How do you factor #6x^2+7x-3#?

1 Answer
Aug 8, 2015

Answer:

Factor: #y = 6x^2 + 7x - 3#

Ans: (3x + 1)(2x - 3)

Explanation:

Use the new AC Method. y = 6(x - p)(x - q)
Converted trinomial: #y' = x^2 + 7x - 18# = (x - p)(x - q)
p' and q' have opposite signs (Rule of signs).
Factor pairs of 18 --> (-1. 18)(-2, 9). This sum is 7 = -b.
Then p' = 2 and q' = -9.
Therefor, #p = (p'/a) = 2/6 = 1/3# and #q' = -9/6 = -3/2#

y = 6(x + 1/3)(2x + 3) = (3x + 1)(2x - 3)