Can the product of two polynomials result in a single term? 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.
 A: 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.
A: 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.
A: 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.
A: 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.
A: 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.
A: 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.
