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The question is as follows: Divide the following Polynomial and place the result into Division Statement.

$$\frac{m^4+n^4}{m^2+n^2}$$

Recently did this on a test and was stumped. A few calulators and classmates later, I'm still stumped.

I know that the division statement is $P(x) = q(x)*d(x) + R$. And I know that the remainder is going to be zero, or 1 degree less than the divisor. In this case, that means a linear remainder.

Trying to use long division did not work, and synthetic division is not possible. Without dividing, I can write the following, just given the information at hand.

$$m^4+n^4=q(x) *(m^2+n^2)+R$$

Any and all help appreciated.

Jyrki Lahtonen
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Isaac K
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  • Also, what's the variable: $m$ or $n$? – Randall Oct 20 '21 at 16:58
  • They're both the variables, the question has two of them. – Isaac K Oct 20 '21 at 16:58
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    Unless I'm way off, there is no such thing as a division algorithm for polynomials in two variables. – Randall Oct 20 '21 at 17:00
  • The only thing I could think of was if the top and bottom are factorable, you could factor them and divide that way. This is why I was so stumped on this question, because far as I know there isn't either. – Isaac K Oct 20 '21 at 17:01
  • I can think of another reason (which happens from time to time). The actual division rather concerns $$\frac{m^4-n^4}{m^2+n^2}$$ – Dietrich Burde Oct 20 '21 at 17:08
  • That would point to a typo on the test, because of course a difference of quartics is easy – Isaac K Oct 20 '21 at 17:11

2 Answers2

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Do you mean something like $$m^4+n^4=(m^2-n^2)(m^2+n^2)+2n^4?$$

That's what you get if you just do long division of $m^2+n^2$ into $m^4+n^4$: you get a quotient of $m^2-n^2$ and a remainder of $2n^4$

$$\require{enclose} \begin{array}{r}\color{red}{m^2}\color{green}{-n^2}\\[-3pt] m^2+n^2\enclose{longdiv}{m^4+n^4}\\[-3pt] \underline{\color{red}{m^4+m^2n^2}}\\[-3pt] -m^2n^2+n^4 \\[-3pt] \underline{\color{green}{-m^2n^2-n^4}}\\[-3pt] 2n^4\end{array}$$

[someone might know how to edit this so it looks better!]

David Quinn
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  • What was your process for long division? The answer is definitely right, but, how? – Isaac K Oct 20 '21 at 17:18
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    Treat $m$ as a variable and $n$ as a constant. (The polynomial is treated as being in $m$ rather than in $x$.)

    However, I find a fault with the problem statement: they should've told you that $m$ is the variable! You would get a different result if $n$ was the variable. Or if $x$ was the variable - in which case this is just a division of two constants, so it is its own quotient and the remainder is $0$...)

     –  Oct 20 '21 at 17:21
  • @IsaacK please see my edited answer – David Quinn Oct 20 '21 at 18:00
  • Solid. I hope you didn't spent 20 minutes writing that out XD. Thanks :) – Isaac K Oct 20 '21 at 18:05
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    I think this works perfectly if only $m$ is a variable. Else, the condition on the degree of the remainder fails. – Randall Oct 20 '21 at 18:07
  • May see also my clumsy version of long division here and here. – Ng Chung Tak Oct 21 '21 at 02:06
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With the concept of homogeneous polynomials, insert the missing terms:

$$\frac{m^4 \color{red}{+0} \color{blue}{m^3 n} \color{red}{+0} \color{blue}{m^2 n^2} \color{red}{+0} \color{blue}{mn^3}+ n^4}{m^2 \color{red}{+0} \color{blue}{mn}+ n^2}$$

then do the long division.

For economy, by letting $(a,b)=(m^2,n^2)$, we may skip those odd terms (of $m$ and $n$):

$$\frac{a^2 \color{red}{+0} \color{blue}{ab}+ b^2}{a+b}= \frac{m^4 \color{red}{+0} \color{blue}{m^2 n^2}+ n^4}{m^2+n^2}$$

Ng Chung Tak
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