It seems that $ 8h^{2}-h-3 $ 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 = -3} $.
$$ a \cdot c = -24 $$Step 3: Find out two numbers that multiply to $ a \cdot c = -24 $ and add to $ b = -1 $.
Step 4: All pairs of numbers with a product of $ -24 $ are:
| PRODUCT = -24 | |
| -1 24 | 1 -24 |
| -2 12 | 2 -12 |
| -3 8 | 3 -8 |
| -4 6 | 4 -6 |
Step 5: Find out which factor pair sums up to $\color{blue}{ b = -1 }$
Step 6: Because none of these pairs will give us a sum of $ \color{blue}{ -1 }$, we conclude the polynomial cannot be factored.