Is it possible to factor #y=2x^2+4x-30#? If so, what are the factors?

1 Answer
Jan 10, 2016

Answer:

y = 2(x - 3)(x + 5)

Explanation:

I use the new AC Method to factor trinomials (Socratic Search).
#y = 2x^2 + 4x - 30 =# 2(x + p)(x + q) (1)
Converted trinomial: #y' = x^2 + 4x - 60 = #(x + p')(x + q')
p' and q' have opposite signs because ac < 0.
Compose factor pairs of (ac = -60) --> (-4, 15)(-6, 10). This sum is
(10 - 6 = 4 = b). Then p' = -6 and q' = 10.
Back to original trinomial (1): #p = (p')/a = -6/2 = -3# and #q = (q')/a = 10/2 = 5#
Factored form: y = 2(x - 3)(x + 5)