How do you find all rational roots for x^4 - 2x^3 - 8x^2 + 10x + 15 = 0?

Aug 8, 2016

This equation has rational roots $- 1 , 3$ and irrational roots $\pm \sqrt{5}$

Explanation:

$f \left(x\right) = {x}^{4} - 2 {x}^{3} - 8 {x}^{2} + 10 x + 15$

By the rational root theorem, any rational zeros of $f \left(x\right)$ must be expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $15$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1 , \pm 3 , \pm 5 , \pm 15$

Trying each in turn, we find:

$f \left(- 1\right) = 1 + 2 - 8 - 10 + 15 = 0$

So $x = - 1$ is a zero and $\left(x + 1\right)$ a factor:

${x}^{4} - 2 {x}^{3} - 8 {x}^{2} + 10 x + 15 = \left(x + 1\right) \left({x}^{3} - 3 {x}^{2} - 5 x + 15\right)$

Looking at the remaining cubic, notice that ratio of the first and second terms is the same as that between the third and fourth terms. So this cubic will factor by grouping:

${x}^{3} - 3 {x}^{2} - 5 x + 15$

$= \left({x}^{3} - 3 {x}^{2}\right) - \left(5 x - 15\right)$

$= {x}^{2} \left(x - 3\right) - 5 \left(x - 3\right)$

$= \left({x}^{2} - 5\right) \left(x - 3\right)$

Hence the remaining zeros are $x = 3$ and $x = \pm \sqrt{5}$