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

1 Answer
Aug 4, 2016

Answer:

#2w^5(4w - 3)(w + 3)#

Explanation:

#f(w) = 2w^5y = 2w^5(4w^2 + 9w - 9)#
Factor y by the new AC Method (Socratic Search)
#y = 4w^2 + 9w - 9 = (w + p)(w + q)#
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 = (p')/a = -3/4# and #q = (q')/a = 12/4 = 3#
Factored form of y:
#y = (w - 3/4)(w + 3) = (4w - 3)(w + 3)#
Factored form of f(w):
#f(w) = 2w^5(4w - 3)(w + 3)#