# How do you find all rational roots for x^5 - 2x^4 + 11x^3 - 22x^2 - 12x + 24 = 0?

Apr 7, 2016

$\pm 1 , 2$

#### Explanation:

We will use the Rational Root Theorem:

If the rational number r/s is a root of a polynomial whose coefficients are integers, then the integer r is a factor of the constant term, and the integer s is a factor of the leading coefficient.

r=24 and s=1

So the rational roots must be factor of 24/1=24:

$\pm 1 , \pm 2 , \pm 3 , \pm 4 , \pm 6 , \pm 8 , \pm 12$

By trying these possible roots, we discover that $\pm 1 , 2$ are roots of the polynom.

If we divide the polynom by $\left(x - 1\right) \left(x + 1\right) \left(x - 2\right)$ we will obtain ${x}^{2} + 12$ which has the roots $\pm i \sqrt{12}$, which are not rational.