# How do you list all possible roots and find all factors of x^3+27?

Dec 10, 2016

${x}^{3} + 27 = \left(x + 3\right) \left(x - \frac{3}{2} - \frac{3 \sqrt{3}}{2} i\right) \left(x - \frac{3}{2} + \frac{3 \sqrt{3}}{2} i\right)$

with corresponding zeros:

$x = - 3$ and $x = \frac{3}{2} \pm \frac{3 \sqrt{3}}{2} i$

#### Explanation:

The sum of cubes identity can be written:

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

We can use this with $a = x$ and $b = 3$ as follows:

${x}^{3} + 27 = {x}^{3} + {3}^{3}$

$\textcolor{w h i t e}{{x}^{3} + 27} = \left(x + 3\right) \left({x}^{2} - 3 x + 9\right)$

So one linear factor is $\left(x + 3\right)$ with corresponding zero $x = - 3$.

We can factor the remaining quadratic using Complex coeffients by completing the square and using the difference of squares identity, which can be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

with $a = \left(x - \frac{3}{2}\right)$ and $b = \frac{3 \sqrt{3}}{2} i$ as follows:

${x}^{2} - 3 x + 9 = \left({x}^{2} - 3 x + \frac{9}{4}\right) + \frac{27}{4}$

$\textcolor{w h i t e}{{x}^{2} - 3 x + 9} = {\left(x - \frac{3}{2}\right)}^{2} - {\left(\frac{3 \sqrt{3}}{2} i\right)}^{2}$

$\textcolor{w h i t e}{{x}^{2} - 3 x + 9} = \left(\left(x - \frac{3}{2}\right) - \frac{3 \sqrt{3}}{2} i\right) \left(\left(x - \frac{3}{2}\right) + \frac{3 \sqrt{3}}{2} i\right)$

$\textcolor{w h i t e}{{x}^{2} - 3 x + 9} = \left(x - \frac{3}{2} - \frac{3 \sqrt{3}}{2} i\right) \left(x - \frac{3}{2} + \frac{3 \sqrt{3}}{2} i\right)$

Hence the remaining zeros are:

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

and the full factorisation can be written:

${x}^{3} + 27 = \left(x + 3\right) \left(x - \frac{3}{2} - \frac{3 \sqrt{3}}{2} i\right) \left(x - \frac{3}{2} + \frac{3 \sqrt{3}}{2} i\right)$

Structly speaking, we have not listed all of the factors of ${x}^{3} + 27$.

The ones we have identified so far are:

${x}^{3} + 27$

$x + 3$

${x}^{2} - 3 x + 9$

$x - \frac{3}{2} - \frac{3 \sqrt{3}}{2} i$

$x - \frac{3}{2} + \frac{3 \sqrt{3}}{2} i$

$1$

In addition the results of multiplying these linear factors give us the other factors:

$\left(x + 3\right) \left(x - \frac{3}{2} - \frac{3 \sqrt{3}}{2} i\right) = {x}^{2} + \left(\frac{3}{2} - \frac{3 \sqrt{3}}{2} i\right) x - \left(\frac{9}{2} + \frac{9 \sqrt{3}}{2} i\right)$

$\left(x + 3\right) \left(x - \frac{3}{2} + \frac{3 \sqrt{3}}{2} i\right) = {x}^{2} + \left(\frac{3}{2} + \frac{3 \sqrt{3}}{2} i\right) x - \left(\frac{9}{2} - \frac{9 \sqrt{3}}{2} i\right)$

I don't think it was intended that you should list all of these factors, but there they are in case.