How do you factor #8w^2 - 18w + 9#?

1 Answer
Aug 1, 2015

Factor #y = 8x^2 - 18x + 9.#(1)

Ans: (4x - 3)(2x - 3)

Explanation:

#y = 8(x - p)(x - q)#
I use the new AC Method to factor trinomials.
Converted #y' = x^2 - 18x + 72 = #(x - p')(x - q')
Factor pairs of (-72) --> ...(4, 18)(6, 12) This sum is 18 = -b.
Then p' = -6 and q' = -12.
Back to equation (1): #p = (p')/a = -6/8 = -4/3# and #q = -12/8 = -3/2.#
Factored form: #y = 8(x - 3/4)(x - 3/2) = (4x - 3)(2x - 3)#