It seems that $ 13a^{2}+15a+40 $ 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 $\color{blue}{ a = 13 }$ by the constant term $\color{blue}{c = 40} $.
$$ a \cdot c = 520 $$Step 3: Find out two numbers that multiply to $ a \cdot c = 520 $ and add to $ b = 15 $.
Step 4: All pairs of numbers with a product of $ 520 $ are:
| PRODUCT = 520 | |
| 1 520 | -1 -520 |
| 2 260 | -2 -260 |
| 4 130 | -4 -130 |
| 5 104 | -5 -104 |
| 8 65 | -8 -65 |
| 10 52 | -10 -52 |
| 13 40 | -13 -40 |
| 20 26 | -20 -26 |
Step 5: Find out which factor pair sums up to $\color{blue}{ b = 15 }$
Step 6: Because none of these pairs will give us a sum of $ \color{blue}{ 15 }$, we conclude the polynomial cannot be factored.