How do you find all the zeros of P(x) = 3x^3 – 6x^2 – 3x – 18?

Aug 7, 2016

$P \left(x\right)$ has zeros: $3$ and $- \frac{1}{2} \pm \frac{\sqrt{7}}{2} i$

Explanation:

$P \left(x\right) = 3 {x}^{3} - 6 {x}^{2} - 3 x - 18 = 3 \left({x}^{3} - 2 {x}^{2} - x - 6\right)$

By the rational root theorem, the only possible rational zeros of ${x}^{3} - 2 {x}^{2} - x - 6$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constnat term $- 6$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

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

Substituting $x = 3$ we find:

${x}^{3} - 2 {x}^{2} - x - 6 = {\left(3\right)}^{3} - 2 {\left(3\right)}^{2} - \left(3\right) - 6 = 27 - 18 - 3 - 6 = 0$

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

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

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

$= \left(x - 3\right) \left({\left(x + \frac{1}{2}\right)}^{2} + \frac{7}{4}\right)$

$= \left(x - 3\right) \left({\left(x + \frac{1}{2}\right)}^{2} - {\left(\frac{\sqrt{7}}{2} i\right)}^{2}\right)$

$= \left(x - 3\right) \left(x + \frac{1}{2} - \frac{\sqrt{7}}{2} i\right) \left(x + \frac{1}{2} + \frac{\sqrt{7}}{2} i\right)$

So the other two zeros are:

$x = - \frac{1}{2} \pm \frac{\sqrt{7}}{2} i$