Factor polynomial $$ -20q^{2}+56q-15 $$
$$ -20q^{2}+56q-15 = -( 2q-5 )( 10q-3 ) $$
Step 1 :
After factoring out $ -1 $ we have:
$$ -20q^{2}+56q-15 = - ~ ( 20q^{2}-56q+15 ) $$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 $\color{blue}{ a = 20 }$ by the constant term $\color{blue}{c = 15} $.
$$ a \cdot c = 300 $$Step 4: Find out two numbers that multiply to $ a \cdot c = 300 $ and add to $ b = -56 $.
Step 5: All pairs of numbers with a product of $ 300 $ are:
| PRODUCT = 300 | |
| 1 300 | -1 -300 |
| 2 150 | -2 -150 |
| 3 100 | -3 -100 |
| 4 75 | -4 -75 |
| 5 60 | -5 -60 |
| 6 50 | -6 -50 |
| 10 30 | -10 -30 |
| 12 25 | -12 -25 |
| 15 20 | -15 -20 |
Step 6: Find out which factor pair sums up to $\color{blue}{ b = -56 }$
| PRODUCT = 300 and SUM = -56 | |
| 1 300 | -1 -300 |
| 2 150 | -2 -150 |
| 3 100 | -3 -100 |
| 4 75 | -4 -75 |
| 5 60 | -5 -60 |
| 6 50 | -6 -50 |
| 10 30 | -10 -30 |
| 12 25 | -12 -25 |
| 15 20 | -15 -20 |
Step 7: Replace middle term $ -56 x $ with $ -6x-50x $:
$$ 20x^{2}-56x+15 = 20x^{2}-6x-50x+15 $$Step 8: Apply factoring by grouping. Factor $ 2x $ out of the first two terms and $ -5 $ out of the last two terms.
$$ 20x^{2}-6x-50x+15 = 2x\left(10x-3\right) -5\left(10x-3\right) = \left(2x-5\right) \left(10x-3\right) $$