Factor polynomial $$ 3x^{2}+8x-16 $$

$$ 3x^{2}+8x-16 = ( 3x-4 )( x+4 ) $$

Step 1: 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:

$$ a = 3 , b = 8 ~ \text{ and } ~ c = -16 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 3 }$ by the constant term $\color{blue}{c = -16} $.

$$ a \cdot c = -48 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = -48 $ and add to $ b = 8 $.

Step 4: 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 5: Find out which factor pair sums up to $\color{blue}{ b = 8 }$

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

Step 6: Replace middle term $ 8 x $ with $ 12x-4x $:

$$ 3x^{2}+8x-16 = 3x^{2}+12x-4x-16 $$

Step 7: Apply factoring by grouping. Factor $ 3x $ out of the first two terms and $ -4 $ out of the last two terms.

$$ 3x^{2}+12x-4x-16 = 3x\left(x+4\right) -4\left(x+4\right) = \left(3x-4\right) \left(x+4\right) $$

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