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