# Division of Polynomials with Division Statement

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.

• Also, what's the variable: $m$ or $n$? Oct 20, 2021 at 16:58
• They're both the variables, the question has two of them. Oct 20, 2021 at 16:58
• Unless I'm way off, there is no such thing as a division algorithm for polynomials in two variables. Oct 20, 2021 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. Oct 20, 2021 at 17:01
• I can think of another reason (which happens from time to time). The actual division rather concerns $$\frac{m^4-n^4}{m^2+n^2}$$ Oct 20, 2021 at 17:08

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!]

• What was your process for long division? The answer is definitely right, but, how? Oct 20, 2021 at 17:18
• 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, 2021 at 17:21
• @IsaacK please see my edited answer Oct 20, 2021 at 18:00
• Solid. I hope you didn't spent 20 minutes writing that out XD. Thanks :) Oct 20, 2021 at 18:05
• I think this works perfectly if only $m$ is a variable. Else, the condition on the degree of the remainder fails. Oct 20, 2021 at 18:07

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 $$n$$):

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