It seems that $ 9a^{2}-15a+106 $ 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 = 106} $.
$$ a \cdot c = 954 $$Step 3: Find out two numbers that multiply to $ a \cdot c = 954 $ and add to $ b = -15 $.
Step 4: All pairs of numbers with a product of $ 954 $ are:
| PRODUCT = 954 | |
| 1 954 | -1 -954 |
| 2 477 | -2 -477 |
| 3 318 | -3 -318 |
| 6 159 | -6 -159 |
| 9 106 | -9 -106 |
| 18 53 | -18 -53 |
Step 5: Find out which factor pair sums up to $\color{blue}{ b = -15 }$
Step 6: Because none of these pairs will give us a sum of $ \color{blue}{ -15 }$, we conclude the polynomial cannot be factored.