Step 1 :

After factoring out $ 8a^{3} $ we have:

$$ 8a^{5}-8a^{4}-48a^{3} = 8a^{3} ( a^{2}-a-6 ) $$

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:

$$ \color{blue}{ b = -1 } ~ \text{ and } ~ \color{red}{ c = -6 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ -1 } $ and multiply to $ \color{red}{ -6 } $.

Step 3: Find out pairs of numbers with a product of $\color{red}{ c = -6 }$.

PRODUCT = -6
-1     6	1     -6
-2     3	2     -3

Step 4: Find out which pair sums up to $\color{blue}{ b = -1 }$

PRODUCT = -6 and SUM = -1
-1     6	1     -6
-2     3	2     -3

Step 5: Put 2 and -3 into placeholders to get factored form.

$$ \begin{aligned} a^{2}-a-6 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ a^{2}-a-6 & = (x + 2)(x -3) \end{aligned} $$

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