Factor polynomial $$ 2x^{2}+4x-8 $$

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

Step 1 :

After factoring out $ 2 $ we have:

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

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:

$$ \color{blue}{ b = 2 } ~ \text{ and } ~ \color{red}{ c = -4 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ 2 } $ and multiply to $ \color{red}{ -4 } $.

Step 3: Find out pairs of numbers with a product of $\color{red}{ c = -4 }$.

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

Step 4: Because none of these pairs will give us a sum of $ \color{blue}{ 2 }$, we conclude the polynomial cannot be factored.

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