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