# How do you use the rational root theorem to find the roots of f(x)=x^4-x-4?

Oct 24, 2015

You can't. $f \left(x\right) = 0$ has no rational roots.

#### Explanation:

By the rational root theorem, any rational root of ${x}^{4} - x - 4 = 0$ is expressible in lowest terms as $\frac{p}{q}$ where $p , q \in \mathbb{Z}$, $q \ne 0$, $p$ a divisor of the constant term $4$ and $q$ a divisor of the coefficient $1$ of the leading term.

So the only possible rational roots are:

$\pm 1$, $\pm 2$, $\pm 4$

Then we find:

$f \left(- 4\right) = 256 + 4 - 4 = 256$
$f \left(- 2\right) = 16 + 2 - 4 = 14$
$f \left(- 1\right) = 1 + 1 - 4 = - 2$
$f \left(1\right) = 1 - 1 - 4 = - 4$
$f \left(2\right) = 16 - 2 - 4 = 10$
$f \left(4\right) = 256 - 4 - 4 = 248$

So $f \left(x\right) = 0$ has no rational roots.

Since $f \left(x\right)$ is continuous, we can tell that $f \left(x\right) = 0$ has Real zeros in $\left(- 2 , - 1\right)$ and $\left(1 , 2\right)$.

Newton's method can then be used to find approximations for the two Real roots. Starting with an approximation ${a}_{0}$, find better approximations by iterating using the formula:

${a}_{i + 1} = {a}_{i} - f \frac{x}{f ' \left(x\right)}$

$f ' \left(x\right) = 4 {x}^{3} - 1$

Starting with ${a}_{0} = - 1.5$, we find:

${a}_{1} = - 1.3232758621$
${a}_{2} = - 1.2853460646$
${a}_{3} = - 1.2837842190$
${a}_{4} = - 1.2837816659$
${a}_{5} = - 1.2837816659$

Starting with ${a}_{0} = 1.5$, we find:

${a}_{1} = 1.535$
${a}_{2} = 1.5337528051$
${a}_{3} = 1.5337511688$
${a}_{4} = 1.5337511688$

graph{x^4-x-4 [-10, 10, -5, 5]}