湖北省武漢六中高三(國際部)數(shù)學復(fù)習課件：13多項式函數(shù)的因式分解

Unit1:PolynomialFunctions

Lesson3:DividingPolynomialsUnit1:PolynomialFunctions

LLongDivisionDoyourememberlongdivision?Let’sreviewitbydividing589by331289271961813willgointo5once3willgointo28ninetimes3willgointo19sixtimesBringdownthe8Bringdownthe9So,589÷3=196witharemainderof1(3×1=3)with2leftoverwith1leftover(3×9=27)with1leftover(3×6=18)LongDivisionDoyourememberlLongDivisionRecapInthepreviousexample:Thedividendwas589WhatisbeingdividedThedivisorwas3WhatisdividingintothedividendThequotientwas196TheanswerTheremainderwas1What’sleftoverDivisionformula:Dividend=divisor×quotient+remainderi.e.589=3×196+1LongDivisionRecapIntheprevDividingPolynomialsbyBinomialsPolynomialsarealgebraicexpressionswithmanytermsx3+2x2–x+5x4-6x3–

4x2+3x–10Binomialsarealgebraicexpressionswithtwotermsx–72x+1WecandivideapolynomialbyabinomialusingthesamelongdivisionprocessweusefornumbersDividingPolynomialsbyBinomiExample1Divide-3x2+2x3+8x–12byx–1Beforewebegin,writethepolynomialinorderofdescendingpowers:2x3–3x2+8x–12Example1Divide-3x2+2x3+8Example1:SolutionDivide2x3byxtoget2x22x2Multiplyx–1by2x2

toget2x3–2x22x3–2x

2Subtract.Bringdownthenextterm–x

2+8xDivide–x2byxtoget-xMultiplyx–1by-x

toget–x2+xSubtract.Bringdownthenextterm–x

–x2+x7x–12Divide7xbyxtoget7Multiplyx–1by7

toget7x–7Subtract.Theremainderis-57+7x–7–5Example1:SolutionDivide2x3Example1:SolutionBasedonthedivisionformula:Dividend=divisor×quotient+remainder–52x3–3x2+8x–12(x–1)(2x2–x+7)=Example1:SolutionBasedonthExample2Divide4x3+9x–12by2x+1Noticethatthere’snox2termBeforewebegin,writein0x2asaplaceholder4x3+0x2+9x–12Example2Divide4x3+9x–12Example2:SolutionDivide4x3by2xtoget2x22x2Multiply2x+1by2x2

toget4x3+2x24x3+2x

2Subtract.Bringdownthenextterm–2x

2+9xDivide–2x2by2xtoget-xMultiply2x+1by-x

toget-2x2-xSubtract.Bringdownthenextterm–x

–2x2–x10x–12Divide10xby2xtoget5Multiply2x+1by5

toget10x+5Subtract.Theremainderis-175+10x+5–17Example2:SolutionDivide4x3Example1:SolutionBasedonthedivisionformula:Dividend=divisor×quotient+remainder–174x3+9x–12(2x+1)(2x2–x+5)=Example1:SolutionBasedonthTheRemainderTheoremWhenaPolynomialFunctionP(x)isdividedbyabinomialax–b,theremainderisacannotbezeroaandbareintegersTheRemainderTheoremWhenaPoExample3VerifytheremaindertheoremusingExamples1&2Example3VerifytheremainderExample3:SolutionInExample1wedivided -3x2+2x3+8x–12byx–1ThereforeAndwewanttofindP(1)Ourremainderwas–5.Example3:SolutionInExampleExample3:SolutionInExample2wedivided4x3+9x–12by2x+1ThereforeAndwewanttofindOurremainderwas–17.Example3:SolutionInExampleSummaryPolynomialscanbedividedbybinomialusingthelongdivisiontechniqueweusefornumbersBeforedividing……writethepolynomialinorderofdescendingpowers(seeExample1)…putazeroinfrontofanymissingterms(seeExample2)WhenaPolynomialFunctionP(x)isdividedbyabinomialax–b,theremainderisP(b/a)knownastheRemainderTheoremSummaryPolynomialscanbediviPracticeProblemsP.91-92#1-3a,7-9bc,10Note:For#1-3a,donotexpressyouranswerinquotientform.IwantyoutowriteyouranswerslikeIdidinExamples1&2(usingthedivisionformula).So,whenyoucheckyouranswersinthebackofthetext,checktheanswerforpartc(notparta)PracticeProblemsP.91-92#1-3Unit1:PolynomialFunctions

