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CHAPTER6.2T
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à 6.2ïFactoring using the Master Product Method
äïPlease factor the following trinomials by the master
êêproduct method.
#âSêêïFactorï3xì + 7x + 2
êêë The master product is 3∙2 = 6
#êêêê 3xì + 7x + 2
#êêêë = 3xì + 6x + x + 2
#êêêë = (3xì + 6x) + (x + 2)
êêêë = 3x(x + 2) + (x + 2)
êêêë = (3x + 1)(x + 2)
éS
#In order to factor the expression, 3xì + 7x + 2, by the master product
method, it is first necessary to find the master product.ïThis number
#is the product of the coefficient of the xì term and the constant term.
The master product is 3∙2 = 6.ïNext, all positive factors of 6 are
written down (1 times 6è2 times 3).ïThen a pair of factors is chosen
that can combine to give the coefficient of the middle term of the
original expression.ïSince 1 + 6 = 7, the pair of factors 1 and 6 is
chosen.
The middle term of the original polynomial is then expressed as a sum.
#êêë 3xì + 7x + 2ï=ï3xì + 6x + x + 2
These four terms are then factored by grouping.
#êêêë 3xì + 6x + x + 2
#êêêè = (3xì + 6x) + (x + 2)
êêêè = 3x(x + 2) + (x + 2)
êêêè = (3x + 1)(x + 2)
#This completes the factorization of the trinaomial, 3xì + 7x + 2, by the
master product method.
1
#êêFactorï2xì + x - 3 by the master product method.
êè A) (2x + 3)(x - 1)êêC) (x - 3)(2x + 1)
êè B) (2x - 1)(x + 3)êêD) å of ç
#üêêêè2xì + x - 3
The master product is 2∙3 = 6.ïFactors of 6 are 1 times 6 and
2 times 3.ïSince the middle term of the original expression can be
obtained by combining 3 and -2, this pair of factors is chosen.
#êêêë 2xì + x - 3
#êêêè = 2xì + 3x - 2x - 3
#êêêè = 2xì + 3x + (-2x) + (-3)
êêêè = x(2x + 3) + (-1)(2x + 3)
êêêè = (2x + 3)(x - 1)
Ç A
2
#êêFactorï5xì + 9x - 2 by the master product method.
êè A) (x - 1)(5x + 2)êêC) (x + 2)(5x - 1)
êè B) (x - 5)(2x + 1)êêD) (x - 1)(2x - 5)
#üêêêè5xì + 9x - 2
The master product is 5(-2) = -10. Factors of 10 are 1 times 10
and 2 times 5.ïSince the middle term of the original expression can be
obtained by combining -1 and 10, this pair of factors is chosen.
#êêêë 5xì + 9x - 2
#êêêè = 5xì + 10x - x - 2
#êêêè = 5xì + 10x + (-x) + (-2)
êêêè = 5x(x + 2) + (-1)(x + 2)
êêêè = (x + 2)(5x - 1)
Ç C
3
#êë Factorï2xì + 11x + 15 by the master product method.
êè A) (2x + 5)(x + 3)êêC) (3x + 2)(x + 5)
êè B) (2x + 3)(x + 5)êêD) (3x + 5)(x + 2)
#üêêêï2xì + 11x + 15
The master product is 2∙15 = 30.ïFactors of 30 are 1 times 30,
2 times 15, 3 times 10 and 6 times 5.ïSince the middle term of the
original expression can be obtained by combining 5 and 6, this pair
of factors is chosen.
#êêêë 2xì + 11x + 15
#êêêè = 2xì + 5x + 6x + 15
êêêè = x(2x + 5) + 3(2x + 5)
êêêè = (2x + 5)(x + 3)
Ç A
4
#êêFactorï6xì - x - 15 by the master product method.
êè A) (3x - 3)(5x + 2)êë C) (3x + 2)(5x - 3)
êè B) (3x - 5)(2x + 3)êë D) (3x + 2)(3x - 5)
#üêêêè6xì - x - 15
The master product is 6(-15) = -90. Factors of 90 are 1 times 90
2 times 45, 3 times 30, 5 times 18, 6 times 15 and 9 times 10.ïSince the
middle term of the original expression can be obtained by combining 9
and -10, this pair of factors is chosen.
