document.write( "Question 620984: Please help it says, showing your work add the following polynomials \r
\n" ); document.write( "\n" ); document.write( "(1/5x^2-1/5x-3/4)+(1/10x^2-1/2x+1/2) \n" ); document.write( "

Algebra.Com's Answer #390516 by KMST(5328) $\"\"$ $\"About$

You can put this solution on YOUR website!
NOTE 1:
\n" ); document.write( "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.
\n" ); document.write( "I do not usually write it that way because it is a waste of time and ink, but to me $\"1%2F5\"$ $\"x%5E2-1%2F5\"$ $\"x-3%2F4\"$ is a sum of 3 terms: $\"1%2F5\"$ $\"x%5E2\"$ +( $\"-1%2F5\"$ $\"x\"$ )+ $\"%28-3%2F4%29\"$
\n" ); document.write( "(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).
\n" ); document.write( "NOTE 2:
\n" ); document.write( "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.
\n" ); document.write( "THE PROBLEM:
\n" ); document.write( "Applying associative and commutative properties, we can rearrange and regroup the terms.
\n" ); document.write( "( $\"1%2F5\"$ $\"x%5E2-1%2F5\"$ $\"x-3%2F4\"$ )+( $\"1%2F10\"$ $\"x%5E2-1%2F2\"$ $\"x%2B1%2F2\"$ )=( $\"1%2F5\"$ $\"x%5E2%2B1%2F10\"$ $\"x%5E2\"$ )+( $\"-1%2F5\"$ $\"x-1%2F2\"$ $\"x\"$ )+ $\"%28-3%2F4%2B1%2F2%29\"$
\n" ); document.write( "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.
\n" ); document.write( "I'll show it baby step by baby step.
\n" ); document.write( "( $\"1%2F5\"$ $\"x%5E2%2B1%2F10\"$ $\"x%5E2\"$ )+( $\"-1%2F5\"$ $\"x-1%2F2\"$ $\"x\"$ )+ $\"%28-3%2F4%2B1%2F2%29\"$ =

$\"x%5E2-7%2F10\"$ $\"x-1%2F4\"$ \n" ); document.write( "

\n" );