SOLUTION: Please help it says, showing your work add the following polynomials (1/5x^2-1/5x-3/4)+(1/10x^2-1/2x+1/2)

Algebra ->  Square-cubic-other-roots -> SOLUTION: Please help it says, showing your work add the following polynomials (1/5x^2-1/5x-3/4)+(1/10x^2-1/2x+1/2)      Log On


   

Question 620984: Please help it says, showing your work add the following polynomials
(1/5x^2-1/5x-3/4)+(1/10x^2-1/2x+1/2)
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
NOTE 1:
It is easier to understand algebra if you consider each minus signs as belonging to the number that immediately follows it. Then, everything is sums, and there is no subtraction.
I do not usually write it that way because it is a waste of time and ink, but to me 1%2F5x%5E2-1%2F5x-3%2F4 is a sum of 3 terms: 1%2F5x%5E2+(-1%2F5x)+%28-3%2F4%29
(If there is no visible number after a minus sign, as in 2-(3x+5), it means there is an invisible number 1. In the case of 2-(3x+5), I see 2+(-1)(3x+5) instead).
NOTE 2:
According to the convention for order of operations 1/5x^2=1%2F5x%5E2, so I have to write it as (1/5)x^2 or as x^2/5.
THE PROBLEM:
Applying associative and commutative properties, we can rearrange and regroup the terms.
(1%2F5x%5E2-1%2F5x-3%2F4)+(1%2F10x%5E2-1%2F2x%2B1%2F2)=(1%2F5x%5E2%2B1%2F10x%5E2)+(-1%2F5x-1%2F2x)+%28-3%2F4%2B1%2F2%29
Each new group can be added because they share the same letter variable part. We take out the letter part as a common factor and end up adding up the number coefficients.
I'll show it baby step by baby step.
(1%2F5x%5E2%2B1%2F10x%5E2)+(-1%2F5x-1%2F2x)+%28-3%2F4%2B1%2F2%29=x%5E2-7%2F10x-1%2F4