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