It seems that $ 7b^{2}-36b+140 $ 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 = 140} $.
$$ a \cdot c = 980 $$Step 3: Find out two numbers that multiply to $ a \cdot c = 980 $ and add to $ b = -36 $.
Step 4: All pairs of numbers with a product of $ 980 $ are:
PRODUCT = 980 | |
1 980 | -1 -980 |
2 490 | -2 -490 |
4 245 | -4 -245 |
5 196 | -5 -196 |
7 140 | -7 -140 |
10 98 | -10 -98 |
14 70 | -14 -70 |
20 49 | -20 -49 |
28 35 | -28 -35 |
Step 5: Find out which factor pair sums up to $\color{blue}{ b = -36 }$
Step 6: Because none of these pairs will give us a sum of $ \color{blue}{ -36 }$, we conclude the polynomial cannot be factored.