# How do you use the rational root theorem to find the roots of f(x)=x^3+x^2-8x-6 ?

Mar 23, 2018

The zeros of $f \left(x\right)$ are $- 3$, $1 + \sqrt{3}$ and $1 - \sqrt{3}$

#### Explanation:

Given:

$f \left(x\right) = {x}^{3} + {x}^{2} - 8 x - 6$

The rational root theorem tells us that any rational zero of $f \left(x\right)$ is expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 6$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1 , \pm 2 , \pm 3 , \pm 6$

Trying each of these in turn, we eventually find:

$f \left(- 3\right) = {\left(\textcolor{b l u e}{- 3}\right)}^{3} + {\left(\textcolor{b l u e}{- 3}\right)}^{2} - 8 \left(\textcolor{b l u e}{- 3}\right) - 6 = - 27 + 9 + 24 - 6 = 0$

So $x = - 3$ is a zero and $\left(x + 3\right)$ a factor:

${x}^{3} + {x}^{2} - 8 x - 6 = \left(x + 3\right) \left({x}^{2} - 2 x - 2\right)$

We can find the zeros of the remaining quadratic by completing the square and using the difference of squares identity:

${A}^{2} - {B}^{2} = \left(A - B\right) \left(A + B\right)$

with$A = \left(x - 1\right)$ and $B = \sqrt{3}$ as follows:

${x}^{2} - 2 x - 2 = {x}^{2} - 2 x + 1 - 3$

$\textcolor{w h i t e}{{x}^{2} - 2 x - 2} = {\left(x - 1\right)}^{2} - {\left(\sqrt{3}\right)}^{2}$

$\textcolor{w h i t e}{{x}^{2} - 2 x - 2} = \left(\left(x - 1\right) - \sqrt{3}\right) \left(\left(x - 1\right) + \sqrt{3}\right)$

$\textcolor{w h i t e}{{x}^{2} - 2 x - 2} = \left(x - 1 - \sqrt{3}\right) \left(x - 1 + \sqrt{3}\right)$

Hence the other two zeros are:

$x = 1 \pm \sqrt{3}$