How do you factor #15b^2 + 32b + 7 = -9#?

1 Answer
Mar 25, 2016

Answer:

(5b + 4)(3b + 4)

Explanation:

#y = 15b^2 + 32b + 16 =# 15(b + p)(b + q)
Use the new AC Method to factor trinomials (Socratic Search).
Converted trinomial: #y' = b^2 + 32b + 240 = #(b + p')(b + q').
p' and q' have same sign because ac > 0.
Compose factor pairs of (ac = 240) -->...(10, 24)(12, 20). This last sum is (32 = b). Then, p' = 12 and q' = 20.
Back to y, #p = (p')/a = 12/15 = 4/5# and #q = (q')/a = 20/15 = 4/3.#
Factored form: #y = 15(b + 4/5)(b + 4/3) = (5b + 4)(3b + 4)#