How do you factor #8x^2-8x+2#?

1 Answer
Aug 9, 2015

Factor: #y = 8x^2 - 8x + 2#

Ans: #2(2x - 1)^2#

Explanation:

Use the new AC Method: y = 8(x - p)(x - q)
Converted trinomial: #y' = x^2 - 8x + 16 =# (x - p')(x - q')
Factor pairs of 16 --> (2, 8)(4, 4) This sum is 8 = -b.
Change this sum to the opposite, Then, p' = -4 and q' = -4.
Therefor, #p = q = -4/8 = -1/2#

#y = 8(x - 1/2)(x - 1/2) = 2(2x - 1)^2#