# Can the degree of the remainder be larger than the degree of the quotient when dividing by a polynomial

A question I got on my homework was if the dividend has a degree of 9, the divisor has a degree of 4, what would the largest possible degree of the remainder be? Since the quotient in this case would have a degree of 5, would the remainder have the same degree? Or is it higher or lower?

• Welcome to MathSX! No. The condition on the remainder $r$ is that $$r=0\enspace\text{ or }\deg r<\deg(\text{divisor}).$$ Oct 17, 2017 at 21:04
• what if you have a equation like x^9 + x^8 - x^7 / x^4 + x^3 there would a remainder of -x^7 but the quotient is x^5 thanks for answering my question btw Oct 17, 2017 at 21:06
• You're wrong: the quotient is $x^5-x^3+x^2-x+1$ and the remainder is $-x^3$. Oct 17, 2017 at 21:21
• I'm sorry, I formatted the question wrong. I was thinking something like this $(x^9 + x^8 - x^7)/( x^4 + x^3)$ Oct 17, 2017 at 21:31
• What does it have to do with remainders? You can simplify the fraction by $x^3$. Oct 17, 2017 at 22:03

Just like with integers, if I divide something by $3$, the remainder is going to be at most $3$, but if the number is $3$ billion, the quotient is going to be a billion. The quotient can be arbitrarily large.
Similarly for polynomials. The quotient can be as big (in degree) as it wants. Given some remainder $r$ of degree less than $\deg(a)$, the polynomial $q(x)a(x)+r(x)$ has quotient $q$ under division by $a$ - where $q$ can be any polynomial, even of degree thirty eight trillion.
The answer to the question posed in the title is in general yes, and the reason follows from the observation of Jack M that not only can the quotient be as big in degree as it wants but it also (same difference) can be as small in degree as it wants, which is what is needed to answer the question in the affirmative. For example if you divide $$x^2+x$$ by $$x^2$$ the quotient is $$1$$ and the remainder is $$x$$.
However the answer for the specific example given in the body of the question is no as Bernard describes in the comments; however it is worthwhile to point out that this depends on the degree of the divisor in relation to the degree of the dividend. Only in the case where the degree of the divisor is less than $$\frac{\deg(\text{dividend})}{2}+1$$ is it not possible for a remainder to have a degree greater than the quotient.