# How do you find all the zeros of  f(x)=x^3+7x^2-6x-72 ?

May 21, 2016

Use the rational root theorem to help find the first zero, then complete the square to find the other two, giving:

$3$, $- 4$, $- 6$

#### Explanation:

$f \left(x\right) = {x}^{3} + 7 {x}^{2} - 6 x - 72$

By the rational root theorem, any rational zeros of $f \left(x\right)$ must be expressible in the form $\frac{p}{q}$ where $p$ is a divisor of the constant term $- 72$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1$, $\pm 2$, $\pm 3$, $\pm 4$, $\pm 6$, $\pm 12$, $\pm 18$, $\pm 24$, $\pm 36$, $\pm 72$

Trying each in turn, the first one that works is:

$f \left(3\right) = {3}^{3} + \left(7 \cdot {3}^{2}\right) - \left(6 \cdot 3\right) - 72 = 27 + 63 - 18 - 72 = 0$

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

${x}^{3} + 7 {x}^{2} - 6 x - 72 = \left(x - 3\right) \left({x}^{2} + 10 x + 24\right)$

One way of factoring the remaining quadratic factor ${x}^{2} + 10 x + 24$ is by completing the square:

${x}^{2} + 10 x + 24$

$= {\left(x + 5\right)}^{2} - 25 + 24$

$= {\left(x + 5\right)}^{2} - {1}^{2}$

$= \left(\left(x + 5\right) - 1\right) \left(\left(x + 5\right) + 1\right)$

$= \left(x + 4\right) \left(x + 6\right)$

Hence zeros:

$x = - 4$ and $x = - 6$