How do you factor #7x^2-30x+8#?

1 Answer
Aug 8, 2015

Answer:

Factor: #y = 7x^2 - 30x + 8#

Ans: (7x - 2)(x - 4)

Explanation:

I use the new AC Method: y = 7(x - p)(x - q)
Converted trinomial #y' = x^2 _ 30x + 56 = #(x - p')(x - q')
p' and q' have same sign (Rule of signs)
Factor pairs of 56 --> (2, 28). This sum is 30 = -b.
Change this sum to the opposite. Then p' = -2 and q' = -28.
Therefor, p = (p')/a = -2/7 and q' = -28/7 = -4

y = 7(x - 2/7)(x - 4) = (7x - 2)(x - 4)