Factor polynomial $$ x^{3}+2x^{2}-48x $$

$$ x^{3}+2x^{2}-48x = x( x-6 )( x+8 ) $$

Step 1 :

After factoring out $ x $ we have:

$$ x^{3}+2x^{2}-48x = x ( x^{2}+2x-48 ) $$

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:

$$ \color{blue}{ b = 2 } ~ \text{ and } ~ \color{red}{ c = -48 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ 2 } $ and multiply to $ \color{red}{ -48 } $.

Step 3: Find out pairs of numbers with a product of $\color{red}{ c = -48 }$.

PRODUCT = -48
-1     48	1     -48
-2     24	2     -24
-3     16	3     -16
-4     12	4     -12
-6     8	6     -8

Step 4: Find out which pair sums up to $\color{blue}{ b = 2 }$

PRODUCT = -48 and SUM = 2
-1     48	1     -48
-2     24	2     -24
-3     16	3     -16
-4     12	4     -12
-6     8	6     -8

Step 5: Put -6 and 8 into placeholders to get factored form.

$$ \begin{aligned} x^{2}+2x-48 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ x^{2}+2x-48 & = (x -6)(x + 8) \end{aligned} $$

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