# How do you use the rational root theorem to find the roots of x^4 - 7x^2 + 12 = 0?

May 12, 2015

If $\frac{p}{q}$ is a root of ${x}^{4} - 7 {x}^{2} + 12 = 0$ written in lowest terms, then $p$ must be a divisor of the constant term $12$ and $q$ must be a divisor of the coefficient ($1$) of the highest order term ${x}^{4}$.

So $q = \pm 1$. We might as well choose $q = 1$, since we can choose the sign of $p$.

$p$ must be a factor of $12$, i.e. $\pm 12$, $\pm 6$, $\pm 4$, $\pm 3$, $\pm 2$ or $\pm 1$.

Since ${x}^{4}$ grows quite fast with larger positive or negative values of $x$, it's probably better to try the possible smaller values of $x$ first.

Trying a few substitutions, we quickly find that $x = \pm 1$ are not solutions and $x = \pm 2$ are solutions. When $| x | = 3$, we find that ${x}^{4} - 7 {x}^{2} + 12 = 81 - 63 + 12 = 30$. Larger values of $x$ will prduce larger values of ${x}^{4} - 7 {x}^{2} + 12$ as the ${x}^{4}$ term becomes dominant.

So the only rational roots are $x = 2$ and $x = - 2$.

May 14, 2015

George C gives a great answer to find that $\pm 2$ are zeros of the polynomial. (Roots of the equation.)

If you need to find all (real or complex) roots, there a bit more work to do.

Because, $2$ and $- 2$ are roots, we know that
$x - 2$ and $x + 2$ are factors of the polynomial on the left..

Divide by the factors using synthetic division or long division. Or divide by the product of these 2 factors, ${x}^{2} - 4$

You'll get

${x}^{2} - 7 x + 12 = \left({x}^{2} - 4\right) \left({x}^{2} - 3\right) = 0$

So the roots are:

$2 , - 2 , \sqrt{3} , - \sqrt{3}$