Step 1 :
After factoring out $ -3y $ we have:
$$ -24y^{3}+90y^{2}-75y = -3y ( 8y^{2}-30y+25 ) $$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 $\color{blue}{ a = 8 }$ by the constant term $\color{blue}{c = 25} $.
$$ a \cdot c = 200 $$Step 4: Find out two numbers that multiply to $ a \cdot c = 200 $ and add to $ b = -30 $.
Step 5: All pairs of numbers with a product of $ 200 $ are:
| PRODUCT = 200 | |
| 1 200 | -1 -200 |
| 2 100 | -2 -100 |
| 4 50 | -4 -50 |
| 5 40 | -5 -40 |
| 8 25 | -8 -25 |
| 10 20 | -10 -20 |
Step 6: Find out which factor pair sums up to $\color{blue}{ b = -30 }$
| PRODUCT = 200 and SUM = -30 | |
| 1 200 | -1 -200 |
| 2 100 | -2 -100 |
| 4 50 | -4 -50 |
| 5 40 | -5 -40 |
| 8 25 | -8 -25 |
| 10 20 | -10 -20 |
Step 7: Replace middle term $ -30 x $ with $ -10x-20x $:
$$ 8x^{2}-30x+25 = 8x^{2}-10x-20x+25 $$Step 8: Apply factoring by grouping. Factor $ 2x $ out of the first two terms and $ -5 $ out of the last two terms.
$$ 8x^{2}-10x-20x+25 = 2x\left(4x-5\right) -5\left(4x-5\right) = \left(2x-5\right) \left(4x-5\right) $$