# How do you use the rational root theorem to find the roots of x^4 + 2x + 2 = 0?

Jun 11, 2015

By the rational root theorem, any rational roots of ${x}^{4} + 2 x + 2 = 0$ must be $\pm 1$ or $\pm 2$, neither of which works. So all the roots are irrational or complex.

#### Explanation:

Given $f \left(x\right) = {x}^{4} + 2 x + 2$

By the rational root theorem, any rational factor $\frac{p}{q}$ in lowest terms must have $p$ is a divisor of the constant term $2$ and $q$ is a divisor of the multiplier ($1$) of the highest order term.

That means that any rational root of $f \left(x\right) = 0$ must be $\pm 1$ or $\pm 2$

$f \left(- 2\right) = 16 - 4 + 2 = 14$
$f \left(- 1\right) = 1 - 2 + 2 = 1$
$f \left(1\right) = 1 + 2 + 2 = 5$
$f \left(2\right) = 16 + 4 + 2 = 22$

So none of $\pm 1$ or $\pm 2$ are roots of $f \left(x\right) = 0$.
So it has no rational roots.

$f ' \left(x\right) = 4 {x}^{3} + 2$

This is zero when $x = - \frac{1}{\sqrt[3]{2}}$ and for no other real values of $x$. So $f \left(x\right)$ has one turning point.

$f \left(- \frac{1}{\sqrt[3]{2}}\right) = {\left(- \frac{1}{\sqrt[3]{2}}\right)}^{4} + 2 \left(- \frac{1}{\sqrt[3]{2}}\right) + 2$

$= \frac{1}{2 \sqrt[3]{2}} - \frac{2}{\sqrt[3]{2}} + 2$

$= - \frac{3}{2} \cdot \frac{1}{\sqrt[3]{2}} + 2$

$> - \frac{3}{2} + 2 = \frac{1}{2} > 0$

So $f \left(x\right) > 0$ for all $x \in \mathbb{R}$

That is $f \left(x\right) = 0$ has no real roots. It has $4$ complex roots.

graph{x^4+2x+2 [-11.5, 8.5, -0.8, 9.2]}