# How do you use the rational root theorem to find the roots of 2x^3 + 7x^2 - 77x - 40?

May 7, 2016

$\left\{\begin{matrix}x = - \frac{1}{2} \\ x = - 8 \\ x = 5\end{matrix}\right.$

#### Explanation:

$f \left(x\right) = 2 {x}^{3} + 7 {x}^{2} - 77 x - 40$

By the rational root theorem, any rational zeros of this polynomial must be expressible in the form $\frac{p}{q}$ for integers $p$, $q$ with $p$ a divisor of the constant term $- 40$ and $q a$ a divisor of the coefficient $2$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{2}$, $\pm 1$, $\pm 2$, $\pm \frac{5}{2}$, $\pm 4$, $\pm 5$, $\pm 8$, $\pm 10$, $\pm 20$, $\pm 40$

This is rather a lot of possibilities to try, but trying each in turn we soon find:

$f \left(- \frac{1}{2}\right) = - \frac{1}{4} + \frac{7}{4} + \frac{77}{2} - 40 = \frac{- 1 + 7 + 154 - 160}{4} = 0$

So $x = - \frac{1}{2}$ is a zero and $\left(2 x + 1\right)$ a factor:

$2 {x}^{3} + 7 {x}^{2} - 77 x - 40 = \left(2 x + 1\right) \left({x}^{2} + 3 x - 40\right)$

To factor the remaining quadaratic find a pair of factors of $40$ that differ by $3$. The pair $8 , 5$ works.

Hence:

${x}^{2} + 3 x - 40 = \left(x + 8\right) \left(x - 5\right)$

So the other two zeros are $x = - 8$ and $x = 5$