# How do you solve -2x^3-8x^2+6x+36?

Jun 13, 2015

$- 2 {x}^{3} - 8 {x}^{2} + 6 x + 36$ can be factored or its zeros found using a combination of the rational roots theorem and by recognising a perfect square trinomial factor.

#### Explanation:

$- 2 {x}^{3} - 8 {x}^{2} + 6 x + 36$

$= - 2 \left({x}^{3} + 4 {x}^{2} - 3 x - 18\right)$

Let $f \left(x\right) = {x}^{3} + 4 {x}^{2} - 3 x - 18$

By the rational roots theorem, any rational root of $f \left(x\right) = 0$ of the form $\frac{p}{q}$ in lowest terms must be such that $p$ is a divisor of $18$ and $q$ is a divisor of $1$. That is:

$x = \pm 1$, $\pm 2$, $\pm 3$, $\pm 6$, $\pm 9$ or $\pm 18$

Trying the first few:

$f \left(1\right) = 1 + 4 - 3 - 18 = - 16$
$f \left(- 1\right) = - 1 + 4 + 3 - 18 = - 12$
$f \left(2\right) = 8 + 16 - 6 - 18 = 0$

So $x = 2$ is a root of $f \left(x\right) = 0$ and $\left(x - 2\right)$ is a factor of $f \left(x\right)$

${x}^{3} + 4 {x}^{2} - 3 x - 18 = \left(x - 2\right) \left({x}^{2} + 6 x + 9\right)$

${x}^{2} + 6 x + 9$ is recognisable as a perfect square trinomial:
It is of the form ${a}^{2} + 2 a b + {b}^{2} = {\left(a + b\right)}^{2}$ with $a = x$ and $b = 3$, so

$\left({x}^{2} + 6 x + 9\right) = {\left(x + 3\right)}^{2}$

Putting this all together we get:

$- 2 {x}^{3} - 8 {x}^{2} + 6 x + 36 = - 2 \left(x - 2\right) \left(x + 3\right) \left(x + 3\right)$

So $- 2 {x}^{3} - 8 {x}^{2} + 6 x + 36 = 0$ has one root $x = 2$ and one repeated root $x = - 3$ with multiplicity $2$.