Factor polynomial $$ 12x^{2}-4x-16 $$

$$ 12x^{2}-4x-16 = 4( 3x-4 )( x+1 ) $$

Step 1 :

After factoring out $ 4 $ we have:

$$ 12x^{2}-4x-16 = 4 ( 3x^{2}-x-4 ) $$

Step 2 :

Step 2: Identify constants $ a $ , $ b $ and $ c $.
$ a $ is a number in front of the $ x^2 $ term $ b $ is a number in front of the $ x $ term and $ c $ is a constant. In this case:

$$ a = 3 , b = -1 ~ \text{ and } ~ c = -4 $$

Step 3: Multiply the leading coefficient $\color{blue}{ a = 3 }$ by the constant term $\color{blue}{c = -4} $.

$$ a \cdot c = -12 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = -12 $ and add to $ b = -1 $.

Step 5: All pairs of numbers with a product of $ -12 $ are:

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

Step 6: Find out which factor 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 7: Replace middle term $ -1 x $ with $ 3x-4x $:

$$ 3x^{2}-x-4 = 3x^{2}+3x-4x-4 $$

Step 8: Apply factoring by grouping. Factor $ 3x $ out of the first two terms and $ -4 $ out of the last two terms.

$$ 3x^{2}+3x-4x-4 = 3x\left(x+1\right) -4\left(x+1\right) = \left(3x-4\right) \left(x+1\right) $$

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