# What are all the possible rational zeros for f(x)=x^3+x^2-8x-6 and how do you find all zeros?

Nov 15, 2017

Possible rational zeros: $\pm 1 , \pm 2 , \pm 3 , \pm 6$

Actual zeros: $- 3$ and $1 \pm \sqrt{3}$

#### Explanation:

Given:

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

By the rational roots 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 $6$ 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 in turn, we find:

$f \left(- 3\right) = - 27 + 9 + 24 - 6 = 0$

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 factor the remaining quadratic by completing the square:

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

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

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

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

Hence the other two zeros of our cubic function are:

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