Factor polynomial $$ 2x^{2}-6x-80 $$

$$ 2x^{2}-6x-80 = 2( x+5 )( x-8 ) $$

Step 1 :

After factoring out $ 2 $ we have:

$$ 2x^{2}-6x-80 = 2 ( x^{2}-3x-40 ) $$

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:

$$ \color{blue}{ b = -3 } ~ \text{ and } ~ \color{red}{ c = -40 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ -3 } $ and multiply to $ \color{red}{ -40 } $.

Step 3: Find out pairs of numbers with a product of $\color{red}{ c = -40 }$.

PRODUCT = -40
-1     40	1     -40
-2     20	2     -20
-4     10	4     -10
-5     8	5     -8

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

PRODUCT = -40 and SUM = -3
-1     40	1     -40
-2     20	2     -20
-4     10	4     -10
-5     8	5     -8

Step 5: Put 5 and -8 into placeholders to get factored form.

$$ \begin{aligned} x^{2}-3x-40 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ x^{2}-3x-40 & = (x + 5)(x -8) \end{aligned} $$

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