# How do you factor 9z^2 + 6z - 8?

Dec 24, 2016

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

#### Explanation:

Given:

$9 {z}^{2} + 6 z - 8$

This example can be factored using an AC method:

Find a pair of factors of $A C = 9 \cdot 8 = 72$ which differ by $B = 6$

The pair $12 , 6$ works.

Use this pair to split the middle term, then factor by grouping:

$9 {z}^{2} + 6 z - 8 = 9 {z}^{2} + 12 z - 6 z - 8$

$\textcolor{w h i t e}{9 {z}^{2} + 6 z - 8} = \left(9 {z}^{2} + 12 z\right) - \left(6 z + 8\right)$

$\textcolor{w h i t e}{9 {z}^{2} + 6 z - 8} = 3 z \left(3 z + 4\right) - 2 \left(3 z + 4\right)$

$\textcolor{w h i t e}{9 {z}^{2} + 6 z - 8} = \left(3 z - 2\right) \left(3 z + 4\right)$

Dec 24, 2016

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

#### Explanation:

Given:

$9 {z}^{2} + 6 z - 8$

We can factor this quadratic by completing the square.

We will also use the difference of squares identity:

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

with $a = \left(3 z + 1\right)$ and $b = 3$

Note that:

${31}^{2} = 961$

So:

${\left(3 z + 1\right)}^{2} = 9 {z}^{2} + 6 z + 1$

[Think what happens when $z = 10$ to see why]

Hence:

$9 {z}^{2} + 6 z - 8 = 9 {z}^{2} + 6 z + 1 - 9$

$\textcolor{w h i t e}{9 {z}^{2} + 6 z - 8} = {\left(3 z + 1\right)}^{2} - {3}^{2}$

$\textcolor{w h i t e}{9 {z}^{2} + 6 z - 8} = \left(\left(3 z + 1\right) - 3\right) \left(\left(3 z + 1\right) + 3\right)$

$\textcolor{w h i t e}{9 {z}^{2} + 6 z - 8} = \left(3 z - 2\right) \left(3 z + 4\right)$