Step 1: 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 2: Multiply the leading coefficient $\color{blue}{ a = 6 }$ by the constant term $\color{blue}{c = -20} $.
$$ a \cdot c = -120 $$Step 3: Find out two numbers that multiply to $ a \cdot c = -120 $ and add to $ b = -7 $.
Step 4: All pairs of numbers with a product of $ -120 $ are:
| PRODUCT = -120 | |
| -1 120 | 1 -120 |
| -2 60 | 2 -60 |
| -3 40 | 3 -40 |
| -4 30 | 4 -30 |
| -5 24 | 5 -24 |
| -6 20 | 6 -20 |
| -8 15 | 8 -15 |
| -10 12 | 10 -12 |
Step 5: Find out which factor pair sums up to $\color{blue}{ b = -7 }$
| PRODUCT = -120 and SUM = -7 | |
| -1 120 | 1 -120 |
| -2 60 | 2 -60 |
| -3 40 | 3 -40 |
| -4 30 | 4 -30 |
| -5 24 | 5 -24 |
| -6 20 | 6 -20 |
| -8 15 | 8 -15 |
| -10 12 | 10 -12 |
Step 6: Replace middle term $ -7 x $ with $ 8x-15x $:
$$ 6x^{2}-7x-20 = 6x^{2}+8x-15x-20 $$Step 7: Apply factoring by grouping. Factor $ 2x $ out of the first two terms and $ -5 $ out of the last two terms.
$$ 6x^{2}+8x-15x-20 = 2x\left(3x+4\right) -5\left(3x+4\right) = \left(2x-5\right) \left(3x+4\right) $$