It seems that $ 6x^{2}-7x-15 $ 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 = -15} $.
$$ a \cdot c = -90 $$Step 3: Find out two numbers that multiply to $ a \cdot c = -90 $ and add to $ b = -7 $.
Step 4: All pairs of numbers with a product of $ -90 $ are:
PRODUCT = -90 | |
-1 90 | 1 -90 |
-2 45 | 2 -45 |
-3 30 | 3 -30 |
-5 18 | 5 -18 |
-6 15 | 6 -15 |
-9 10 | 9 -10 |
Step 5: Find out which factor pair sums up to $\color{blue}{ b = -7 }$
Step 6: Because none of these pairs will give us a sum of $ \color{blue}{ -7 }$, we conclude the polynomial cannot be factored.