Factor polynomial $$ 9x^{2}-39x-30 $$

$$ 9x^{2}-39x-30 = 3( x-5 )( 3x+2 ) $$

Step 1 :

After factoring out $ 3 $ we have:

$$ 9x^{2}-39x-30 = 3 ( 3x^{2}-13x-10 ) $$

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:

$$ a = 3 , b = -13 ~ \text{ and } ~ c = -10 $$

Step 3: Multiply the leading coefficient $\color{blue}{ a = 3 }$ by the constant term $\color{blue}{c = -10} $.

$$ a \cdot c = -30 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = -30 $ and add to $ b = -13 $.

Step 5: All pairs of numbers with a product of $ -30 $ are:

PRODUCT = -30
-1     30	1     -30
-2     15	2     -15
-3     10	3     -10
-5     6	5     -6

Step 6: Find out which factor pair sums up to $\color{blue}{ b = -13 }$

PRODUCT = -30 and SUM = -13
-1     30	1     -30
-2     15	2     -15
-3     10	3     -10
-5     6	5     -6

Step 7: Replace middle term $ -13 x $ with $ 2x-15x $:

$$ 3x^{2}-13x-10 = 3x^{2}+2x-15x-10 $$

Step 8: Apply factoring by grouping. Factor $ x $ out of the first two terms and $ -5 $ out of the last two terms.

$$ 3x^{2}+2x-15x-10 = x\left(3x+2\right) -5\left(3x+2\right) = \left(x-5\right) \left(3x+2\right) $$

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