Factor polynomial $$ 12m^{2}+20m-8 $$

$$ 12m^{2}+20m-8 = 4( 3m-1 )( m+2 ) $$

Step 1 :

After factoring out $ 4 $ we have:

$$ 12m^{2}+20m-8 = 4 ( 3m^{2}+5m-2 ) $$

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:

$$ a = 3 , b = 5 ~ \text{ and } ~ c = -2 $$

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

$$ a \cdot c = -6 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = -6 $ and add to $ b = 5 $.

Step 5: All pairs of numbers with a product of $ -6 $ are:

PRODUCT = -6
-1     6	1     -6
-2     3	2     -3

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

PRODUCT = -6 and SUM = 5
-1     6	1     -6
-2     3	2     -3

Step 7: Replace middle term $ 5 x $ with $ 6x-x $:

$$ 3x^{2}+5x-2 = 3x^{2}+6x-x-2 $$

Step 8: Apply factoring by grouping. Factor $ 3x $ out of the first two terms and $ -1 $ out of the last two terms.

$$ 3x^{2}+6x-x-2 = 3x\left(x+2\right) -1\left(x+2\right) = \left(3x-1\right) \left(x+2\right) $$

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