How do you factor #18b^2+15bd+2d^2#?

1 Answer
Sep 15, 2016

Answer:

(6b + d)(3b + 2d)

Explanation:

Consider b as variable and d as a constant, and factor the trinomial by the new AC Method (Socratic Search)
#y = 18b^2 + 15db + 2d^2 =# 18(b + p)(b + q)
Converted trinomial: #y' = b^2 + 15db + 36d^2 = #(b + p')(b + q').
Find p' and q' that have same sign (ac > 0).
Compose factor pairs of (ac = 36d^2) --> (2d, 18d)(3d, 12d). This last sum is (3d + 12d = 15d = b). Then p' = 3d and q' = 12d.
Back to y, we get: #p = (p')/a = (3d)/18 = d/6#, and #q = (q')/a = (12d)/18 = (2d)/3#
Factored form:
#y = 18(b + d/6)(b + (2d)/3) = (6b + d)(3b + 2d)#