It seems that $ 8x^{2}-150x-5 $ 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 $\color{blue}{ a = 8 }$ by the constant term $\color{blue}{c = -5} $.
$$ a \cdot c = -40 $$Step 3: Find out two numbers that multiply to $ a \cdot c = -40 $ and add to $ b = -150 $.
Step 4: All pairs of numbers with a product of $ -40 $ are:
| PRODUCT = -40 | |
| -1 40 | 1 -40 |
| -2 20 | 2 -20 |
| -4 10 | 4 -10 |
| -5 8 | 5 -8 |
Step 5: Find out which factor pair sums up to $\color{blue}{ b = -150 }$
Step 6: Because none of these pairs will give us a sum of $ \color{blue}{ -150 }$, we conclude the polynomial cannot be factored.