Step 1 :
After factoring out $ 3 $ we have:
$$ 21x^{2}-24x-36 = 3 ( 7x^{2}-8x-12 ) $$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:
Step 3: Multiply the leading coefficient $\color{blue}{ a = 7 }$ by the constant term $\color{blue}{c = -12} $.
$$ a \cdot c = -84 $$Step 4: Find out two numbers that multiply to $ a \cdot c = -84 $ and add to $ b = -8 $.
Step 5: 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 6: 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 7: Replace middle term $ -8 x $ with $ 6x-14x $:
$$ 7x^{2}-8x-12 = 7x^{2}+6x-14x-12 $$Step 8: Apply factoring by grouping. Factor $ x $ out of the first two terms and $ -2 $ out of the last two terms.
$$ 7x^{2}+6x-14x-12 = x\left(7x+6\right) -2\left(7x+6\right) = \left(x-2\right) \left(7x+6\right) $$