It seems that $ 4x^{2}-8x-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 $\color{blue}{ a = 4 }$ by the constant term $\color{blue}{c = -15} $.
$$ a \cdot c = -60 $$Step 3: Find out two numbers that multiply to $ a \cdot c = -60 $ and add to $ b = -8 $.
Step 4: All pairs of numbers with a product of $ -60 $ are:
PRODUCT = -60 | |
-1 60 | 1 -60 |
-2 30 | 2 -30 |
-3 20 | 3 -20 |
-4 15 | 4 -15 |
-5 12 | 5 -12 |
-6 10 | 6 -10 |
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.