallrealrootsforeachpolynomialequationa)𝑓 𝑥 = 2𝑥0 − 3𝑥( − 𝑥 − 2Possiblevaluesof𝑏are:±1,±2Possiblevaluesforare:±1,±2Possiblevaluesfor9

:= ±1,± 5

(, ±2

𝑓 2 = 2(2)0 − 3 2 ( − 2 − 2 = 0, ∴ 𝑥 − 2isafactorof𝑓(𝑥)b)𝑔 𝑥 = 8𝑥0 + 125Hint:Thisisadifferenceofcubesà𝑎0 + 𝑏0 = (𝑎 + 𝑏) 𝑎( − 𝑎𝑏 + 𝑏(

Solution(s):(2,0)

Solution(s):

A−52 , 0)B

L5–2.4–FamiliesofPolynomialFunctionsLessonMHF4UJensenInthissection,youwilldetermineequationsforafamilyofpolynomialfunctionsfromasetofzeros.Givenadditionalinformation,youwilldetermineanequationforaparticularmemberofthefamily.Part1:Investigation1)a)Howarethegraphsofthefunctionssimilarandhowaretheydifferent?

Same Different• 𝑥-intercepts(zeros)• equationshavesamedegree

• 𝑦-intercepts• stretchorcompressionfactors• vertices

b)Describetherelationshipbetweenthegraphsoffunctionsoftheform𝑦 = 𝑘(𝑥 − 1)(𝑥 + 2),where𝑘 ∈ ℝTheyhavethesame𝑥-intercepts.

2)a)Examinethefollowingfunctions.Howaretheysimilar?Howaretheydifferent?

i) 𝑦 = −2(𝑥 − 1)(𝑥 + 3)(𝑥 − 2)ii) 𝑦 = −(𝑥 − 1)(𝑥 + 3)(𝑥 − 2)iii) 𝑦 = (𝑥 − 1)(𝑥 + 3)(𝑥 − 2)iv) 𝑦 = 2(𝑥 − 1)(𝑥 + 3)(𝑥 − 2)

b)Predicthowthegraphsofthefunctionswillbesimilarandhowtheywillbedifferent.Theywillhavethesame𝑥-interceptsbuttheirshapeanddirectionwillbedifferentduetothesignandvalueoftheleadingcoefficient.c)Usetechnologytohelpyousketchthegraphsofallfourfunctionsonthesamesetofaxes.

Afamilyoffunctionsisasetoffunctionsthathavethesamecharacteristics.Polynomialfunctionswiththesamezerosaresaidtobelongtothesamefamily.Thegraphsofpolynomialfunctionsthatbelongtothesamefamilyhavethesame𝑥-interceptsbuthavedifferent𝑦-intercepts(unless0isoneofthe𝑥-intercepts).Anequationforthefamilyofpolynomialfunctionswithzeros𝑎/, 𝑎1, 𝑎2,…, 𝑎5is:

𝑦 = 𝑘(𝑥 − 𝑎/)(𝑥 − 𝑎1)(𝑥 − 𝑎2)… (𝑥 − 𝑎5),where𝑘 ∈ ℝ, 𝑘 ≠ 0

Part2:RepresentaFamilyofFunctionsAlgebraically1)Thezerosofafamilyofquadraticfunctionsare2and-3.a)Determineanequationforthisfamilyoffunctions.𝑦 = 𝑘(𝑥 − 2)(𝑥 + 3)b)Writeequationsfortwofunctionsthatbelongtothisfamily𝑦 = 8(𝑥 − 2)(𝑥 + 3)𝑦 = −3(𝑥 − 2)(𝑥 + 3)c)Determineanequationforthememberofthefamilythatpassesthroughthepoint(1,4).𝑦 = 𝑘(𝑥 − 2)(𝑥 + 3)4 = 𝑘(1 − 2)(1 + 3)4 = 𝑘(−1)(4)4 = −4𝑘−1 = 𝑘𝑦 = −(𝑥 − 2)(𝑥 + 3)2)Thezerosofafamilyofcubicfunctionsare-2,1,and3.a)Determineanequationforthisfamily.𝑦 = 𝑘(𝑥 + 2)(𝑥 − 1)(𝑥 − 3)b)Determineanequationforthememberofthefamilywhosegraphhasa𝑦-interceptof-15.−15 = 𝑘(0 + 2)(0 − 1)(0 − 3)−15 = 𝑘(2)(−1)(−3)−15 = 6𝑘𝑘 = −2.5

d)SketchagraphofthefunctionNegativeleadingcoefficientandodddegreesoitwillextendfromQ2toQ4Part3:DetermineanEquationforaFunctionFromaGraph3)Determineanequationforthequarticfunctionrepresentedbythisgraph.

