SOLUTION: The directions are: Factor each polynomial, if possible. If the polynomial cannot be factored using intergers, write prime. The problem is: 12q^2+34q-28

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: The directions are: Factor each polynomial, if possible. If the polynomial cannot be factored using intergers, write prime. The problem is: 12q^2+34q-28      Log On


   

Question 308194: The directions are:
Factor each polynomial, if possible. If the polynomial cannot be factored using intergers, write prime.
The problem is:
12q^2+34q-28
Answer by edjones(8007) About Me  (Show Source):
You can put this solution on YOUR website!
12q^2+34q-28
=2(6q^2+17q-14)
6*-14=-84
2 factors of -84 whose sum is 17 are 21 and -4
2(6q^2+21q-4q-14)
=2(3q(2q+7)-2(2q+7))
=2(3q-2)(2q+7)
.
Ed