# How do you solve t^3-13t-12=0 ?

Dec 22, 2016

The solutions are $t = - 1$, $t = 4$ and $t = - 3$.

#### Explanation:

Given:

$f \left(t\right) = {t}^{3} - 13 t - 12$

Notice that $12 + 1 = 13$, so is there some value like $t = \pm 1$ which will give $f \left(t\right) = 0$ ?

We find:

$f \left(- 1\right) = - 1 + 13 - 12 = 0$

So $t = - 1$ is a zero and $\left(t + 1\right)$ a factor:

${t}^{3} - 13 t - 12 = \left(t + 1\right) \left({t}^{2} - t - 12\right)$

To factor ${t}^{2} - t - 12$ find a pair of factors of $12$ which differ by $1$.

The pair $4 , 3$ works. Hence we find:

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

So the other two zeros are:

$t = 4 \text{ }$ and $\text{ } t = - 3$