Factor polynomial $$ 8n^{2}-6n-9 $$

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

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:

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

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

$$ a \cdot c = -72 $$

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 $ 6x-12x $:

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

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

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

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