How do you use the rational roots theorem to find all possible zeros of x^4-9x^2+20?

Jul 28, 2016

Zeros: $x = \pm 2$ and $x = \pm \sqrt{5}$

Explanation:

Given $f \left(x\right) = {x}^{4} - 9 {x}^{2} + 20$, the rational roots theorem tells you that any rational zeros must be expressible in the form $\frac{p}{q}$ for integers $p$, $q$ with $p$ a divisor of the constant term $20$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1 , \pm 2 , \pm 4 , \pm 5 , \pm 10 , \pm 20$

Evaluating $f \left(x\right)$ for each in turn, this allows you to find the two rational zeros:

$f \left(2\right) = f \left(- 2\right) = 16 - 36 + 20 = 0$

You can then divide $f \left(x\right)$ by $\left(x - 2\right) \left(x + 2\right) = {x}^{2} - 4$ to get ${x}^{2} - 5$, which has zeros $x = \pm \sqrt{5}$.

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Alternatively:

Note that ${x}^{4} - 9 {x}^{2} + 20$ is a quadratic in ${x}^{2}$. That is:

${x}^{4} - 9 {x}^{2} + 20 = {\left({x}^{2}\right)}^{2} - 9 \left({x}^{2}\right) + 20$

Note also that $4 + 5 = 9$ and $4 \times 5 = 20$

So we find:

${x}^{4} - 9 {x}^{2} + 20$

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

$= \left({x}^{2} - {2}^{2}\right) \left({x}^{2} - {\left(\sqrt{5}\right)}^{2}\right)$

$= \left(x - 2\right) \left(x + 2\right) \left(x - \sqrt{5}\right) \left(x + \sqrt{5}\right)$

Hence zeros:

$x = \pm 2$ and $x = \pm \sqrt{5}$