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I have trouble understanding how to divide $x^4 + y^4$ by $f_1 = x^2 + y$ and $f_2 = x^2 y + 1$ using the ordering $ y \leq x$ and separately for $ x \leq y$.

Please help! I went to the tutoring center but none of the specialists understand this and I have never done this before in high school.

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If you had to divide $x^4+81$ by $x^2+3$, would you be able to do that? I think that dividing $x^4+y^4$ by $x^2+y$ using the ordering $y\le x$ amounts to the same thing, only with $y$ in place of $3$. – Gerry Myerson Nov 5 '12 at 22:43
Yes. I completely understand how to do single variable polynomial division but not for 2 variables. I really need some guidance – Lorraine Jane Nov 5 '12 at 22:44
Well, I told you what to do --- just do the division you know how to do, but with $y$ in place of 3. So ... do it! – Gerry Myerson Nov 5 '12 at 23:00
Do you have any worked examples from lecture notes or from a text that you could point us to? – Gerry Myerson Nov 5 '12 at 23:04
I got $x^4 + 81 = (x^2 - 3)(x^2 + 3) + 90$ so you're saying that's its $x^4 + y^4 = (x^2 -y)(x^2 + y) + y^4 + y^2$??? I just guessed that $90 = 3^4 + 3^2$ but I need to know what's really going on – Lorraine Jane Nov 5 '12 at 23:08
up vote 2 down vote accepted

Note: I am making an assumption here about exactly what is wanted, but I think it the likeliest possibility for a pre-calculus algebra course.

For the order $y\le x$, pretend that $y$ is a constant. Then you get this long division:

$$\require{enclose}\begin{align} x^2-\phantom{-x^2}y\phantom{+y^4+y^2}\\ x^2+y\enclose{longdiv}{x^4+\phantom{-x^2y}+\phantom{y^4+}y^4}\\ \underline{x^4+\phantom{-}x^2y\,\phantom{+y^4+y^2}}\\ -x^2y+\phantom{y^4+}y^4\\ \underline{-x^2y-\,\phantom{y^4+}y^2}\\ y^4+y^2 \end{align}$$

Thus, $x^4+y^4=\left(x^2+y\right)\left(x^2-y\right)+\left(y^4+y^2\right)$, with quotient $x^2-y$ and remainder $y^4+y^2$.

For the order $x\le y$ you treat $x$ as if it were a constant:

$$\require{enclose}\begin{align} y^3-\phantom-x^2y^2+x^4y-\phantom{-y}x^6\;\phantom{+x^8+x^4}\\ y+x^2\enclose{longdiv}{y^4+\phantom{-x^2y^3-x^4y^2+-x^6y}+\phantom{x^8+}x^4}\\ \underline{y^4+\phantom-x^2y^3\;\phantom{-x^4y^2+-x^6y+x^8+x^4}}\\ -x^2y^3\,\phantom{-x^4y^2+-x^6y+x^8+x^4}\\ \underline{-x^2y^3-x^4y^2\phantom{+-x^6y+x^8+x^4}}\\ x^4y^2\phantom{+-x^6y+x^8+x^4}\\ \underline{x^4y^2+\phantom-x^6y\phantom{+x^8+x^4}\;\;}\\ -x^6y+\;\phantom{x^8+}x^4\\ \underline{-x^6y-\;\phantom{x^8+}x^8}\\ x^8+x^4 \end{align}$$

That is, $y^4+x^4=\left(y+x^2\right)\left(y^3-x^2y^2+x^4y-x^6\right)+\left(x^8+x^4\right)$, with quotient $$y^3-x^2y^2+x^4y-x^6$$ and remainder $x^8+x^4$.

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We were getting there, slowly, in the comments. – Gerry Myerson Nov 5 '12 at 23:25
@Gerry: So I (now) see. I wasn’t following them: getting that monster formatted while doing the calculation was taking all of my concentration. – Brian M. Scott Nov 5 '12 at 23:26
I see that you agree with Gerry for one ordering and did the other on your own. But wait your remainder for the second ordering is too big. $x^8$ is too big can that happen? – Lorraine Jane Nov 5 '12 at 23:28
@Desperate: Yes, it can. (It did!) Remember, I’m thinking of $x$ as a constant in that calculation, so it’s just like ending up with larger coefficients in the remainder than you started with in the divident and divisor. – Brian M. Scott Nov 5 '12 at 23:35
Brian, no worries, what you're doing is probably helping more than what I was doing. – Gerry Myerson Nov 5 '12 at 23:42

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