# How do you factorize -9z^2+24z-32 over C?

May 22, 2017

$- 9 {z}^{2} + 24 z - 32 = \left(x - \frac{4}{3} + \frac{\sqrt{15}}{3} i\right) \left(x - \frac{4}{3} - \frac{\sqrt{15}}{3} i\right)$

#### Explanation:

If by "factorize...over C", you mean the set of complex numbers, then here are the following steps:

Step 1. Set the equation equal to zero
$- 9 {z}^{2} + 24 z - 32 = 0$

Step 2. Plug the coefficients into the quadratic equation
$z = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$ where $a = - 9$, $b = 24$, and $c = - 32$

This gives

$z = \frac{- 24 \pm \sqrt{{24}^{2} - 4 \left(- 9\right) \left(- 31\right)}}{2 \left(- 9\right)}$

Step 3. Simplify
$z = \frac{- 24 \pm \sqrt{576 - 1116}}{- 18}$
$z = \frac{- 24 \pm \sqrt{- 540}}{-} 18$
$z = \frac{- 24 \pm \sqrt{- 36 \cdot 15}}{-} 18$
$z = \frac{- 24 \pm 6 \sqrt{- 15}}{-} 18$
$z = \frac{- 24 \pm 6 \sqrt{15} i}{-} 18$
$z = - \frac{24}{-} 18 \pm \frac{6 \sqrt{15} i}{-} 18$
$z = \frac{4}{3} \pm \frac{- \sqrt{15}}{3} i$

Step 3. Break into two parts

$z = \frac{4}{3} - \frac{\sqrt{15}}{3} i$ and $z = \frac{4}{3} + \frac{\sqrt{15}}{3} i$

$z - \frac{4}{3} + \frac{\sqrt{15}}{3} i = 0$ and $z - \frac{4}{3} - \frac{\sqrt{15}}{3} i = 0$

Step 4. Multiply these terms together for the final answer
$- 9 {z}^{2} + 24 z - 32 = \left(z - \frac{4}{3} + \frac{\sqrt{15}}{3} i\right) \left(z - \frac{4}{3} - \frac{\sqrt{15}}{3} i\right)$