#êêêë 6xì - x - 15
#êêêè = 6xì + (-x) + (-15)
#êêêè = 6xì + (-10x) + (9x) + (-15)
êêêè = 2x(3x + (-5)) + 3(3x + (-5))
êêêè = (3x - 5)(2x + 3)
Ç B
5
#êêFactorïxì - 5x - 14 by the master product method.
êè A) (7x - 2)(x - 1)êêC) (x - 2)(x + 7)
êè B) (2x - 7)(x + 1)êêD) (x - 7)(x + 2)
#üêêêèxì - 5x - 14
The master product is 1(-14) = -14. Factors of 14 are 1 times 14
and 2 times 7.ïSince the middle term of the original expression can be
obtained by combining 2 and -7, this pair of factors is chosen.
#êêêë xì - 5x - 14
#êêêè = xì + (-5x) + (-14)
#êêêè = xì + (-7x) + (2x) + (-14)
êêêè = x(x + (-7)) + 2(x + (-7))
êêêè = (x - 7)(x + 2)
Ç D
6
#êë Factorïxì + 7x + 12 by the master product method.
êè A) (3x + 1)(4x + 4)êë C) (4x + 3)(x + 1)
êè B) (x + 3)(x + 4)êê D) (3x - 1)(x + 4)
ü
#êêêêxì + 7x + 12
#êêêë= xì + 3x + 4x + 12
êêêë= x(x + 3) + 4(x + 3)
êêêë= (x + 3)(x + 4)
Ç B
7
#êëFactorïaì + 2ab - 15bì by the master product method.
êè A) (3a + 5b)(a - 1)êë C) (a + 5b)(a - 3b)
êè B) (5a - b)(3a + b)êë D) (a - b)(a + 5b)
ü
#êêêêaì + 2ab - 15bì
#êêêë= aì + 5ab - 3ab - 15bì
#êêêë= aì + 5ab + (-3ab) + (-15bì)
êêêë= a(a + 5b) + (-3b)(a + 5b)
êêêë= (a + 5b)(a - 3b)
Ç C
8
#êëFactor 2xmì + 9xm - 5xïby the master product method.
êè A) x(m + 5)(2m - 1)êë C) x(5m - 1)(2m + 5)
êè B) x(2m + 5)(m - 1)êë D) x(m - 5)(m - 2)
ü First take out the greatest common factor and then factor by
#êgrouping.êë2xmì + 9xm - 5x
#êêêë= x(2mì + 9m - 5)
#êêêë= x(2mì + 10m + (-m) + (-5))
êêêë= x(2m(m + 5) + (-1)(m + 5))
êêêë= x(m + 5)(2m - 1)
Ç A
9
#êïFactor 2bìxì - 2bìx - 60bìïby the master product method.
#êè A) 2bì(x - 6)(x + 5)êëC) 2bì(6x + 1)(x + 5)
#êè B) 2bì(5x - 1)(x - 6)êè D) 2bì(x - 5)(6x + 5)
ü
#êêêê2bìxì - 2bìx - 60bì
#êêêë= 2bì(xì - x - 30)
#êêêë= 2bì(xì + (-x) + (-30))
#êêêë= 2bì(xì + (-6x) + 5x + (-30)
#êêêë= 2bì(x(x + (-6)) + 5(x + (-6)))
#êêêë= 2bì(x - 6)(x + 5)
Ç A
10
#êë Factor xì - x - 20ïby the master product method.
êè A) (4x - 1)(x - 5)êêC) (5x - 1)(x + 4)
êè B) (5x + 4)(x - 1)êêD) (x + 4)(x - 5)
ü
#êêêêïxì - x - 20
#êêêê= xì + 4x + (-5x) + (-20)
êêêê= x(x + 4) + (-5)(x + 4)
êêêê= (x + 4)(x - 5)
Ç D