It seems that $ 4x^{2}-x-16 $ 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 = -16} $.
$$ a \cdot c = -64 $$Step 3: Find out two numbers that multiply to $ a \cdot c = -64 $ and add to $ b = -1 $.
Step 4: All pairs of numbers with a product of $ -64 $ are:
| PRODUCT = -64 | |
| -1 64 | 1 -64 |
| -2 32 | 2 -32 |
| -4 16 | 4 -16 |
| -8 8 | 8 -8 |
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.