Assume that the polynomials that we multiply consist of more than one term.

I don't think we can get a result containing only a single term, but I don't know how to prove it.


6 Answers 6


Yes. Take the polynomial $2X^2+X \in \mathbb{Z}_4[X]$. Then, $(2X^2+X)(2X^2+X)=X^2$.

If you are on an integral domain, this cannot happen. This is because the coefficients of the top degree and the lowest degree will be non-zero.

Therefore, if you are talking about polynomials with rational coefficients or integer coefficients, the answer is: No.

Bottom line: If you don't know what an integral domain is, you are probably talking about a polynomial with rational coefficients, to which the answer to your question is: No, that can't happen.

If you know what an integral domain is, then the answer to your question is: In general, yes, it can happen. But it can't happen in an integral domain.

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    $\begingroup$ (Since the question is tagged 'algebra-precalculus', it is a safe bet that talking about integral domains is probably going to confuse the poster more that necessary!) $\endgroup$ Jun 9, 2016 at 23:00
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    $\begingroup$ @MarianoSuárez-Alvarez I see your point, and I'll add clarification. Thank you for the suggestion. But answers are also for future reference, and people who stumble here may or may not be confused by this. $\endgroup$
    – Aloizio Macedo
    Jun 9, 2016 at 23:02
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    $\begingroup$ $\mathbb Z_4$ is the ring of integers mod $4$. $\endgroup$ Jun 9, 2016 at 23:09
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    $\begingroup$ @paulpaul1076: I see you didn't get an answer that tells you anything if you don't already know it. Well, $\mathbb{Z}_4$ is like non-negative integers except that after every operation (addition, subtraction or multiplication) you take the (positive) remainder when divided by $4$. So here $1-2 = 3$ because the ordinary $-1$ is $3$ more than a multiple of $4$. This structure has only $4$ things, namely $0,1,2,3$. You can convince yourself that it satisfies $a+b = b+a$ and $ab = ba$ and $(a+b)·c = a·c+b·c$ and $0+a = a$ and $1·a = a$ and $a-a = 0$. Note though that here $2·2 = 0 = 0·0$. $\endgroup$
    – user21820
    Jun 10, 2016 at 3:42
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    $\begingroup$ @user21820 thanks a lot for the comment. I know that mod 4 is an operation that breaks up the set of integers into 4 equivalence classes [0],[1],[2] and [3] and have verified that $(2x^2+x)(2x^2+x)=4x^4+4x^3+x^2$, so you only need to look at the last term, because the first 2 are divisible by 4. I am new to math so I wasn't familiar with the notation $\mathbb{Z}_4$ $\endgroup$
    – Pavel
    Jun 10, 2016 at 4:25

I assume that you speak about polynomials with real coefficients.

Suppose that the polynomials are $P(x)=ax^p+\cdots+bx^q$, $Q(x)=cx^r+\cdots+dx^s$, where

  • $a,b,c,d\neq 0$
  • $p>q\ge 0$ and $r>s\ge 0$
  • Of course, the terms omitted between have intemediate degree.

Then, the product $P\cdot Q$ has at least two terms: $acx^{p+r}$ and $bdx^{q+s}$. Note that $ac$ and $bd$ are not zero.

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    $\begingroup$ It's also important to note that $p+r \neq q+s$ (more specifically, $p+r \gt q+s$), so these two terms cannot be combined into one. $\endgroup$ Jun 11, 2016 at 14:45

If you multiply two monomials you will get a monomial.

The only root of a monomial is x=0.
Polynomials (excluding monomials) have at least one non-zero root (possibly complex and possibly in addition to 0, but at least one non-zero root).
If we multiply one polynomial by another, the roots of the of the original polynomials are roots of the product.

If the original polynomial has a non-zero root, the product has non-zero roots, and is not a monomial.


If $p$ is a non-zero polynomial, let us call width of $p$ the difference between the degree of the top monomial appearing in $p$ and the degree of the lowest monomial. For example, the width of $4x^8+2x^5-2x^4$ iss $8-4=4$ and the width of a monomial $7x^{19}$ is $19-19=0$.

You can easily show that if $p$ and $q$ are two non-zero polynomials, then the width of the product $pq$ is equal to the sum of the width of $p$ and the width of $q$.

Using this, you can check easily that the only way for two polynomials to have a product that is a monomial is that they both be monomials themselves.

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    $\begingroup$ Dually to your comment, I'd say you should make explicit for which polynomials this is true. $\endgroup$
    – quid
    Jun 9, 2016 at 23:07
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    $\begingroup$ While the logic boils down to the same as another answer, this is better, due to inventing a term, 'width', which is easy to understand, and makes stating the result easy. This is more like how math is done in the real world. $\endgroup$
    – DarenW
    Jun 11, 2016 at 6:40

It fails iff the coefficient ring has zero-divisors, i.e. $\,ab=0\,$ for $\,a,b\neq 0.\,$ Indeed if so then

$$ (ax+b)\,(bx+a)\, =\, (a^2\!+b^2)\, x$$

Conversely, if there are no zero-divisors, then the coefficients of the least and greatest degree terms of the product do not vanish (as above), yielding at least two terms in the product.

  • $\begingroup$ Note $\ $ Nontrivial commutative rings with no zero-divisors are known as (integral) domains. Equivalently they are nontrivial rings where nonzero elements are cancellable. $\endgroup$ Jun 10, 2016 at 17:27
  • $\begingroup$ Does your calculation require the ring to be commutative? Or is there some reason for $ab=0$ to imply $ba=0$ even if $ab\ne ba$ in general? $\endgroup$ Jun 10, 2016 at 18:26
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    $\begingroup$ @Mario The problem is tagged algebra-precalculus, so it seems reasonable to (implicitly) assume the coefficient ring is commutative, associative etc. I fear that the answer is already too abstract as is, without introducing further ring-theoretic terminology unfamiliar to most algebra-precalc students. $\endgroup$ Jun 10, 2016 at 19:22

Assuming we're taking polynomials over a unique factorization domain, this is impossible. Indeed, if the result is only one monomial $X_{1}^{e_1} \dots X_k^{e_k}$, this uniquely factorizes into the product of $e_1$ copies of $X_1$, $e_2$ copies of $X_2$, and so forth. Hence any expression of the monomial as the product of two polynomials will have those polynomials actually be monomials.

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    $\begingroup$ There is no need to have unique factorization for this. $\endgroup$ Jun 9, 2016 at 22:58

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