It seems that $ 3r^{2}-8r+7 $ 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 $\color{blue}{ a = 3 }$ by the constant term $\color{blue}{c = 7} $.
$$ a \cdot c = 21 $$Step 3: Find out two numbers that multiply to $ a \cdot c = 21 $ and add to $ b = -8 $.
Step 4: All pairs of numbers with a product of $ 21 $ are:
PRODUCT = 21 | |
1 21 | -1 -21 |
3 7 | -3 -7 |
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.