It seems that $ 6x^{2}+8x-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 $\color{blue}{ a = 6 }$ by the constant term $\color{blue}{c = -3} $.
$$ a \cdot c = -18 $$Step 3: Find out two numbers that multiply to $ a \cdot c = -18 $ and add to $ b = 8 $.
Step 4: All pairs of numbers with a product of $ -18 $ are:
| PRODUCT = -18 | |
| -1 18 | 1 -18 |
| -2 9 | 2 -9 |
| -3 6 | 3 -6 |
Step 5: Find out which factor pair sums up to $\color{blue}{ b = 8 }$
Step 6: Because none of these pairs will give us a sum of $ \color{blue}{ 8 }$, we conclude the polynomial cannot be factored.