It seems that $ 19r^{2}+15r+6 $ 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 $\color{blue}{ a = 19 }$ by the constant term $\color{blue}{c = 6} $.
$$ a \cdot c = 114 $$Step 3: Find out two numbers that multiply to $ a \cdot c = 114 $ and add to $ b = 15 $.
Step 4: All pairs of numbers with a product of $ 114 $ are:
| PRODUCT = 114 | |
| 1 114 | -1 -114 |
| 2 57 | -2 -57 |
| 3 38 | -3 -38 |
| 6 19 | -6 -19 |
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.