Factor polynomial $$ 90v^{2}-181v+90 $$
Answer
Explanation
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 $\color{blue}{ a = 90 }$ by the constant term $\color{blue}{c = 90} $.
$$ a \cdot c = 8100 $$Step 3: Find out two numbers that multiply to $ a \cdot c = 8100 $ and add to $ b = -181 $.
Step 4: All pairs of numbers with a product of $ 8100 $ are:
| PRODUCT = 8100 | |
| 1 8100 | -1 -8100 |
| 2 4050 | -2 -4050 |
| 3 2700 | -3 -2700 |
| 4 2025 | -4 -2025 |
| 5 1620 | -5 -1620 |
| 6 1350 | -6 -1350 |
| 9 900 | -9 -900 |
| 10 810 | -10 -810 |
| 12 675 | -12 -675 |
| 15 540 | -15 -540 |
| 18 450 | -18 -450 |
| 20 405 | -20 -405 |
| 25 324 | -25 -324 |
| 27 300 | -27 -300 |
| 30 270 | -30 -270 |
| 36 225 | -36 -225 |
| 45 180 | -45 -180 |
| 50 162 | -50 -162 |
| 54 150 | -54 -150 |
| 60 135 | -60 -135 |
| 75 108 | -75 -108 |
| 81 100 | -81 -100 |
| 90 90 | -90 -90 |
Step 5: Find out which factor pair sums up to $\color{blue}{ b = -181 }$
| PRODUCT = 8100 and SUM = -181 | |
| 1 8100 | -1 -8100 |
| 2 4050 | -2 -4050 |
| 3 2700 | -3 -2700 |
| 4 2025 | -4 -2025 |
| 5 1620 | -5 -1620 |
| 6 1350 | -6 -1350 |
| 9 900 | -9 -900 |
| 10 810 | -10 -810 |
| 12 675 | -12 -675 |
| 15 540 | -15 -540 |
| 18 450 | -18 -450 |
| 20 405 | -20 -405 |
| 25 324 | -25 -324 |
| 27 300 | -27 -300 |
| 30 270 | -30 -270 |
| 36 225 | -36 -225 |
| 45 180 | -45 -180 |
| 50 162 | -50 -162 |
| 54 150 | -54 -150 |
| 60 135 | -60 -135 |
| 75 108 | -75 -108 |
| 81 100 | -81 -100 |
| 90 90 | -90 -90 |
Step 6: Replace middle term $ -181 x $ with $ -81x-100x $:
$$ 90x^{2}-181x+90 = 90x^{2}-81x-100x+90 $$Step 7: Apply factoring by grouping. Factor $ 9x $ out of the first two terms and $ -10 $ out of the last two terms.
$$ 90x^{2}-81x-100x+90 = 9x\left(10x-9\right) -10\left(10x-9\right) = \left(9x-10\right) \left(10x-9\right) $$