# How do you use the rational roots theorem to find all possible zeros of P(x) = x^5 -2x^4 +4x -8?

Apr 26, 2016

The only Real zero is $x = 2$

#### Explanation:

By the rational roots theorem, any rational zeros of $P \left(x\right)$ will be expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 8$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1$, $\pm 2$, $\pm 4$, $\pm 8$

Trying each in turn, we find:

$P \left(2\right) = 32 - 32 + 8 - 8 = 0$

So $x = 2$ is a zero and $\left(x - 2\right)$ a factor of $P \left(x\right)$:

${x}^{5} - 2 {x}^{4} + 4 x - 8 = \left(x - 2\right) \left({x}^{4} + 4\right)$

Note that ${x}^{4} \ge 0$ for all Real values of $x$ so ${x}^{4} + 4 \ge 4$ has no Real zeros.

It does have Complex zeros $x = 1 \pm i$ and $x = - 1 \pm i$, being the $4$th roots of $- 4$.

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Bonus

Note that although ${x}^{4} + 4$ has no Real zeros or linear factors, it is possible to factor it into a product of two quadratics with Real coefficients.

In fact, we find:

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

This is an instance of a nice identity for factoring quartics:

$\left({a}^{2} - k a b + {b}^{2}\right) \left({a}^{2} + k a b + {b}^{2}\right) = {a}^{4} + \left(2 - {k}^{2}\right) {a}^{2} {b}^{2} + {b}^{4}$

Note in particular that if we put $k = \sqrt{2}$ we find:

$\left({a}^{2} - \sqrt{2} a b + {b}^{2}\right) \left({a}^{2} + \sqrt{2} a b + {b}^{2}\right) = {a}^{4} + {b}^{4}$

In our case we used this with $k = \sqrt{2}$, $a = x$ and $b = \sqrt{2}$