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CHAPTER6.3T
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à 6.3ïFactoring by the Trial and Error method.
äïPlease factor the following trinomials by the trial and
êêerror method.
#âêFactorï2xì - 11x - 6 by the trial and error method.
Factors of 2 are 1 and 2.ïFactors of 6 are 1 and 6 or 2 and 3.ïSince
there is only one pair of factors of 2, the two first entries
are 2x and x, i.e.ï(2x _)(x _).ïSince the middle term must add up to
eleven, the two second entries are 1 and 6, i.e. (2xï1)(xï6).
Choosing the signs to be first "+" then "-".
#êêë2xì - 11x - 6 = (2x + 1)(x - 6).
éS
#In order to factor the trinomial, 2xì - 11x - 6, byïthe trial and error
method, it is necessary to determine the entries in the expression
(_ + _)(_ + _) that will multiply out to give the original polynomial.
From the foil method of multiplying we know that the product of the first
#entries must be 2xì.ïTherefore the two first entries should be 2x and x
i.e. (2x + _)(x + _).ïWe also know from the foil method that the product
of the last entries must be 6.ïThe possible factors of 6 are 1 and 6 or
2 and 3.ïThese two pairs of factors generate the following possible
trials.ïJust ignore the signs for the moment.
#è(2xè6)(xè1)êThe terms inêè2xìè2xè6xï6
#è(2xè1)(xè6)êç productsê 2xìï12xè xï6
#è(2xè2)(xè3)êareêêë2xìè6xè2xï6
#è(2xè3)(xè2)êêêê 2xìè4xè3xï6
Only one of ç four trials will work and we look for the middle term
combination that can combine to give 11x.ïSince 12x and x can combine
to give 11x, the second trial is chosen.è(2xè1)(xè6)
Next, since the product of the two last terms is negative six, the signs
on the 1 and 6 must be either -1 and +6 or +1 and -6.ïThese two trials
and their products are shown below.
#êêë (2x - 1)(x + 6)ï=ï2xì + 11x - 6
#êêë (2x + 1)(x - 6)ï=ï2xì - 11x - 6
Since the second of ç trials multiplies out to give the original
polynomial, (2x + 1)(x - 6) is the correct factorization of
#2xì - 11x - 6. You can see why this is called the trial and error method
1
#êêêèFactorïxì + x - 12.
êèA)ï(x - 12)(x + 1)êë C)ï(x - 3)(x + 4)
êèB)ï(x - 4)(x + 3)êêD)ï(x - 1)(x + 12)
#üêêêèxì + x - 12
#Factors of xìèFactors of 12èTrials
#è x∙xêë1∙12êï(xè1)(xè12) = xìï12xèxï12
#êêë2∙6êè(xè2)(xè6)ï= xìè6xï2xï12
#êêë3∙4êè(xè3)(xè4)ï= xìè3xï4xï12
3x and 4x combine to give the middle term x.ïThe signs on 3 and 4 are
-3 and +4.ïThe correct factorization isï(x - 3)(x + 4).
Ç C
2
#êêêèFactorïxì + 8x + 15.
êèA)ï(x - 3)(x - 5)êêC)ï(x + 7)(x - 8)
êèB)ï(x + 1)(x - 15)êë D)ï(x + 3)(x + 5)
#üêêêèxì + 8x + 15
#Factors of xìèFactors of 15èTrials
#è x∙xêë1∙15êï(xè1)(xè15) = xìï15xèxï15
#êêë3∙5êè(xè3)(xè5)ï= xìè5xï3xï15
5x and 3x combine to give the middle term 8x.ïThe signs on 5 and 3 are
+5 and +3.ïThe correct factorization isï(x + 3)(x + 5).
Ç D
3
#êêêèFactorïxì - 7x + 10.
