It seems that $ 7v^{2}-5v-20 $ 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 = -20} $.
$$ a \cdot c = -140 $$Step 3: Find out two numbers that multiply to $ a \cdot c = -140 $ and add to $ b = -5 $.
Step 4: All pairs of numbers with a product of $ -140 $ are:
PRODUCT = -140 | |
-1 140 | 1 -140 |
-2 70 | 2 -70 |
-4 35 | 4 -35 |
-5 28 | 5 -28 |
-7 20 | 7 -20 |
-10 14 | 10 -14 |
Step 5: Find out which factor pair sums up to $\color{blue}{ b = -5 }$
Step 6: Because none of these pairs will give us a sum of $ \color{blue}{ -5 }$, we conclude the polynomial cannot be factored.