Factor polynomial $$ 4x^{2}-4x-48 $$

$$ 4x^{2}-4x-48 = 4( x+3 )( x-4 ) $$

Step 1 :

After factoring out $ 4 $ we have:

$$ 4x^{2}-4x-48 = 4 ( x^{2}-x-12 ) $$

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:

$$ \color{blue}{ b = -1 } ~ \text{ and } ~ \color{red}{ c = -12 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ -1 } $ and multiply to $ \color{red}{ -12 } $.

Step 3: Find out pairs of numbers with a product of $\color{red}{ c = -12 }$.

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

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

PRODUCT = -12 and SUM = -1
-1     12	1     -12
-2     6	2     -6
-3     4	3     -4

Step 5: Put 3 and -4 into placeholders to get factored form.

$$ \begin{aligned} x^{2}-x-12 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ x^{2}-x-12 & = (x + 3)(x -4) \end{aligned} $$

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