How do you factor completely 6a^3+3a^2-18a?

Jul 4, 2018

$3 a \left(a + 2\right) \left(2 a - 3\right)$

Explanation:

Given that

$6 {a}^{3} + 3 {a}^{2} - 18 a$

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

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

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

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

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

Jul 4, 2018

$3 a \left(a + 2\right) \left(2 a - 3\right)$

Explanation:

To start, we notice that all terms have a $3 a$ in common, so we can factor that out to get

$3 a \left(2 {a}^{2} + a - 6\right)$

Next, we can factor by grouping. We can rewrite $a$ as $4 a - 3 a$. We now have

$3 a \left(\textcolor{b l u e}{2 {a}^{2} + 4 a} + \textcolor{p u r p \le}{- 3 a - 6}\right)$

We can factor a $2 a$ out of the blue term and a $- 3$ out of the purple term. We now have

$3 a \left(2 a \left(a + 2\right) - 3 \left(a + 2\right)\right)$

Both terms have an $a + 2$ in common, so we can factor that out. We will get

$3 a \left(a + 2\right) \left(2 a - 3\right)$

Hope this helps!