# Question #9ed95

Feb 7, 2018

$x = 4$

#### Explanation:

By inspection you can see that ${\left(4\right)}^{3} - {\left(4\right)}^{2} = 64 - 16 = 48$.

If we rearrange we get ${x}^{3} - {x}^{2} - 48 = 0$.

Knowing that $x = 4$ is a zero means that $x - 4$ is a factor.

Dividing we have:

$\frac{{x}^{3} - {x}^{2} - 48}{x - 4} = {x}^{2} + 3 x + 12$

So we can say that:

${x}^{3} - {x}^{2} - 48 = \left(x - 4\right) \left({x}^{2} + 3 x + 12\right)$

so

${x}^{3} - {x}^{2} - 48 = 0$

is equivalent to

$\left(x - 4\right) \left({x}^{2} + 3 x + 12\right) = 0$

We already known $x = 4$ is a solution.

${x}^{2} + 3 x + 12 = 0$ has no real solutions since the discriminant is less than 0: $9 - 4 \left(12\right) < 0$.

The only solution is $x = 4$, which I found by inspection, but we could also find by the rational root theorem.