how do i divide x^2+3x-1 by 2x^5+3x^3+2x^2-x+1?

8 years agoSo, this division is the same as creating the fraction: $$ \frac{2x^5 + 3x^3 + 2x^2 - x + 1}{x^2 + 3x - 1} $$

8 years agoEr, sorry, other way. $$ \frac{x^2 + 3x - 1}{2x^5 + 3x^3 + 2x^2 - x + 1} $$

8 years agoThis 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.

8 years agoThat was a bit too long, sorry. It breaks up into: \[\begin{align}\frac{x^2}{2x^5 + 3x^3 + 2x^2 -x + 1} + \frac{3x}{2x^5 + 3x^3 + 2x^2 -x + 1} \\+ \frac{-1}{2x^5 + 3x^3 + 2x^2 -x + 1}\end{align}\]

8 years agoUse long division?

8 years agoAre 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.

8 years agoThat would indeed much nicer.

8 years ago