Step 1 :
After factoring out $ 3 $ we have:
$$ 3x^{2}+3x-36 = 3 ( x^{2}+x-12 ) $$Step 2 :
Step 2: Identify constants $ \color{blue}{ b }$ and $\color{red}{ c }$. ( $ \color{blue}{ b }$ is a number in front of the $ x $ term and $ \color{red}{ c } $ is a constant). In our case:
$$ \color{blue}{ b = 1 } ~ \text{ and } ~ \color{red}{ c = -12 }$$Now we must discover two numbers that sum up to $ \color{blue}{ 1 } $ and multiply to $ \color{red}{ -12 } $.
Step 3: Find out pairs of numbers with a product of $\color{red}{ c = -12 }$.
PRODUCT = -12 | |
-1 12 | 1 -12 |
-2 6 | 2 -6 |
-3 4 | 3 -4 |
Step 4: Find out which pair sums up to $\color{blue}{ b = 1 }$
PRODUCT = -12 and SUM = 1 | |
-1 12 | 1 -12 |
-2 6 | 2 -6 |
-3 4 | 3 -4 |
Step 5: Put -3 and 4 into placeholders to get factored form.
$$ \begin{aligned} x^{2}+x-12 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ x^{2}+x-12 & = (x -3)(x + 4) \end{aligned} $$