# How do you find the zeros of g(x)=33x^2-9x-24?

Jun 24, 2017

$x = 1 , x = - \frac{24}{33}$

#### Explanation:

$\text{note that " g(1)=0rArr(x-1)" is a factor}$

$g \left(x\right) = \textcolor{red}{33 x} \left(x - 1\right) \textcolor{m a \ge n t a}{+ 33 x} - 9 x - 24$

$\textcolor{w h i t e}{g \left(x\right)} = \textcolor{red}{33 x} \left(x - 1\right) \textcolor{red}{+ 24} \left(x - 1\right) \textcolor{m a \ge n t a}{+ 24} - 24$

$\textcolor{w h i t e}{g \left(x\right)} = \textcolor{red}{33 x} \left(x - 1\right) \textcolor{red}{+ 24} \left(x - 1\right) + 0$

$\Rightarrow g \left(x\right) = \left(x - 1\right) \left(\textcolor{red}{33 x + 24}\right) = 0$

$\text{equate each factor to zero}$

$x - 1 = 0 \Rightarrow x = 1 \leftarrow \text{ is a zero}$

$33 x + 24 = 0 \Rightarrow x = - \frac{24}{33} \leftarrow \text{ is a zero}$