Step 1 :

After factoring out $ 2 $ we have:

$$ 2y^{2}-16y-40 = 2 ( y^{2}-8y-20 ) $$

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:

$$ \color{blue}{ b = -8 } ~ \text{ and } ~ \color{red}{ c = -20 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ -8 } $ and multiply to $ \color{red}{ -20 } $.

Step 3: Find out pairs of numbers with a product of $\color{red}{ c = -20 }$.

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

Step 4: Find out which pair sums up to $\color{blue}{ b = -8 }$

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

Step 5: Put 2 and -10 into placeholders to get factored form.

$$ \begin{aligned} y^{2}-8y-20 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ y^{2}-8y-20 & = (x + 2)(x -10) \end{aligned} $$

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