êèA)ï(x - 2)(x - 5)êêC)ï(x - 2)(x + 5)
êèB)ï(x - 5)(x + 2)êêD)ï(x + 2)(x + 5)
#üêêêèxì - 7x + 10
#Factors of xìèFactors of 10
è x∙xêë1∙10êïChoose the trial (xï2)(xï5). Since the
êêë2∙5êèproduct of the two last terms is plus 10
êêêêè the signs on 2 and 5 must be -2 and -5.
êêêêêê(x - 2)(x - 5)
Ç A
4
#êêêèFactorïxì + 4x - 21.
êèA)ï(x - 7)(x + 3)êêC)ï(x - 3)(x + 7)
êèB)ï(x - 21)(x + 1)êë D)ï(x - 1)(x + 21)
#üêêêèxì + 4x - 21
#Factors of xìèFactors of 21
è x∙xêë1∙21êïChoose the trial (xï3)(xï7). Since the
êêë3∙7êèproduct of the two last terms is - 21
êêêêè the signs on 3 and 7 must be -3 and +7.
êêêêêê(x - 3)(x + 7)
Ç C
5
#êêêèFactorïxì + x - 30.
êèA)ï(x - 3)(x + 10)êë C)ï(x - 6)(x + 5)
êèB)ï(x - 5)(x + 6)êêD)ï(x + 30)(x - 1)
#üêêêèxì + x - 30
#è Factors of xìèFactors of 30
êïx∙xêë1∙30
êêêè2∙15êïChooseï(x - 5)(x + 6)
êêêè3∙10
êêêè5∙6
Ç B
6
#êêêèFactorïxì - 15x + 56
êèA)ï(x + 2)(x + 28)êë C)ï(x + 7)(x + 8)
êèB)ï(x + 4)(x + 14)êë D)ï(x - 7)(x - 8)
#üêêêèxì - 15x + 56
#è Factors of xìèFactors of 56
êïx∙xêë1∙56
êêêè2∙28êïChooseï(x - 7)(x - 8)
êêêè4∙14
êêêè7∙8
Ç D
7
#êêêèFactorï2xì + 11x + 12
êèA)ï(2x + 3)(x + 4)êë C)ï(2x + 12)(x + 1)
êèB)ï(2x + 6)(x + 2)êë D)ï(2x - 3)(2x - 4)
#üêêêï2xì + 11x + 12
#è Factors of 2xìïFactors of 12
ê 2x∙xêë1∙12
êêêè2∙6êèChooseï(2x + 3)(x + 4)
êêêè3∙4
Ç A
8
#êêêèFactorï3xì - 11x - 20
êèA)ï(3x + 2)(x - 10)êëC)ï(3x + 4)(x - 5)
êèB)ï(3x + 1)(x - 20)êëD)ï(x + 4)(3x - 5)
#üêêêï3xì - 11x - 20
#è Factors of 3xìïFactors of 20
ê 3x∙xêë1∙20
êêêè2∙10êïChooseï(3x + 4)(x - 5)
êêêè4∙5
Ç C
9
#êêêèFactorï6xì - 13x - 5
êèA)ï(3x + 1)(2x - 5)êëC)ï(6x - 5)(x - 2)
êèB)ï(3x + 5)(x - 2)êë D)ï(3x - 5)(2x + 1)
#üêêêï6xì - 13x - 5
#è Factors of 6xìïFactors of 5
ê 6x∙xêë1∙5
ê 2x∙3xêêêèChooseï(3x + 1)(2x - 5)
Ç A
10
#êêêïFactorï6xì - 5xy - 4yì
êèA)ï(2x - 4y)(3x + y)êè C)ï(6x + 4y)(x - y)
êèB)ï(2x + y)(3x - 4y)êè D)ï(2x + 2y)(3x + 2y)
#üêêêï6xì - 5xy - 4yì
#è Factors of 6xìïFactors of 4yì
ê 6x∙xêë4y∙y
ê 2x∙3xêè 2y∙2yê Chooseï(2x + y)(3x - 4y)
Ç B