It seems that $ 4x^{2}+20x+81 $ 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 $\color{blue}{ a = 4 }$ by the constant term $\color{blue}{c = 81} $.
$$ a \cdot c = 324 $$Step 3: Find out two numbers that multiply to $ a \cdot c = 324 $ and add to $ b = 20 $.
Step 4: All pairs of numbers with a product of $ 324 $ are:
| PRODUCT = 324 | |
| 1 324 | -1 -324 |
| 2 162 | -2 -162 |
| 3 108 | -3 -108 |
| 4 81 | -4 -81 |
| 6 54 | -6 -54 |
| 9 36 | -9 -36 |
| 12 27 | -12 -27 |
| 18 18 | -18 -18 |
Step 5: Find out which factor pair sums up to $\color{blue}{ b = 20 }$
Step 6: Because none of these pairs will give us a sum of $ \color{blue}{ 20 }$, we conclude the polynomial cannot be factored.