It seems that $ -7y^{2}+2y-8 $ cannot be factored out.
Step 1 :
After factoring out $ -1 $ we have:
$$ -7y^{2}+2y-8 = - ~ ( 7y^{2}-2y+8 ) $$Step 2 :
Step 2: 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 3: Multiply the leading coefficient $\color{blue}{ a = 7 }$ by the constant term $\color{blue}{c = 8} $.
$$ a \cdot c = 56 $$Step 4: Find out two numbers that multiply to $ a \cdot c = 56 $ and add to $ b = -2 $.
Step 5: All pairs of numbers with a product of $ 56 $ are:
PRODUCT = 56 | |
1 56 | -1 -56 |
2 28 | -2 -28 |
4 14 | -4 -14 |
7 8 | -7 -8 |
Step 6: Find out which factor pair sums up to $\color{blue}{ b = -2 }$
Step 7: Because none of these pairs will give us a sum of $ \color{blue}{ -2 }$, we conclude the polynomial cannot be factored.