Step 1 :
After factoring out $ 6 $ we have:
$$ 12q^{2}-18q-12 = 6 ( 2q^{2}-3q-2 ) $$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:
Step 3: Multiply the leading coefficient $\color{blue}{ a = 2 }$ by the constant term $\color{blue}{c = -2} $.
$$ a \cdot c = -4 $$Step 4: Find out two numbers that multiply to $ a \cdot c = -4 $ and add to $ b = -3 $.
Step 5: All pairs of numbers with a product of $ -4 $ are:
PRODUCT = -4 | |
-1 4 | 1 -4 |
-2 2 | 2 -2 |
Step 6: Find out which factor pair sums up to $\color{blue}{ b = -3 }$
PRODUCT = -4 and SUM = -3 | |
-1 4 | 1 -4 |
-2 2 | 2 -2 |
Step 7: Replace middle term $ -3 x $ with $ x-4x $:
$$ 2x^{2}-3x-2 = 2x^{2}+x-4x-2 $$Step 8: Apply factoring by grouping. Factor $ x $ out of the first two terms and $ -2 $ out of the last two terms.
$$ 2x^{2}+x-4x-2 = x\left(2x+1\right) -2\left(2x+1\right) = \left(x-2\right) \left(2x+1\right) $$