179 
… 6.2‹Factoring Trinomials by the Master Product Method
„‹Please factor the following trinomials by the master
ˆˆproduct method.
#ƒSˆˆ‹Factor‹3x + 7x + 2
ˆˆ‰ The master product is 3ω2 = 6
#ˆˆˆˆ 3x + 7x + 2
#ˆˆˆ‰ = 3x + 6x + x + 2
#ˆˆˆ‰ = (3x + 6x) + (x + 2)
ˆˆˆ‰ = 3x(x + 2) + (x + 2)
ˆˆˆ‰ = (3x + 1)(x + 2)
‚S
#In order to factor the expression, 3x + 7x + 2 by the master product
method it is first necessary to find the master product.‹This number
#is the product of the coefficient of the x term and the constant term.
The master product is 3ω2 = 6.‹Next, all positive factors of 6 are
written down (1 times 6Š2 times 3).‹Then a pair of factors is chosen
that can combine to give the coefficient of the middle term of the
original expression.‹Since 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.
#ˆˆ‰ 3x + 7x + 2‹=‹3x + 6x + x + 2
These four terms are then factored by grouping.
#ˆˆˆ‰ 3x + 6x + x + 2
#ˆˆˆŠ = (3x + 6x) + (x + 2)
ˆˆˆŠ = 3x(x + 2) + (x + 2)
ˆˆˆŠ = (3x + 1)(x + 2)

#This completes the factorization of the trinaomial, 3x + 7x + 2, by the
master product method.
 1
#ˆˆFactor‹2x + x - 3 by the master product method.


ˆŠ A) (2x + 3)(x - 1)ˆˆC) (x - 3)(2x + 1)
ˆŠ B) (2x - 1)(x + 3)ˆˆD) † of ‡
#ˆˆˆŠ2x + x - 3
The master product is 2ω3 = 6.‹Factors of 6 are 1 times 6 and
2 times 3.‹Since the middle term of the original expression can be
obtained by combining 3 and -2, this pair of factors is chosen.
#ˆˆˆ‰ 2x + x - 3
#ˆˆˆŠ = 2x + 3x - 2x - 3
#ˆˆˆŠ = 2x + 3x + (-2x) + (-3)
ˆˆˆŠ = x(2x + 3) + (-1)(2x + 3)
ˆˆˆŠ = (2x + 3)(x - 1)
€ A
 2
#ˆˆFactor‹5x + 9x - 2 by the master product method.


ˆŠ A) (x - 1)(5x + 2)ˆˆC) (x + 2)(5x - 1)
ˆŠ B) (x - 5)(2x + 1)ˆˆD) (x - 1)(2x - 5)
#ˆˆˆŠ5x + 9x - 2
The master product is 5(-2) = -10. Factors of 10 are 1 times 10
and 2 times 5.‹Since the middle term of the original expression can be
obtained by combining -1 and 10, this pair of factors is chosen.
#ˆˆˆ‰ 5x + 9x - 2
#ˆˆˆŠ = 5x + 10x - x - 2
#ˆˆˆŠ = 5x + 10x + (-x) + (-2)
ˆˆˆŠ = 5x(x + 2) + (-1)(x + 2)
ˆˆˆŠ = (x + 2)(5x - 1)
€ C
 3
#ˆ‰ Factor‹2x + 11x + 15 by the master product method.


ˆŠ A) (2x + 5)(x + 3)ˆˆC) (3x + 2)(x + 5)
ˆŠ B) (2x + 3)(x + 5)ˆˆD) (3x + 5)(x + 2)
#ˆˆˆ‹2x + 11x + 15
The master product is 2ω15 = 30.‹Factors of 30 are 1 times 30,
2 times 15, 3 times 10 and 6 times 5.‹Since the middle term of the
original expression can be obtained by combining 5 and 6, this pair
of factors is chosen.
#ˆˆˆ‰ 2x + 11x + 15
#ˆˆˆŠ = 2x + 5x + 6x + 15
ˆˆˆŠ = x(2x + 5) + 3(2x + 5)
ˆˆˆŠ = (2x + 5)(x + 3)
€ A
 4
#ˆˆFactor‹6x - x - 15 by the master product method.


