How do you factor #9x ^ { 2} + 21x - 18#?

2 Answers
Jul 21, 2017

To factor #9x2+21x−18#

Factor out the #3#: #3(3x^2+7x-6)#
Continue factoring: #3(3x-2)(x+3)#

Sep 10, 2017

Answer:

#3(3x - 2)(x + 3)#

Explanation:

#f(x) = 9x^2 + 21x - 18 = 3(3x^2 + 7x - 6)#

Factor the trinomial in parentheses by the new AC Method (Socratic, Google Search)

#y = 3x^2 + 7x - 6 = 3(x + p)(x + q)#

Factor converted trinomial:
#y' = x^2 + 7x - 18 =# (x + p')(x + q').

Find # p' and q' #

knowing the sum #(b = 7)# and the product #(ac = - 18).#

They are: #p' = -2 and q' = 9 rarr ["sum " = 7 and " product " = - 18]#

Back to y:

#p = (p')/a = -2/3# , and #q = (q')/a = 9/3 = 3#

Factored form of #f(x):#

#f(x) = 3(3)(x - 2/3)(x + 3) = 3(3x - 2)(x + 3)#