Question 620984
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 {{{1/5}}}{{{x^2-1/5}}}{{{x-3/4}}} is a sum of 3 terms: {{{1/5}}}{{{x^2}}}+({{{-1/5}}}{{{x}}})+{{{(-3/4)}}}
(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={{{1/5x^2}}}, 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.
({{{1/5}}}{{{x^2-1/5}}}{{{x-3/4}}})+({{{1/10}}}{{{x^2-1/2}}}{{{x+1/2}}})=({{{1/5}}}{{{x^2+1/10}}}{{{x^2}}})+({{{-1/5}}}{{{x-1/2}}}{{{x}}})+{{{(-3/4+1/2)}}}
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.
({{{1/5}}}{{{x^2+1/10}}}{{{x^2}}})+({{{-1/5}}}{{{x-1/2}}}{{{x}}})+{{{(-3/4+1/2)}}}={{{(1/5+1/10)x^2+(-1/5-1/2)x+(-3/4+2/4)=(2/10+1/10)x^2+(-2/10-5/10)x+(-1/4)=3/10}}}{{{x^2-7/10}}}{{{x-1/4}}}