It seems that $ 101x^{2}+10x+5175 $ cannot be factored out.
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 $\color{blue}{ a = 101 }$ by the constant term $\color{blue}{c = 5175} $.
$$ a \cdot c = 522675 $$Step 3: Find out two numbers that multiply to $ a \cdot c = 522675 $ and add to $ b = 10 $.
Step 4: All pairs of numbers with a product of $ 522675 $ are:
| PRODUCT = 522675 | |
| 1 522675 | -1 -522675 |
| 3 174225 | -3 -174225 |
| 5 104535 | -5 -104535 |
| 9 58075 | -9 -58075 |
| 15 34845 | -15 -34845 |
| 23 22725 | -23 -22725 |
| 25 20907 | -25 -20907 |
| 45 11615 | -45 -11615 |
| 69 7575 | -69 -7575 |
| 75 6969 | -75 -6969 |
| 101 5175 | -101 -5175 |
| 115 4545 | -115 -4545 |
| 207 2525 | -207 -2525 |
| 225 2323 | -225 -2323 |
| 303 1725 | -303 -1725 |
| 345 1515 | -345 -1515 |
| 505 1035 | -505 -1035 |
| 575 909 | -575 -909 |
Step 5: Find out which factor pair sums up to $\color{blue}{ b = 10 }$
Step 6: Because none of these pairs will give us a sum of $ \color{blue}{ 10 }$, we conclude the polynomial cannot be factored.