# What are all the possible rational zeros for f(x)=x^4-7x^2+10 and how do you find all zeros?

Nov 20, 2016

$f \left(x\right)$ only has irrational zeros: $\pm \sqrt{5}$ and $\pm \sqrt{2}$

#### Explanation:

Given:

$f \left(x\right) = {x}^{4} - 7 {x}^{2} + 10$

By the rational root theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $10$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

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

However, note that we can factor $f \left(x\right)$ as a quadratic in ${x}^{2}$ then using the difference of squares identity to find:

${x}^{4} - 7 {x}^{2} + 10 = \left({x}^{2} - 5\right) \left({x}^{2} - 2\right)$

$\textcolor{w h i t e}{{x}^{4} - 7 {x}^{2} + 10} = \left({x}^{2} - {\left(\sqrt{5}\right)}^{2}\right) \left({x}^{2} - {\left(\sqrt{2}\right)}^{2}\right)$

$\textcolor{w h i t e}{{x}^{4} - 7 {x}^{2} + 10} = \left(x - \sqrt{5}\right) \left(x + \sqrt{5}\right) \left(x - \sqrt{2}\right) \left(x + \sqrt{2}\right)$

So the only zeros of $f \left(x\right)$ are irrational: $\pm \sqrt{5}$, $\pm \sqrt{2}$