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anonymous
 5 years ago
how do i divide x^2+3x1 by 2x^5+3x^3+2x^2x+1?
anonymous
 5 years ago
how do i divide x^2+3x1 by 2x^5+3x^3+2x^2x+1?

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shadowfiend
 5 years ago
Best ResponseYou've already chosen the best response.0So, this division is the same as creating the fraction: $$ \frac{2x^5 + 3x^3 + 2x^2  x + 1}{x^2 + 3x  1} $$

shadowfiend
 5 years ago
Best ResponseYou've already chosen the best response.0Er, sorry, other way. $$ \frac{x^2 + 3x  1}{2x^5 + 3x^3 + 2x^2  x + 1} $$

shadowfiend
 5 years ago
Best ResponseYou've already chosen the best response.0This breaks up into: $$\frac{x^2}{2x^5 + 3x^3 + 2x^2 x + 1} + \frac{3x}{2x^5 + 3x^3 + 2x^2 x + 1} + \frac{1}{2x^5 + 3x^3 + 2x^2 x + 1} $$ You can do the division on each individual part, but you'll end up with negative exponents (or fractions) everywhere.

shadowfiend
 5 years ago
Best ResponseYou've already chosen the best response.0That was a bit too long, sorry. It breaks up into: \[\begin{align}\frac{x^2}{2x^5 + 3x^3 + 2x^2 x + 1} + \frac{3x}{2x^5 + 3x^3 + 2x^2 x + 1} \\+ \frac{1}{2x^5 + 3x^3 + 2x^2 x + 1}\end{align}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Are we sure the OP has the division right side up? If so, the first term of the quotient (from long division) is 1/2 x^(3), and the quotient goes on for an infinite number of terms, I think. Dividing the other way round makes more sense to me.

shadowfiend
 5 years ago
Best ResponseYou've already chosen the best response.0That would indeed much nicer.
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