Factor polynomial $$ 7m^{2}+8m-12 $$

$$ 7m^{2}+8m-12 = ( 7m-6 )( m+2 ) $$

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:

$$ a = 7 , b = 8 ~ \text{ and } ~ c = -12 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 7 }$ by the constant term $\color{blue}{c = -12} $.

$$ a \cdot c = -84 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = -84 $ and add to $ b = 8 $.

Step 4: All pairs of numbers with a product of $ -84 $ are:

PRODUCT = -84
-1     84	1     -84
-2     42	2     -42
-3     28	3     -28
-4     21	4     -21
-6     14	6     -14
-7     12	7     -12

Step 5: Find out which factor pair sums up to $\color{blue}{ b = 8 }$

PRODUCT = -84 and SUM = 8
-1     84	1     -84
-2     42	2     -42
-3     28	3     -28
-4     21	4     -21
-6     14	6     -14
-7     12	7     -12

Step 6: Replace middle term $ 8 x $ with $ 14x-6x $:

$$ 7x^{2}+8x-12 = 7x^{2}+14x-6x-12 $$

Step 7: Apply factoring by grouping. Factor $ 7x $ out of the first two terms and $ -6 $ out of the last two terms.

$$ 7x^{2}+14x-6x-12 = 7x\left(x+2\right) -6\left(x+2\right) = \left(7x-6\right) \left(x+2\right) $$

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