It seems that $ y^{2}+16y+65 $ cannot be factored out.
Step 1: Identify constants $ \color{blue}{ b }$ and $\color{red}{ c }$. ( $ \color{blue}{ b }$ is a number in front of the $ x $ term and $ \color{red}{ c } $ is a constant). In our case:
$$ \color{blue}{ b = 16 } ~ \text{ and } ~ \color{red}{ c = 65 }$$Now we must discover two numbers that sum up to $ \color{blue}{ 16 } $ and multiply to $ \color{red}{ 65 } $.
Step 2: Find out pairs of numbers with a product of $\color{red}{ c = 65 }$.
| PRODUCT = 65 | |
| 1 65 | -1 -65 |
| 5 13 | -5 -13 |
Step 3: Because none of these pairs will give us a sum of $ \color{blue}{ 16 }$, we conclude the polynomial cannot be factored.