Step 1 :
After factoring out $ x $ we have:
$$ 2x^{3}+9x^{2}-5x = x ( 2x^{2}+9x-5 ) $$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 = -5} $.
$$ a \cdot c = -10 $$Step 4: Find out two numbers that multiply to $ a \cdot c = -10 $ and add to $ b = 9 $.
Step 5: All pairs of numbers with a product of $ -10 $ are:
PRODUCT = -10 | |
-1 10 | 1 -10 |
-2 5 | 2 -5 |
Step 6: Find out which factor pair sums up to $\color{blue}{ b = 9 }$
PRODUCT = -10 and SUM = 9 | |
-1 10 | 1 -10 |
-2 5 | 2 -5 |
Step 7: Replace middle term $ 9 x $ with $ 10x-x $:
$$ 2x^{2}+9x-5 = 2x^{2}+10x-x-5 $$Step 8: Apply factoring by grouping. Factor $ 2x $ out of the first two terms and $ -1 $ out of the last two terms.
$$ 2x^{2}+10x-x-5 = 2x\left(x+5\right) -1\left(x+5\right) = \left(2x-1\right) \left(x+5\right) $$