Step 1 :
After factoring out $ 2 $ we have:
$$ 20x^{2}+22x-12 = 2 ( 10x^{2}+11x-6 ) $$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 $\color{blue}{ a = 10 }$ by the constant term $\color{blue}{c = -6} $.
$$ a \cdot c = -60 $$Step 4: Find out two numbers that multiply to $ a \cdot c = -60 $ and add to $ b = 11 $.
Step 5: All pairs of numbers with a product of $ -60 $ are:
| PRODUCT = -60 | |
| -1 60 | 1 -60 |
| -2 30 | 2 -30 |
| -3 20 | 3 -20 |
| -4 15 | 4 -15 |
| -5 12 | 5 -12 |
| -6 10 | 6 -10 |
Step 6: Find out which factor pair sums up to $\color{blue}{ b = 11 }$
| PRODUCT = -60 and SUM = 11 | |
| -1 60 | 1 -60 |
| -2 30 | 2 -30 |
| -3 20 | 3 -20 |
| -4 15 | 4 -15 |
| -5 12 | 5 -12 |
| -6 10 | 6 -10 |
Step 7: Replace middle term $ 11 x $ with $ 15x-4x $:
$$ 10x^{2}+11x-6 = 10x^{2}+15x-4x-6 $$Step 8: Apply factoring by grouping. Factor $ 5x $ out of the first two terms and $ -2 $ out of the last two terms.
$$ 10x^{2}+15x-4x-6 = 5x\left(2x+3\right) -2\left(2x+3\right) = \left(5x-2\right) \left(2x+3\right) $$