How do you factor completely #18x^2-30x-3x+5#?

1 Answer
Feb 16, 2016

Answer:

y = (6x - 1)(3x - 5)

Explanation:

I use the systematic new AC Method to factor trinomials
#f(x) = 18x^2 - 33x + 5 = #18(x + p)(x + q)
Converted trinomial: #f'(x) = x^2 - 33x + 90 =# (x + p')(x + q').
p' and q' have same sign since ac > 0.
Factor pairs of (ac = 90) --> (2, 45)(3, 30). This sum is 33 = -b. Then the opposite sum (-3, -30) gives p' = -3 and q' = -30.
Back to original trinomial: #p = (p')/a = -3/18 = -1/6# and #q = (q')/a = -30/18 = -5/3.#
Factored form: #y = 18(x - 1/6)(x - 5/3) = (6x - 1)(3x - 5)#