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

Sep 15, 2016

(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)
$y = 18 {b}^{2} + 15 \mathrm{db} + 2 {d}^{2} =$ 18(b + p)(b + q)
Converted trinomial: $y ' = {b}^{2} + 15 \mathrm{db} + 36 {d}^{2} =$(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: $p = \frac{p '}{a} = \frac{3 d}{18} = \frac{d}{6}$, and $q = \frac{q '}{a} = \frac{12 d}{18} = \frac{2 d}{3}$
Factored form:
$y = 18 \left(b + \frac{d}{6}\right) \left(b + \frac{2 d}{3}\right) = \left(6 b + d\right) \left(3 b + 2 d\right)$