It seems that $ 7x^{2}-4x-18 $ 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 $\color{blue}{ a = 7 }$ by the constant term $\color{blue}{c = -18} $.
$$ a \cdot c = -126 $$Step 3: Find out two numbers that multiply to $ a \cdot c = -126 $ and add to $ b = -4 $.
Step 4: All pairs of numbers with a product of $ -126 $ are:
| PRODUCT = -126 | |
| -1 126 | 1 -126 |
| -2 63 | 2 -63 |
| -3 42 | 3 -42 |
| -6 21 | 6 -21 |
| -7 18 | 7 -18 |
| -9 14 | 9 -14 |
Step 5: Find out which factor pair sums up to $\color{blue}{ b = -4 }$
Step 6: Because none of these pairs will give us a sum of $ \color{blue}{ -4 }$, we conclude the polynomial cannot be factored.