# How do you factor 6c³ + 11c² -10c?

May 23, 2015

$6 {c}^{3} + 11 {c}^{2} - 10 c = c \left(6 {c}^{2} + 11 x - 10\right)$

$= c \left(2 c + 5\right) \left(3 c - 2\right)$

I first noticed that all the terms were divisible by $c$, so separated that out as a factor.

I then looked for a factorisation of the form:

$6 {c}^{2} + 11 x - 10 = \left(2 c + a\right) \left(3 c + b\right)$

$= 6 {c}^{2} + \left(3 a + 2 b\right) c + a b$

So I looked for $a$ and $b$ such that $3 a + 2 b = 11$ and $a b = - 10$

I could tell that $a$ must be odd, because otherwise both $2 c$ and $a$ would be even and all the resulting coefficients would be even.

So $a = \pm 1$ or $a = \pm 5$.

In order to get $3 a + b = 11$ then the best choice to try seemed to be $a = 5$, which forces $b = \frac{- 10}{a} = \frac{- 10}{5} = - 2$.

So try $\left(2 c + 5\right) \left(3 c - 2\right)$ - works.