Factor polynomial $$ 9p^{2}+6p-8 $$

$$ 9p^{2}+6p-8 = ( 3p-2 )( 3p+4 ) $$

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:

$$ a = 9 , b = 6 ~ \text{ and } ~ c = -8 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 9 }$ by the constant term $\color{blue}{c = -8} $.

$$ a \cdot c = -72 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = -72 $ and add to $ b = 6 $.

Step 4: All pairs of numbers with a product of $ -72 $ are:

PRODUCT = -72
-1     72	1     -72
-2     36	2     -36
-3     24	3     -24
-4     18	4     -18
-6     12	6     -12
-8     9	8     -9

Step 5: Find out which factor pair sums up to $\color{blue}{ b = 6 }$

PRODUCT = -72 and SUM = 6
-1     72	1     -72
-2     36	2     -36
-3     24	3     -24
-4     18	4     -18
-6     12	6     -12
-8     9	8     -9

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

$$ 9x^{2}+6x-8 = 9x^{2}+12x-6x-8 $$

Step 7: Apply factoring by grouping. Factor $ 3x $ out of the first two terms and $ -2 $ out of the last two terms.

$$ 9x^{2}+12x-6x-8 = 3x\left(3x+4\right) -2\left(3x+4\right) = \left(3x-2\right) \left(3x+4\right) $$

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