Factor polynomial $$ -12n^{2}-8n+64 $$

$$ -12n^{2}-8n+64 = ( -4 )( n-2 )( 3n+8 ) $$

Step 1 :

After factoring out $ -4 $ we have:

$$ -12n^{2}-8n+64 = -4 ( 3n^{2}+2n-16 ) $$

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:

$$ a = 3 , b = 2 ~ \text{ and } ~ c = -16 $$

Step 3: Multiply the leading coefficient $\color{blue}{ a = 3 }$ by the constant term $\color{blue}{c = -16} $.

$$ a \cdot c = -48 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = -48 $ and add to $ b = 2 $.

Step 5: All pairs of numbers with a product of $ -48 $ are:

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

Step 6: Find out which factor 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 7: Replace middle term $ 2 x $ with $ 8x-6x $:

$$ 3x^{2}+2x-16 = 3x^{2}+8x-6x-16 $$

Step 8: Apply factoring by grouping. Factor $ x $ out of the first two terms and $ -2 $ out of the last two terms.

$$ 3x^{2}+8x-6x-16 = x\left(3x+8\right) -2\left(3x+8\right) = \left(x-2\right) \left(3x+8\right) $$

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