Factor polynomial $$ 2x^{2}+10x-48 $$

$$ 2x^{2}+10x-48 = 2( x-3 )( x+8 ) $$

Step 1 :

After factoring out $ 2 $ we have:

$$ 2x^{2}+10x-48 = 2 ( x^{2}+5x-24 ) $$

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:

$$ \color{blue}{ b = 5 } ~ \text{ and } ~ \color{red}{ c = -24 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ 5 } $ and multiply to $ \color{red}{ -24 } $.

Step 3: Find out pairs of numbers with a product of $\color{red}{ c = -24 }$.

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

Step 4: Find out which pair sums up to $\color{blue}{ b = 5 }$

PRODUCT = -24 and SUM = 5
-1     24	1     -24
-2     12	2     -12
-3     8	3     -8
-4     6	4     -6

Step 5: Put -3 and 8 into placeholders to get factored form.

$$ \begin{aligned} x^{2}+5x-24 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ x^{2}+5x-24 & = (x -3)(x + 8) \end{aligned} $$

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