Factor polynomial $$ 18x^{2}+96x-72 $$

$$ 18x^{2}+96x-72 = 6( 3x-2 )( x+6 ) $$

Step 1 :

After factoring out $ 6 $ we have:

$$ 18x^{2}+96x-72 = 6 ( 3x^{2}+16x-12 ) $$

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:

$$ a = 3 , b = 16 ~ \text{ and } ~ c = -12 $$

Step 3: Multiply the leading coefficient $\color{blue}{ a = 3 }$ by the constant term $\color{blue}{c = -12} $.

$$ a \cdot c = -36 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = -36 $ and add to $ b = 16 $.

Step 5: All pairs of numbers with a product of $ -36 $ are:

PRODUCT = -36
-1     36	1     -36
-2     18	2     -18
-3     12	3     -12
-4     9	4     -9
-6     6	6     -6

Step 6: Find out which factor pair sums up to $\color{blue}{ b = 16 }$

PRODUCT = -36 and SUM = 16
-1     36	1     -36
-2     18	2     -18
-3     12	3     -12
-4     9	4     -9
-6     6	6     -6

Step 7: Replace middle term $ 16 x $ with $ 18x-2x $:

$$ 3x^{2}+16x-12 = 3x^{2}+18x-2x-12 $$

Step 8: Apply factoring by grouping. Factor $ 3x $ out of the first two terms and $ -2 $ out of the last two terms.

$$ 3x^{2}+18x-2x-12 = 3x\left(x+6\right) -2\left(x+6\right) = \left(3x-2\right) \left(x+6\right) $$

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