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

##### 1 Answer
Feb 16, 2016

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

#### Explanation:

I use the systematic new AC Method to factor trinomials
$f \left(x\right) = 18 {x}^{2} - 33 x + 5 =$18(x + p)(x + q)
Converted trinomial: $f ' \left(x\right) = {x}^{2} - 33 x + 90 =$ (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: $p = \frac{p '}{a} = - \frac{3}{18} = - \frac{1}{6}$ and $q = \frac{q '}{a} = - \frac{30}{18} = - \frac{5}{3.}$
Factored form: $y = 18 \left(x - \frac{1}{6}\right) \left(x - \frac{5}{3}\right) = \left(6 x - 1\right) \left(3 x - 5\right)$