It seems that $ 2n^{2}+44n+121 $ 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 $\color{blue}{ a = 2 }$ by the constant term $\color{blue}{c = 121} $.
$$ a \cdot c = 242 $$Step 3: Find out two numbers that multiply to $ a \cdot c = 242 $ and add to $ b = 44 $.
Step 4: All pairs of numbers with a product of $ 242 $ are:
PRODUCT = 242 | |
1 242 | -1 -242 |
2 121 | -2 -121 |
11 22 | -11 -22 |
Step 5: Find out which factor pair sums up to $\color{blue}{ b = 44 }$
Step 6: Because none of these pairs will give us a sum of $ \color{blue}{ 44 }$, we conclude the polynomial cannot be factored.