Factor polynomial $$ 3b^{2}+18b-48 $$

$$ 3b^{2}+18b-48 = 3( b-2 )( b+8 ) $$

Step 1 :

After factoring out $ 3 $ we have:

$$ 3b^{2}+18b-48 = 3 ( b^{2}+6b-16 ) $$

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:

$$ \color{blue}{ b = 6 } ~ \text{ and } ~ \color{red}{ c = -16 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ 6 } $ and multiply to $ \color{red}{ -16 } $.

Step 3: Find out pairs of numbers with a product of $\color{red}{ c = -16 }$.

PRODUCT = -16
-1     16	1     -16
-2     8	2     -8
-4     4	4     -4

Step 4: Find out which pair sums up to $\color{blue}{ b = 6 }$

PRODUCT = -16 and SUM = 6
-1     16	1     -16
-2     8	2     -8
-4     4	4     -4

Step 5: Put -2 and 8 into placeholders to get factored form.

$$ \begin{aligned} b^{2}+6b-16 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ b^{2}+6b-16 & = (x -2)(x + 8) \end{aligned} $$

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