ˆŠ A) (3x - 3)(5x + 2)ˆ‰ C) (3x + 2)(5x - 3)
ˆŠ B) (3x - 5)(2x + 3)ˆ‰ D) (3x + 2)(3x - 5)
#ˆˆˆŠ6x - 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.‹Since the
middle term of the original expression can be obtained by combining 9
and -10, this pair of factors is chosen.
#ˆˆˆ‰ 6x - x - 15
#ˆˆˆŠ = 6x + (-x) + (-15)
#ˆˆˆŠ = 6x + (-10x) + (9x) + (-15)
ˆˆˆŠ = 2x(3x + (-5)) + 3(3x + (-5))
ˆˆˆŠ = (3x - 5)(2x + 3)
€ B
 5
#ˆˆFactor‹x - 5x - 14 by the master product method.


ˆŠ A) (7x - 2)(x - 1)ˆˆC) (x - 2)(x + 7)
ˆŠ B) (2x - 7)(x + 1)ˆˆD) (x - 7)(x + 2)
#ˆˆˆŠx - 5x - 14
The master product is 1(-14) = -14. Factors of 14 are 1 times 14
and 2 times 7.‹Since the middle term of the original expression can be
obtained by combining 2 and -7, this pair of factors is chosen.
#ˆˆˆ‰ x - 5x - 14
#ˆˆˆŠ = x + (-5x) + (-14)
#ˆˆˆŠ = x + (-7x) + (2x) + (-14)
ˆˆˆŠ = x(x + (-7)) + 2(x + (-7))
ˆˆˆŠ = (x - 7)(x + 2)
€ D
 6
#ˆ‰ Factor‹x + 7x + 12 by the master product method.


ˆŠ A) (3x + 1)(4x + 4)ˆ‰ C) (4x + 3)(x + 1)
ˆŠ B) (x + 3)(x + 4)ˆˆ D) (3x - 1)(x + 4)


#ˆˆˆˆx + 7x + 12
#ˆˆˆ‰= x + 3x + 4x + 12
ˆˆˆ‰= x(x + 3) + 4(x + 3)
ˆˆˆ‰= (x + 3)(x + 4)
€ B
 7
#ˆ‰Factor‹a + 2ab - 15b by the master product method.


ˆŠ A) (3a + 5b)(a - 1)ˆ‰ C) (a + 5b)(a - 3b)
ˆŠ B) (5a - b)(3a + b)ˆ‰ D) (a - b)(a + 5b)

#ˆˆˆˆa + 2ab - 15b
#ˆˆˆ‰= a + 5ab - 3ab - 15b
#ˆˆˆ‰= a + 5ab + (-3ab) + (-15b)
ˆˆˆ‰= a(a + 5b) + (-3b)(a + 5b)
ˆˆˆ‰= (a + 5b)(a - 3b)
€ C
 8
#ˆ‰Factor 2xm + 9xm - 5x‹by the master product method.


ˆŠ A) x(m + 5)(2m - 1)ˆ‰ C) x(5m - 1)(2m + 5)
ˆŠ B) x(2m + 5)(m - 1)ˆ‰ D) x(m - 5)(m - 2)
 First take out the greatest common factor and then factor by
#ˆgrouping.ˆ‰2xm + 9xm - 5x
#ˆˆˆ‰= x(2m + 9m - 5)
#ˆˆˆ‰= x(2m + 10m + (-m) + (-5))
ˆˆˆ‰= x(2m(m + 5) + (-1)(m + 5))
ˆˆˆ‰= x(m + 5)(2m - 1)
€ A
 9
#ˆ‹Factor 2bx - 2bx - 60b‹by the master product method.


#ˆŠ A) 2b(x - 6)(x + 5)ˆ‰C) 2b(6x + 1)(x + 5)
#ˆŠ B) 2b(5x - 1)(x - 6)ˆŠ D) 2b(x - 5)(6x + 5)

#ˆˆˆˆ2bx - 2bx - 60b
#ˆˆˆ‰= 2b(x - x - 30)
#ˆˆˆ‰= 2b(x + (-x) + (-30))
#ˆˆˆ‰= 2b(x + (-6x) + 5x + (-30)
#ˆˆˆ‰= 2b(x(x + (-6)) + 5(x + (-6)))
#ˆˆˆ‰= 2b(x - 6)(x + 5)
€ A
 10
#ˆ‰ Factor x - x - 20‹by the master product method.


ˆŠ A) (4x - 1)(x - 5)ˆˆC) (5x - 1)(x + 4)
ˆŠ B) (5x + 4)(x - 1)ˆˆD) (x + 4)(x - 5)

#ˆˆˆˆ‹x - x - 20
#ˆˆˆˆ= x + 4x + (-5x) + (-20)
ˆˆˆˆ= x(x + 4) + (-5)(x + 4)
ˆˆˆˆ= (x + 4)(x - 5)

€ D