It seems that $ a^{2}+7a-1090 $ 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:
$$ \color{blue}{ b = 7 } ~ \text{ and } ~ \color{red}{ c = -1090 }$$Now we must discover two numbers that sum up to $ \color{blue}{ 7 } $ and multiply to $ \color{red}{ -1090 } $.
Step 2: Find out pairs of numbers with a product of $\color{red}{ c = -1090 }$.
| PRODUCT = -1090 | |
| -1 1090 | 1 -1090 |
| -2 545 | 2 -545 |
| -5 218 | 5 -218 |
| -10 109 | 10 -109 |
Step 3: Because none of these pairs will give us a sum of $ \color{blue}{ 7 }$, we conclude the polynomial cannot be factored.