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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

1 Answer 1

up vote 1 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 (sorry for the awkward formatting):

$$\begin{array}{rrr|rr} &&&x^2&-&y\\ \hline x^2&+&y&x^4&+&&+&y^4\\ &&&x^4&+&x^2y\\ \hline &&&&&-x^2y&+&y^4\\ &&&&&-x^2y&-&y^2\\ \hline &&&&&&&y^4+y^2 \end{array}$$

Thus, $x^4+y^4=(x^2+y)(x^2-y)+(y^4+y^2)$, 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:

$$\begin{array}{rrr|rr} &&&y^3&-&x^2y^2&+&x^4y&-&x^6\\ \hline y&+&x^2&y^4&+&&&&&&+&x^4\\ &&&y^4&+&x^2y^3\\ \hline &&&&&-x^2y^3\\ &&&&&-x^2y^3&-&x^4y^2\\ \hline &&&&&&&x^4y^2\\ &&&&&&&x^4y^2&+&x^6y\\ \hline &&&&&&&&&-x^6y&+&x^4\\ &&&&&&&&&-x^6y&-&x^8\\ \hline &&&&&&&&&&&x^8+x^4 \end{array}$$

That is, $y^4+x^4=(y+x^2)(y^3-x^2y^2+x^4y-x^6)+(x^8+x^4)$, 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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