# How do you find all the zeros of f(x)=x^3 -x^2 -4x -6?

Apr 1, 2016

Use the rational root theorem and completing the square to find:

$x = 3$

$x = - 1 \pm i$

#### Explanation:

By the rational root theorem, any rational zeros of this $f \left(x\right)$ must be expressible in the form $\frac{p}{q}$ with integers $p , q$ with $p$ a divisor of the constant term $- 6$ and $q$ a divisor of the coefficient $1$ of the leading term.

So the only possible rational roots are:

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

Trying each in turn, we find:

$f \left(3\right) = 27 - 9 - 12 - 6 = 0$

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

${x}^{3} - {x}^{2} - 4 x - 6 = \left(x - 3\right) \left({x}^{2} + 2 x + 2\right)$

The remaining quadratic factor is of the form $a {x}^{2} + b x + c$ with $a = 1$, $b = 2$ and $c = 2$. This has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {2}^{2} - \left(4 \cdot 1 \cdot 2\right) = 4 - 8 = - 4$

Since this is negative, the quadratic has no Real zeros and no linear factors with Real coefficients.

We can factor it with Complex coefficients by completing the square:

${x}^{2} + 2 x + 2$

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

$= {\left(x + 1\right)}^{2} - {i}^{2}$

$= \left(\left(x + 1\right) - i\right) \left(\left(x + 1\right) + i\right)$

$= \left(x + 1 - i\right) \left(x + 1 + i\right)$

Putting it all together:

${x}^{3} - {x}^{2} - 4 x - 6$

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

$= \left(x - 3\right) \left(x + 1 - i\right) \left(x + 1 + i\right)$

Hence zeros:

$x = 3$

$x = - 1 \pm i$