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 $\color{blue}{ a = 12 }$ by the constant term $\color{blue}{c = -35} $.
$$ a \cdot c = -420 $$Step 3: Find out two numbers that multiply to $ a \cdot c = -420 $ and add to $ b = -32 $.
Step 4: All pairs of numbers with a product of $ -420 $ are:
| PRODUCT = -420 | |
| -1 420 | 1 -420 |
| -2 210 | 2 -210 |
| -3 140 | 3 -140 |
| -4 105 | 4 -105 |
| -5 84 | 5 -84 |
| -6 70 | 6 -70 |
| -7 60 | 7 -60 |
| -10 42 | 10 -42 |
| -12 35 | 12 -35 |
| -14 30 | 14 -30 |
| -15 28 | 15 -28 |
| -20 21 | 20 -21 |
Step 5: Find out which factor pair sums up to $\color{blue}{ b = -32 }$
| PRODUCT = -420 and SUM = -32 | |
| -1 420 | 1 -420 |
| -2 210 | 2 -210 |
| -3 140 | 3 -140 |
| -4 105 | 4 -105 |
| -5 84 | 5 -84 |
| -6 70 | 6 -70 |
| -7 60 | 7 -60 |
| -10 42 | 10 -42 |
| -12 35 | 12 -35 |
| -14 30 | 14 -30 |
| -15 28 | 15 -28 |
| -20 21 | 20 -21 |
Step 6: Replace middle term $ -32 x $ with $ 10x-42x $:
$$ 12x^{2}-32x-35 = 12x^{2}+10x-42x-35 $$Step 7: Apply factoring by grouping. Factor $ 2x $ out of the first two terms and $ -7 $ out of the last two terms.
$$ 12x^{2}+10x-42x-35 = 2x\left(6x+5\right) -7\left(6x+5\right) = \left(2x-7\right) \left(6x+5\right) $$