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