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## Factoring Quadratics 1.10 - Solution

We want to solve the given equation by factoring. To do so, we need to rewrite the linear coefficient $\text{-} 21$ as the sum of two numbers whose product is ${\color{#0000FF}{a}}{\color{#009600}{c}}.$ $\begin{gathered} {\color{#0000FF}{2}}x^2-21x{\color{#009600}{\,-\,36}}=0 \\ \Downarrow \\ {\color{#0000FF}{a}}{\color{#009600}{c}}={\color{#0000FF}{2}}({\color{#009600}{\text{-} 36}})=\text{-} 72 \end{gathered}$ Therefore, we need to find two numbers whose sum is $\text{-} 21$ and whose product is $\text{-} 72.$ In this case these numbers are $\text{-} 24$ and $3.$ With this, we can factor the left-hand side and solve the equation by using the Zero Product Property.

### Factoring

Let's start by rewriting $\text{-} 21x$ as $\text{-} 24x+3x.$
$2x^2-21x-36=0$
$2x^2-24x+3x-36=0$
Factor out $2x\ \&\ 3$
$2x(x-12)+3x-36=0$
$2x(x-12)+3(x-12)=0$
$(x-12)(2x+3)=0$

### Solving

To solve this equation, we will apply the Zero Product Property.
$(x-12)(2x+3)=0$
$\begin{array}{lc}x-12=0 & \text{(I)}\\ 2x+3=0 & \text{(II)}\end{array}$
$\begin{array}{l}x=12 \\ 2x+3=0 \end{array}$
$\begin{array}{l}x=12 \\ 2x=\text{-}3 \end{array}$
$\begin{array}{l}x_1=12 \\ x_2=\frac{\text{-} 3}{2} \end{array}$
$\begin{array}{l}x_1=12 \\ x_2=\text{-} \frac{3}{2} \end{array}$