Tosketchagraph:

• Ploty-intercept• Plotx-intercepts• Usedegreeandleading

coefficienttodetermineendbehaviour

The𝑥-interceptsare−3,−/1,1,and2

𝑦 = 𝑘(𝑥 + 3)(2𝑥 + 1)(𝑥 − 1)(𝑥 − 2)Thegraphpassesthroughthepoint(-1,-6)−6 = 𝑘(−1 + 3)(2(−1) + 1)(−1 − 1)(−1 − 2)−6 = 𝑘(2)(−1)(−2)(−3)−6 = −12𝑘𝑘 = 0.5𝑦 = 0.5(𝑥 + 3)(2𝑥 + 1)(𝑥 − 1)(𝑥 − 2)

L6–2.5–SolvingInequalitiesLessonMHF4UJensenInthissection,youwilllearnthemeaningofapolynomialinequalityandexaminemethodsforsolvingpolynomialinequalities.Part1:IntrotoInequalitiesTask:ReadthefollowingonyourownExaminethegraphof𝑦 = 𝑥$ + 4𝑥 − 12.The𝑥-interceptsare6and-2.Thesecorrespondtothezerosofthefunction𝑦 = 𝑥$ + 4𝑥 − 12.Notethatthefactoredformversionofthefunctionis𝑦 = (𝑥 +6)(𝑥 − 2).Bymovingfromlefttorightalongthe𝑥-axis,wecanmakethefollowingobservations:

• Thefunctionispositivewhen𝑥 < −6sincethe𝑦-valuesarepositive

• Thefunctionisnegativewhen−6 < 𝑥 < 2sincethe𝑦-valuesarenegative

• Thefunctionispositivewhen𝑥 > 2sincethe𝑦-valuesarepositive.

Thezeros-6and2dividethe𝑥-axisintothreeintervals.Ineachinterval,thefunctioniseitherpositiveornegative.Theinformationcanbesummarizedinatable:Interval 𝑥 < −6 −6 < 𝑥 < 2 𝑥 > 2SignofFunction + − +

-1 0 1 2 3 4 5 6 7 8 9 10 -1 0 1 2 3 4 5 6 7 8 9 10

PolynomialInequalitiesApolynomialinequalityresultswhentheequalsigninapolynomialequationisreplacedwithaninequalitysymbol.Therealzerosofapolynomialfunction,or𝑥-interceptsofthecorrespondinggraph,dividethe𝑥-axisintointervalsthatcanbeusedtosolveapolynomialinequality.Part1:InequalitiesandNumberLinesExample1:Writeaninequalitythatcorrespondstothevaluesof𝑥shownoneachnumberlinea) b)Part2:SolveanInequalitygiventheGraphExample2:Usethegraphofthefunction𝑓(𝑥)toanswerthefollowinginequalities…𝑓 𝑥 = 0.1(𝑥 − 1)(𝑥 + 3)(𝑥 − 4)

a)𝑓(𝑥) < 0𝑓(𝑥) < 0when:𝑥 < −3or1 < 𝑥 < 4 (−∞,−3) ∪ (1, 4)b)𝑓(𝑥) ≥ 0𝑓(𝑥) ≥ 0when:−3 ≤ 𝑥 ≤ 1or𝑥 ≥ 4 [−3, 1] ∪ [4,∞)

𝑥 ≤ 4

OR

(−∞, 4]

3 ≤ 𝑥 < 9

OR

[3, 9)

Part2:SolveLinearInequalitiesNote:Solvinglinearinequalitiesisthesameassolvinglinearequations.However,whenbothsidesofaninequalityaremultipliedordividedbyanegativenumber,theinequalitysignmustbereversed.Example3:Solveeachinequalitya)𝑥 − 8 ≥ 3 b)−4 − 2𝑥 < 12Part2:SolveInequalitiesofDegree2andHigher

Example4:Solveeachpolynomialinequalityalgebraicallya)2𝑥$ + 3𝑥 − 9 > 0Method1:Graphtheinequality2𝑥$ + 3𝑥 − 9 > 0when… 𝑥 < −3or𝑥 > 1.5 (−∞,−3) ∪ (1.5,∞)

Stepsforsolvingpolynomialinequalitiesalgebraically:

1) Useinverseoperationstomovealltermstoonesideoftheinequality2) Factorthepolynomialtodeterminethezerosofthecorrespondingequation3) Findtheinterval(s)wherethefunctionispositiveornegativebyeither:

a. Graphingthefunctionusingthezeros,leadingcoefficient,anddegreeb. Makeafactortableandtestvaluesineachinterval

whenisitabovethe𝑥-axis?

Method2:FactorTable(signchart)

Tomakeafactortable:

• Use𝑥-interceptsandverticalasymptotestodivideintointervals• Useatestpointwithineachintervaltofindthesignofeachfactor• Determinetheoverallsignoftheproductbymultiplyingsignsofeachfactor

withineachinterval.

b)−2𝑥= − 6𝑥$ + 12𝑥 ≤ −16Method1:Graphtheinequality

Solution:−4 ≤ 𝑥 ≤ −1or𝑥 ≥ 2 −4,−1 ∪ [2,∞)

Method2:FactorTable(signchart)

c)𝑥= + 4𝑥$ + 6𝑥 < −24Part2:ApplicationsofInequalities3)Theprice,𝑝,indollars,ofastock𝑡yearsafter1999canbemodeledbythefunction𝑝 𝑡 = 0.5𝑡= −5.5𝑡$ + 14𝑡.Whenwillthepriceofthestockbemorethan$90?