Factor polynomial $$ x^{2}+2x-8 $$

$$ x^{2}+2x-8 = ( x-2 )( x+4 ) $$

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:

$$ \color{blue}{ b = 2 } ~ \text{ and } ~ \color{red}{ c = -8 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ 2 } $ and multiply to $ \color{red}{ -8 } $.

Step 2: Find out pairs of numbers with a product of $\color{red}{ c = -8 }$.

PRODUCT = -8
-1     8	1     -8
-2     4	2     -4

Step 3: Find out which pair sums up to $\color{blue}{ b = 2 }$

PRODUCT = -8 and SUM = 2
-1     8	1     -8
-2     4	2     -4

Step 4: Put -2 and 4 into placeholders to get factored form.

$$ \begin{aligned} x^{2}+2x-8 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ x^{2}+2x-8 & = (x -2)(x + 4) \end{aligned} $$

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