I am working with an 8th grader.
In their book, Algebra, Chen and Gelfand ask this problem:
Problem 97. Is it possible when multiplying two polynomials that, after the collection of similar (like) terms, all terms except one vanish? (Do not count the case when each of the polynomials has only one monomial.)
A similar question (also from the same book) was asked a while ago.
I guess the answer to this one is similar to that one.
I tried to explain it by considering two polynomials $A: a_0 + a_1x$ and $B: b_0 + b_1x$ and then considering the cases where the product has only one term (either $a_0b_0$, $(a_0b_1 + a_1b_0)x$, or $a_1b_1x^2$) without making both $A$ and $B$ monomials or either of them the zero polynomial (that is at least three of $a_0, a_1, b_0, b_1$ are nonzero). In each of the cases, we showed that it is impossible to do so.
Is our answer correct?
I guess the result can be generalized for polynomials in single variable (assuming that the answer is correct) using the standard form: $\displaystyle \sum_0^n a_ix^i \cdot \sum_0^m b_ix^i$.
Best Answer
Yes, you are correct. Along the same lines we show the general case.
If $A$ is not a monomial then it can be written as $\sum_r^n a_ix^i$ with $0\leq r<n$, $a_r\not=0$ and $a_n\not=0$. Let $B=\sum_s^m b_ix^i$ with $0\leq s\leq m$, $b_s\not=0$ and $b_m\not=0$. Notice that $B$ is monomial when $m=s$.
The product $AB$ contains a unique monomial of maximum degree $a_nb_mx^{n+m}$ and a unique monomial of minimum degree $a_rb_sx^{r+s}$. Those two monomials are non-zero because $a_nb_m\not=0$, $a_rb_s\not=0$, and distinct because $r+s<n+m$. Therefore the product $AB$, after the collection of similar terms, has at least two terms.