Factor polynomial $$ 12t^{2}-18t-84 $$

$$ 12t^{2}-18t-84 = 6( 2t-7 )( t+2 ) $$

Step 1 :

After factoring out $ 6 $ we have:

$$ 12t^{2}-18t-84 = 6 ( 2t^{2}-3t-14 ) $$

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:

$$ a = 2 , b = -3 ~ \text{ and } ~ c = -14 $$

Step 3: Multiply the leading coefficient $\color{blue}{ a = 2 }$ by the constant term $\color{blue}{c = -14} $.

$$ a \cdot c = -28 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = -28 $ and add to $ b = -3 $.

Step 5: All pairs of numbers with a product of $ -28 $ are:

PRODUCT = -28
-1     28	1     -28
-2     14	2     -14
-4     7	4     -7

Step 6: Find out which factor pair sums up to $\color{blue}{ b = -3 }$

PRODUCT = -28 and SUM = -3
-1     28	1     -28
-2     14	2     -14
-4     7	4     -7

Step 7: Replace middle term $ -3 x $ with $ 4x-7x $:

$$ 2x^{2}-3x-14 = 2x^{2}+4x-7x-14 $$

Step 8: Apply factoring by grouping. Factor $ 2x $ out of the first two terms and $ -7 $ out of the last two terms.

$$ 2x^{2}+4x-7x-14 = 2x\left(x+2\right) -7\left(x+2\right) = \left(2x-7\right) \left(x+2\right) $$

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