# How do you factor 8w^7+18w^6-18w^5?

Aug 4, 2016

$2 {w}^{5} \left(4 w - 3\right) \left(w + 3\right)$

#### Explanation:

$f \left(w\right) = 2 {w}^{5} y = 2 {w}^{5} \left(4 {w}^{2} + 9 w - 9\right)$
Factor y by the new AC Method (Socratic Search)
$y = 4 {w}^{2} + 9 w - 9 = \left(w + p\right) \left(w + q\right)$
Converted trinomial y' = w^2 + 9w - 36 = (x + p')(x + q').
p' and q' have opposite signs because ac < 0. Compose factor pairs of (ac = -36)-->...(-3, 12). This sum is 9 = b. Then p' = -3 and q' = 12.
Back to trinomial y, $p = \frac{p '}{a} = - \frac{3}{4}$ and $q = \frac{q '}{a} = \frac{12}{4} = 3$
Factored form of y:
$y = \left(w - \frac{3}{4}\right) \left(w + 3\right) = \left(4 w - 3\right) \left(w + 3\right)$
Factored form of f(w):
$f \left(w\right) = 2 {w}^{5} \left(4 w - 3\right) \left(w + 3\right)$