Factor polynomial $$ -7x^{2}-51x+40 $$
$$ -7x^{2}-51x+40 = -( 7x-5 )( x+8 ) $$
Step 1 :
After factoring out $ -1 $ we have:
$$ -7x^{2}-51x+40 = - ~ ( 7x^{2}+51x-40 ) $$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 $\color{blue}{ a = 7 }$ by the constant term $\color{blue}{c = -40} $.
$$ a \cdot c = -280 $$Step 4: Find out two numbers that multiply to $ a \cdot c = -280 $ and add to $ b = 51 $.
Step 5: All pairs of numbers with a product of $ -280 $ are:
| PRODUCT = -280 | |
| -1 280 | 1 -280 |
| -2 140 | 2 -140 |
| -4 70 | 4 -70 |
| -5 56 | 5 -56 |
| -7 40 | 7 -40 |
| -8 35 | 8 -35 |
| -10 28 | 10 -28 |
| -14 20 | 14 -20 |
Step 6: Find out which factor pair sums up to $\color{blue}{ b = 51 }$
| PRODUCT = -280 and SUM = 51 | |
| -1 280 | 1 -280 |
| -2 140 | 2 -140 |
| -4 70 | 4 -70 |
| -5 56 | 5 -56 |
| -7 40 | 7 -40 |
| -8 35 | 8 -35 |
| -10 28 | 10 -28 |
| -14 20 | 14 -20 |
Step 7: Replace middle term $ 51 x $ with $ 56x-5x $:
$$ 7x^{2}+51x-40 = 7x^{2}+56x-5x-40 $$Step 8: Apply factoring by grouping. Factor $ 7x $ out of the first two terms and $ -5 $ out of the last two terms.
$$ 7x^{2}+56x-5x-40 = 7x\left(x+8\right) -5\left(x+8\right) = \left(7x-5\right) \left(x+8\right) $$