# How do you factor 27-8t^3?

Jan 30, 2016

$= 27 - 8 {t}^{3} = \left(- 2 t + 3\right) \left(4 {t}^{2} + 6 t + 9\right) = \left(- 2 t + 3\right) \left(2 t + \frac{3}{2} - \frac{3 \sqrt{3}}{2} i\right) \left(2 t + \frac{3}{2} + \frac{3 \sqrt{3}}{2} i\right)$

#### Explanation:

The difference of cubes formula states that
${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$
(try expanding the right hand side to verify this)

Applying the above formula, we have

$27 - 8 {t}^{3} = {3}^{3} - {\left(2 t\right)}^{3}$

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

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

If we are limiting ourselves to the reals, then we are done, as the discriminant ${6}^{2} - 4 \left(4\right) \left(9\right)$ of $4 {t}^{2} + 6 t + 9$ is less than zero. If we are willing to use complex numbers, then by the quadratic formula, $4 {t}^{2} + 6 t + 9$ has the roots

$t = \frac{- 6 \pm \sqrt{- 108}}{8}$

$= \frac{- 6 \pm 6 \sqrt{3} i}{8}$

$= - \frac{3}{4} \pm \frac{3 \sqrt{3}}{4} i$

and so, multiplying by $4$ to obtain the correct coefficient for $t$, we have the complete factorization as

$27 - 8 {t}^{3} = \left(- 2 t + 3\right) \left(2 t + \frac{3}{2} - \frac{3 \sqrt{3}}{2} i\right) \left(2 t + \frac{3}{2} + \frac{3 \sqrt{3}}{2} i\right)$