Product of Polynomials is Zero How would I prove that if two polynomials (over reals) have a product of zero, then one of those polynomials must be zero?
A different proof.  All nonzero polynomials have a high order term $a_nx^n$ with $a_n\ne0$, $n\ge 0$ integer. (High order means that all terms in the polynomial have an exponent $\le n$; for example in $a+bx+cx^2$ the high order term is $cx^2$.) Suppose both of your polynomials have high order terms, $a_nx^n$ and $b_mx^m$ respectively, with $a_n\ne 0$ and $b_m\ne 0$. Then the term $a_nb_mx^{n+m}$ appears in the product, and what is more it is the only term where $x$ appears with exponent $n+m$.  However, by our hypothesis $a_nb_m$ has to be zero.  The product of two nonzero reals is nonzero.  This is the desired contradiction.
The proof is easy using any computer language, versus using a mathematical language.  Note also, this proof resolves as something being true, not something which must be further observed to be true, as would be the case with a proof resembling    0 = p(a) * p(b) ...

proof =  ( (0==p(a)*p(b)) == (0==p(a)||0==p(b)) ) ? true : false;

This method examines the product as well as the terms.