# How do you find all the zeros of f(x) = 9x^2 + 6x - 8?

Aug 5, 2016

The zeros of $f \left(x\right)$ are $x = \frac{2}{3}$ and $x = - \frac{4}{3}$

#### Explanation:

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

Use an AC method to factor $f \left(x\right)$:

Find a pair of factors of $A C = 9 \cdot 8 = 72$ with difference $B = 6$.

The pair $12 , 6$ works.

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

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

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

$= \left(9 {x}^{2} + 12 x\right) - \left(6 x + 8\right)$

$= 3 x \left(3 x + 4\right) - 2 \left(3 x + 4\right)$

$= \left(3 x - 2\right) \left(3 x + 4\right)$

Hence the zeros of $f \left(x\right)$ are $x = \frac{2}{3}$ and $x = - \frac{4}{3}$