Lesson3:DividingPolynomialsUnit1:PolynomialFunctions

LLongDivisionDoyourememberlongdivision?Let’sreviewitbydividing589by331289271961813willgointo5once3willgointo28ninetimes3willgointo19sixtimesBringdownthe8Bringdownthe9So,589÷3=196witharemainderof1(3×1=3)with2leftoverwith1leftover(3×9=27)with1leftover(3×6=18)LongDivisionDoyourememberlLongDivisionRecapInthepreviousexample:Thedividendwas589WhatisbeingdividedThedivisorwas3WhatisdividingintothedividendThequotientwas196TheanswerTheremainderwas1What’sleftoverDivisionformula:Dividend=divisor×quotient+remainderi.e.589=3×196+1LongDivisionRecapIntheprevDividingPolynomialsbyBinomialsPolynomialsarealgebraicexpressionswithmanytermsx3+2x2–x+5x4-6x3–

4x2+3x–10Binomialsarealgebraicexpressionswithtwotermsx–72x+1WecandivideapolynomialbyabinomialusingthesamelongdivisionprocessweusefornumbersDividingPolynomialsbyBinomiExample1Divide-3x2+2x3+8x–12byx–1Beforewebegin,writethepolynomialinorderofdescendingpowers:2x3–3x2+8x–12Example1Divide-3x2+2x3+8Example1:SolutionDivide2x3byxtoget2x22x2Multiplyx–1by2x2

toget2x3–2x22x3–2x

2Subtract.Bringdownthenextterm–x

2+8xDivide–x2byxtoget-xMultiplyx–1by-x

toget–x2+xSubtract.Bringdownthenextterm–x

–x2+x7x–12Divide7xbyxtoget7Multiplyx–1by7

toget7x–7Subtract.Theremainderis-57+7x–7–5Example1:SolutionDivide2x3Example1:SolutionBasedonthedivisionformula:Dividend=divisor×quotient+remainder–52x3–3x2+8x–12(x–1)(2x2–x+7)=Example1:SolutionBasedonthExample2Divide4x3+9x–12by2x+1Noticethatthere’snox2termBeforewebegin,writein0x2asaplaceholder4x3+0x2+9x–12Example2Divide4x3+9x–12Example2:SolutionDivide4x3by2xtoget2x22x2Multiply2x+1by2x2

toget4x3+2x24x3+2x

2Subtract.Bringdownthenextterm–2x

2+9xDivide–2x2by2xtoget-xMultiply2x+1by-x

toget-2x2-xSubtract.Bringdownthenextterm–x

–2x2–x10x–12Divide10xby2xtoget5Multiply2x+1by5

toget10x+5Subtract.Theremainderis-175+10x+5–17Example2:SolutionDivide4x3Example1:SolutionBasedonthedivisionformula:Dividend=divisor×quotient+remainder–174x3+9x–12(2x+1)(2x2–x+5)=Example1:SolutionBasedonthTheRemainderTheoremWhenaPolynomialFunctionP(x)isdividedbyabinomialax–b,theremainderisacannotbezeroaandbareintegersTheRemainderTheoremWhenaPoExample3VerifytheremaindertheoremusingExamples1&2Example3VerifytheremainderExample3:SolutionInExample1wedivided -3x2+2x3+8x–12byx–1ThereforeAndwewanttofindP(1)Ourremainderwas–5.Example3:SolutionInExampleExample3:SolutionInExample2wedivided4x3+9x–12by2x+1