It seems that $ 5x^{2}+8x+20 $ 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 $\color{blue}{ a = 5 }$ by the constant term $\color{blue}{c = 20} $.
$$ a \cdot c = 100 $$Step 3: Find out two numbers that multiply to $ a \cdot c = 100 $ and add to $ b = 8 $.
Step 4: All pairs of numbers with a product of $ 100 $ are:
| PRODUCT = 100 | |
| 1 100 | -1 -100 |
| 2 50 | -2 -50 |
| 4 25 | -4 -25 |
| 5 20 | -5 -20 |
| 10 10 | -10 -10 |
Step 5: Find out which factor pair sums up to $\color{blue}{ b = 8 }$
Step 6: Because none of these pairs will give us a sum of $ \color{blue}{ 8 }$, we conclude the polynomial cannot be factored.