It seems that $ 20x^{2}-6x-9 $ 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 $\color{blue}{ a = 20 }$ by the constant term $\color{blue}{c = -9} $.
$$ a \cdot c = -180 $$Step 3: Find out two numbers that multiply to $ a \cdot c = -180 $ and add to $ b = -6 $.
Step 4: All pairs of numbers with a product of $ -180 $ are:
PRODUCT = -180 | |
-1 180 | 1 -180 |
-2 90 | 2 -90 |
-3 60 | 3 -60 |
-4 45 | 4 -45 |
-5 36 | 5 -36 |
-6 30 | 6 -30 |
-9 20 | 9 -20 |
-10 18 | 10 -18 |
-12 15 | 12 -15 |
Step 5: Find out which factor pair sums up to $\color{blue}{ b = -6 }$
Step 6: Because none of these pairs will give us a sum of $ \color{blue}{ -6 }$, we conclude the polynomial cannot be factored.