# How do you factor #6x^2 - 11x +3?

$f \left(x\right) = 6 {x}^{2} - 11 x + 3 = \left(x - p\right) \left(x - q\right)$. Find p and q.
Converted trinomial:$f ' \left(x\right) = {x}^{2} - 11 x + 18 = \left(x - p '\right) \left(x - q '\right)$(a.c = 18)
Find p' and q' by composing factor pairs of a.c = 18: (1, 18)(2, 9). This last sum is 11 = -b. Then p' = -2 and q'= -9. We get: $p = \frac{p '}{a} = - \frac{2}{6} = - \frac{1}{3}$and $q = \frac{q '}{a} = - \frac{9}{6} = - \frac{3}{2}$.
$f \left(x\right) = \left(x - \frac{1}{3}\right) \left(x - \frac{3}{2}\right) = \left(3 x - 1\right) \left(2 x - 3\right)$
$f \left(x\right) = 6 {x}^{2} - 9 x - 2 x + 3 = 6 {x}^{2} - 11 x + 3$. OK