It seems that $ 9x^{2}-91x+232 $ 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 $\color{blue}{ a = 9 }$ by the constant term $\color{blue}{c = 232} $.
$$ a \cdot c = 2088 $$Step 3: Find out two numbers that multiply to $ a \cdot c = 2088 $ and add to $ b = -91 $.
Step 4: All pairs of numbers with a product of $ 2088 $ are:
| PRODUCT = 2088 | |
| 1 2088 | -1 -2088 |
| 2 1044 | -2 -1044 |
| 3 696 | -3 -696 |
| 4 522 | -4 -522 |
| 6 348 | -6 -348 |
| 8 261 | -8 -261 |
| 9 232 | -9 -232 |
| 12 174 | -12 -174 |
| 18 116 | -18 -116 |
| 24 87 | -24 -87 |
| 29 72 | -29 -72 |
| 36 58 | -36 -58 |
Step 5: Find out which factor pair sums up to $\color{blue}{ b = -91 }$
Step 6: Because none of these pairs will give us a sum of $ \color{blue}{ -91 }$, we conclude the polynomial cannot be factored.