# How do you use the rational root theorem to find the roots of P(x) = x^4 - x^3 - 15x^2 + 3x + 36?

Jul 11, 2015

Use the rational roots theorem to help find rational roots $- 3$ and $4$, then divide $P \left(x\right)$ by $\left(x + 3\right) \left(x - 4\right)$ to reduce the problem to a simple quadratic with roots $\pm \sqrt{3}$

#### Explanation:

By the rational root theorem, all rational roots of $P \left(x\right)$ must be of the form $\frac{p}{q}$ in lowest terms for some integers $p$, $q$ where $p$ is a divisor of the constant term $36$ of $P \left(x\right)$ and $q$ is a divisor of the coefficient $1$ of the highest order term ${x}^{4}$.

$36 = {2}^{2} \cdot {3}^{2}$

So any rational roots must be one of:

$\pm 1$, $\pm 2$, $\pm 3$, $\pm 4$, $\pm 6$, $\pm 9$, $\pm 12$, $\pm 18$ or $\pm 36$

$P \left(1\right) = 1 - 1 - 15 + 3 + 36 = 24$
$P \left(- 1\right) = 1 + 1 - 15 - 3 + 36 = 20$
$P \left(2\right) = 16 - 8 - 60 + 6 + 36 = - 10$
$P \left(- 2\right) = 16 + 8 - 60 - 6 + 36 = - 6$
$P \left(3\right) = 81 - 27 - 135 + 9 + 36 = - 36$
$P \left(- 3\right) = 81 + 27 - 135 - 9 + 36 = 0$
$P \left(4\right) = 256 - 64 - 240 + 12 + 36 = 0$

So $x = - 3$ and $x = 4$ are roots and $\left(x + 3\right)$ and $\left(x - 4\right)$ are factors.

If we divide $P \left(x\right)$ by $\left(x + 3\right) \left(x - 4\right)$ then the constant term of the resulting polynomial will be $- 3$. So the only possible remaining rational roots would be $\pm 1$ or $\pm 3$. We have eliminated $\pm 1$ and $3$, but it is possible that $- 3$ is a repeated root? No, because if it were, it would be possible to divide by $\left(x + 3\right)$ again, leaving a linear factor $\left(x - 1\right)$ and we already know that $1$ is not a root.

So we have found all of the rational roots. What else is there?

Let's divide $P \left(x\right)$ by $\left(x + 3\right) \left(x - 4\right)$ to find out ...

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

$P \left(x\right) = {x}^{4} - {x}^{3} - 15 {x}^{2} + 3 x + 36$

$= \left({x}^{4} - {x}^{3} - 12 {x}^{2}\right) - \left(3 {x}^{2} - 3 x - 36\right)$

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

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

So $\frac{P \left(x\right)}{{x}^{2} - x - 12} = {x}^{2} - 3$

which has roots $x = \pm \sqrt{3}$