Factor polynomial $$ x^{2}+3x-108 $$

$$ x^{2}+3x-108 = ( x-9 )( x+12 ) $$

Step 1: 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:

$$ \color{blue}{ b = 3 } ~ \text{ and } ~ \color{red}{ c = -108 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ 3 } $ and multiply to $ \color{red}{ -108 } $.

Step 2: Find out pairs of numbers with a product of $\color{red}{ c = -108 }$.

PRODUCT = -108
-1     108	1     -108
-2     54	2     -54
-3     36	3     -36
-4     27	4     -27
-6     18	6     -18
-9     12	9     -12

Step 3: Find out which pair sums up to $\color{blue}{ b = 3 }$

PRODUCT = -108 and SUM = 3
-1     108	1     -108
-2     54	2     -54
-3     36	3     -36
-4     27	4     -27
-6     18	6     -18
-9     12	9     -12

Step 4: Put -9 and 12 into placeholders to get factored form.

$$ \begin{aligned} x^{2}+3x-108 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ x^{2}+3x-108 & = (x -9)(x + 12) \end{aligned} $$

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