Factor polynomial $$ x^{2}+4x-117 $$

$$ x^{2}+4x-117 = ( x-9 )( x+13 ) $$

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:

$$ \color{blue}{ b = 4 } ~ \text{ and } ~ \color{red}{ c = -117 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ 4 } $ and multiply to $ \color{red}{ -117 } $.

Step 2: Find out pairs of numbers with a product of $\color{red}{ c = -117 }$.

PRODUCT = -117
-1     117	1     -117
-3     39	3     -39
-9     13	9     -13

Step 3: Find out which pair sums up to $\color{blue}{ b = 4 }$

PRODUCT = -117 and SUM = 4
-1     117	1     -117
-3     39	3     -39
-9     13	9     -13

Step 4: Put -9 and 13 into placeholders to get factored form.

$$ \begin{aligned} x^{2}+4x-117 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ x^{2}+4x-117 & = (x -9)(x + 13) \end{aligned} $$

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