Factor polynomial $$ 6x^{2}+21x-90 $$

$$ 6x^{2}+21x-90 = 3( 2x-5 )( x+6 ) $$

Step 1 :

After factoring out $ 3 $ we have:

$$ 6x^{2}+21x-90 = 3 ( 2x^{2}+7x-30 ) $$

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:

$$ a = 2 , b = 7 ~ \text{ and } ~ c = -30 $$

Step 3: Multiply the leading coefficient $\color{blue}{ a = 2 }$ by the constant term $\color{blue}{c = -30} $.

$$ a \cdot c = -60 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = -60 $ and add to $ b = 7 $.

Step 5: All pairs of numbers with a product of $ -60 $ are:

PRODUCT = -60
-1     60	1     -60
-2     30	2     -30
-3     20	3     -20
-4     15	4     -15
-5     12	5     -12
-6     10	6     -10

Step 6: Find out which factor pair sums up to $\color{blue}{ b = 7 }$

PRODUCT = -60 and SUM = 7
-1     60	1     -60
-2     30	2     -30
-3     20	3     -20
-4     15	4     -15
-5     12	5     -12
-6     10	6     -10

Step 7: Replace middle term $ 7 x $ with $ 12x-5x $:

$$ 2x^{2}+7x-30 = 2x^{2}+12x-5x-30 $$

Step 8: Apply factoring by grouping. Factor $ 2x $ out of the first two terms and $ -5 $ out of the last two terms.

$$ 2x^{2}+12x-5x-30 = 2x\left(x+6\right) -5\left(x+6\right) = \left(2x-5\right) \left(x+6\right) $$

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