# How do you factor completely 8z^3 +27?

Aug 21, 2016

$8 {x}^{3} + 27 = \left(2 x + 3\right) \left(4 {x}^{2} - 6 x + 9\right)$

#### 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 use this with $a = 2 z$ and $b = 3$ to find:

$8 {x}^{3} + 27$

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

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

$= \left(2 x + 3\right) \left(4 {x}^{2} - 6 x + 9\right)$

Note that the remaining quadratic expression has no linear factors with Real coefficients. You can tell this from its discriminant:

$\Delta = {\left(- 6\right)}^{2} - 4 \left(4\right) \left(9\right) = 36 - 144 = - 108 < 0$