# How do you find all the real and complex roots of (x^3 + 9)(x^2 - 4) = 0?

Mar 2, 2016

$x = - \sqrt[3]{9} , \pm 2 , \frac{\sqrt[3]{9}}{2} \pm \frac{3 \sqrt[6]{3}}{2} i$

#### Explanation:

We can split this up into two separate equations:

${x}^{3} + 9 = 0 \text{ "" and "" } {x}^{2} - 4 = 0$

Let's first solve the second equation (it's simpler):

${x}^{2} - 4 = 0$

${x}^{2} = 4$

$x = \pm 2$

These are the first $2$ of what should be $5$ total solutions.

Onto the next equation, we may be tempted to solve like so:

${x}^{3} + 9 = 0$

${x}^{3} = - 9$

$x = \sqrt[3]{- 9}$

However, this doesn't deal with the two remaining zeros we should have. We must try to find a way to find the two leftover complex roots in ${x}^{3} + 9 = 0$.

We can factor ${x}^{3} + 9$ as a sum of cubes, where the terms being cubed are $x$ and ${3}^{\frac{2}{3}}$.

${x}^{3} + 9 = 0$

$\left(x + {3}^{\frac{2}{3}}\right) \left({x}^{2} - {3}^{\frac{2}{3}} x + {3}^{\frac{4}{3}}\right) = 0$

Solving the linear term gives us $x = - {3}^{\frac{2}{3}} = - \sqrt[3]{9}$. To find the remaining two roots, use the quadratic formula on the quadratic factor.

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

$x = \frac{{3}^{\frac{2}{3}} \pm \sqrt{- 3 \left({3}^{\frac{4}{3}}\right)}}{2}$

$x = \frac{{3}^{\frac{2}{3}} \pm \sqrt{{3}^{\frac{7}{3}}} i}{2}$

$x = {3}^{\frac{2}{3}} / 2 \pm {3}^{\frac{7}{6}} / 2 i$

This can also be written as

$x = \frac{\sqrt[3]{9}}{2} \pm \frac{3 \sqrt[6]{3}}{2} i$

or

$x \approx 1.0400 \pm 1.8014 i$

depending on preference.