Step 1 :
After factoring out $ 4 $ we have:
$$ 12p^{2}-40p-32 = 4 ( 3p^{2}-10p-8 ) $$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 $\color{blue}{ a = 3 }$ by the constant term $\color{blue}{c = -8} $.
$$ a \cdot c = -24 $$Step 4: Find out two numbers that multiply to $ a \cdot c = -24 $ and add to $ b = -10 $.
Step 5: All pairs of numbers with a product of $ -24 $ are:
PRODUCT = -24 | |
-1 24 | 1 -24 |
-2 12 | 2 -12 |
-3 8 | 3 -8 |
-4 6 | 4 -6 |
Step 6: Find out which factor pair sums up to $\color{blue}{ b = -10 }$
PRODUCT = -24 and SUM = -10 | |
-1 24 | 1 -24 |
-2 12 | 2 -12 |
-3 8 | 3 -8 |
-4 6 | 4 -6 |
Step 7: Replace middle term $ -10 x $ with $ 2x-12x $:
$$ 3x^{2}-10x-8 = 3x^{2}+2x-12x-8 $$Step 8: Apply factoring by grouping. Factor $ x $ out of the first two terms and $ -4 $ out of the last two terms.
$$ 3x^{2}+2x-12x-8 = x\left(3x+2\right) -4\left(3x+2\right) = \left(x-4\right) \left(3x+2\right) $$