It seems that $ 9x^{2}-19x+14 $ 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 = 14} $.
$$ a \cdot c = 126 $$Step 3: Find out two numbers that multiply to $ a \cdot c = 126 $ and add to $ b = -19 $.
Step 4: All pairs of numbers with a product of $ 126 $ are:
| PRODUCT = 126 | |
| 1 126 | -1 -126 |
| 2 63 | -2 -63 |
| 3 42 | -3 -42 |
| 6 21 | -6 -21 |
| 7 18 | -7 -18 |
| 9 14 | -9 -14 |
Step 5: Find out which factor pair sums up to $\color{blue}{ b = -19 }$
Step 6: Because none of these pairs will give us a sum of $ \color{blue}{ -19 }$, we conclude the polynomial cannot be factored.