How do you factor 4c^3-2c^2-6c?

Aug 30, 2016

Answer:

$4 {c}^{3} - 2 {c}^{2} - 6 c = c \left(2 c - 3\right) \left(2 c + 2\right)$

Explanation:

Note that all of the terms are divisible by $c$, so we can separate that out as a factor first.

We can then factor the remaining quadratic by completing the square and using the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

with $a = \left(2 c - \frac{1}{2}\right)$ and $b = \frac{5}{2}$ as follows:

$4 {c}^{3} - 2 {c}^{2} - 6 c$

$= c \left(4 {c}^{2} - 2 c - 6\right)$

$= c \left({\left(2 c - \frac{1}{2}\right)}^{2} - \frac{1}{4} - 6\right)$

$= c \left({\left(2 c - \frac{1}{2}\right)}^{2} - \frac{25}{4}\right)$

$= c \left({\left(2 c - \frac{1}{2}\right)}^{2} - {\left(\frac{5}{2}\right)}^{2}\right)$

$= c \left(\left(2 c - \frac{1}{2}\right) - \frac{5}{2}\right) \left(\left(2 c - \frac{1}{2}\right) + \frac{5}{2}\right)$

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