# How do you use the rational roots theorem to find all possible zeros of f(x) = x^3 - 12x - 16?

Jun 18, 2016

The zeros of this $f \left(x\right)$ are all rational:

$x = - 2$ with multiplicity $2$

$x = 4$

#### Explanation:

$f \left(x\right) = {x}^{3} - 12 x - 16$

By the rational roots 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 $- 16$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1$, $\pm 2$, $\pm 4$, $\pm 8$, $\pm 16$

Also note that the signs of the coefficients of $f \left(x\right)$ follow the pattern $+ - -$, with one change of sign. Hence $f \left(x\right)$ has exactly one positive zero.

So start trying the positive possibilities first:

$f \left(1\right) = 1 - 12 - 16 = - 27$

$f \left(2\right) = 8 - 24 - 16 = - 32$

$f \left(4\right) = 64 - 48 - 16 = 0$

So $x = 4$ is a zero and $\left(x - 4\right)$ a factor:

${x}^{3} - 12 x - 16 = \left(x - 4\right) \left({x}^{2} + 4 x + 4\right) = \left(x - 4\right) \left(x + 2\right) \left(x + 2\right)$

So $x = - 2$ is the other zero of $f \left(x\right)$, with multiplicity $2$