# How do you factor 6z^3-3z^2-30z?

Oct 3, 2015

$3 z \left(2 z - 5\right) \left(z + 2\right)$

#### Explanation:

Factor the the GCF, which is $3 z$

$3 z \left(2 {z}^{2} - 1 z - 10\right)$

To see if the quadratic can be factored any further, find the two roots ${r}_{1}$ and ${r}_{2}$, if they exist the quadratic will be able to be written in the form of $\left(z - {r}_{1}\right) \left(z - {r}_{2}\right)$.

The roots are $z = \frac{1 \pm \sqrt{1 - 4 \cdot 2 \cdot \left(- 10\right)}}{4}$, from that we get

$z = \frac{1 \pm \sqrt{1 - \left(- 80\right)}}{4} = \frac{1 \pm \sqrt{81}}{4}$
$z = \frac{1 \pm 9}{4}$

So we have
${r}_{1} = \frac{1 + 9}{4} = \frac{10}{4} = \frac{5}{2}$
${r}_{2} = \frac{1 - 9}{4} = - \frac{8}{4} = - 2$

Now, since the parabola there has a two as the leading coefficient and we have a root who's a fraction two, we can say that the factored form is

$3 z \left(2 z - 5\right) \left(z + 2\right)$