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 $\color{blue}{ a = 40 }$ by the constant term $\color{blue}{c = -6} $.
$$ a \cdot c = -240 $$Step 3: Find out two numbers that multiply to $ a \cdot c = -240 $ and add to $ b = 1 $.
Step 4: All pairs of numbers with a product of $ -240 $ are:
| PRODUCT = -240 | |
| -1 240 | 1 -240 |
| -2 120 | 2 -120 |
| -3 80 | 3 -80 |
| -4 60 | 4 -60 |
| -5 48 | 5 -48 |
| -6 40 | 6 -40 |
| -8 30 | 8 -30 |
| -10 24 | 10 -24 |
| -12 20 | 12 -20 |
| -15 16 | 15 -16 |
Step 5: Find out which factor pair sums up to $\color{blue}{ b = 1 }$
| PRODUCT = -240 and SUM = 1 | |
| -1 240 | 1 -240 |
| -2 120 | 2 -120 |
| -3 80 | 3 -80 |
| -4 60 | 4 -60 |
| -5 48 | 5 -48 |
| -6 40 | 6 -40 |
| -8 30 | 8 -30 |
| -10 24 | 10 -24 |
| -12 20 | 12 -20 |
| -15 16 | 15 -16 |
Step 6: Replace middle term $ 1 x $ with $ 16x-15x $:
$$ 40x^{2}+x-6 = 40x^{2}+16x-15x-6 $$Step 7: Apply factoring by grouping. Factor $ 8x $ out of the first two terms and $ -3 $ out of the last two terms.
$$ 40x^{2}+16x-15x-6 = 8x\left(5x+2\right) -3\left(5x+2\right) = \left(8x-3\right) \left(5x+2\right) $$