# In polynomial division, the remainder's degree is always less than that of the divisor, but when dividing $x^3+y^3$ by $x+5$, it isn't

I'm just a 9th grader trying to self-study, so if the question sounds silly to you please excuse me.

$$P(x) = x^3 + y^3$$ is divided by something like $$g(x) = x + 5;$$ the degree of the polynomial is $$3$$. Doesn't the degree of the remainder always be less than that of the divisor. Can someone explain what is happening?

• I believe the problem is simply that you're dealing with a polynomial in two variables, $x,y$ for $P$. That the remainder has a lesser degree than the divisor only really holds in the case of polynomials in one variable, I believe. – Eevee Trainer Feb 20 at 7:17
• To rewrite my previous (now deleted) comment because I'm dumb and it's been a while since I've dealt with this. Basically I noticed you have $P(x)$, but $P(x,y)$ would be more appropriate if $y$ was a variable. If $y$ is not a variable, i.e. some constant, then your remainder (which I got to be $y^3 - 125$) would be of degree zero as intended in the one-variable case. Of course if we have multiple variables a lot of nice stuff we have simply flies out the window. Such is math. – Eevee Trainer Feb 20 at 7:23
• The degree of that result viewed as a polynomial in x is less than 3. – William Elliot Feb 20 at 7:53
• So, y is not a variable but just a constant that we do not know the value of.Since y is a constant its degree will be 0 and not 3. – Aditya Bharadwaj Feb 20 at 8:12
• If you wish to learn how to extend the division algorithm to multivariate polynomails then search on "grobner basis". – Bill Dubuque Feb 20 at 21:52

As clarified by the OP in the comments that $$y$$ is a constant. I would rewrite $$y$$ as $$a$$ so that it looks more constant-like for the sake of convenience.
Now $$P(x)=x^3+a^3$$ and $$g(x)=x+5$$. By Euclid's Division Lemma for Polynomials you do have $$\text{deg}\ r(x)=0$$ or $$\text{deg }r(x)\lt\text{deg }g(x)$$.
You may confirm this by long division method which gives you the following result: $$\dfrac{x^3+a^3}{x+5}=x^2+5x-25+\dfrac{125+a^3}{x+5}$$ or $$r(x)=125+a^3$$ and $$q(x)=x^2+5x-25$$ which clearly satisfies $$\text{deg }g(x)=1\gt\text{deg }r(x)=0$$. So there is no discrepancy.