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: Perhaps simpler: 
Suppose $p(x) q(x) = 0$, with neither $p(x)$ nor $q(x)$ identically zero.  WLOG, let the first polynomial have $n$ roots and the second one $m$ roots. Let $a \in \mathbb R$ be any real number other than these roots (which are finite).  Now $p(a) \neq 0$ as $a$ is not a root, and similarly $q(a) \neq 0$.  But $p(a)q(a) = 0$, so we have a contradiction as two non-zero real numbers are giving a zero product.
A: Hint: any polynomial which is not zero has only a finite number of roots, and the product has a root if and only if one of the factors has a root.
A: Let $p(x),q(x) \in \mathbb R[x]$ be two non zero polynomials of degree $m$ and $n$ respectively. We can write $p$ and $q$ in this fashion :
$$p(x) = \sum_{i=0}^{m}a_ix^i$$
$$q(x) = \sum_{j=0}^{n}b_jx^j$$
Since $p$ and $q$ are both non zero of order $m$ and $n$ respectively, then the last coefficients $a_m$ and $b_n$ must be non zero. Now let's take their product :
$$p(x)\cdot q(x)=\sum_{i=0}^{m}\sum_{j=0}^{n}a_ib_jx^ix^j=\sum_{i=0}^{m}\sum_{j=0}^{n}a_ib_jx^{i+j}$$
This resulting polynomial is zero if and only if all of its coefficients are zero, namely $a_ib_j$ for all $i$ and $j$. Since $a_m\neq0$ and $b_n\neq0$, then the last coefficient of this new polynomial, namely $a_nb_m$, is not zero. Therefore $p(x)\cdot q(x)\neq 0$.
Take the contraposition and you're done.
A: The product of two polynomials of degree $n,m \ge 1$ has degree $n+m\ge 1\,.$ A polynomial with positive degree cannot be zero since by the division algorithm for polynomials a nonzero polynomial has only finitely many roots. The case where one of the degrees is zero isn't any different.
A: $\bullet \;$Let $p(x)=\sum_{j=0}^na_jx^j$ where $n\geq 1$ and $a_n\ne 0.$  For $x\ne 0$ we have $p(x)=a_nx^n(1+\sum_{j=0}^{n-1}a_j/x^{n-j}a_n).$  Then $$(\;|x|>1 \land|x|>\max_{0\leq j\leq n-1} 2 n |a_j/a_n|\;)\implies$$ $$\implies \max_{0\leq j\leq n-1}|a_j/x^{n-j}a_n|\leq \max_{0\leq j\leq n-1}|a_j/x a_n|<1/2 n\implies $$ $$\implies |p(x)|\geq |a_nx^n|\cdot |1-\sum_{j=0}^{n-1}|a_j/x^{n-j}a_n|\geq |a_nx^n|\cdot |1-\sum_{j=0}^{n-1}1/2 n|=$$ $$=|a_nx^n/2|\ne 0.$$  
$\bullet \;$ With $p$ as above and with $q(x)=\sum_{i=0}^mb_ix^i$ with $b_m\ne 0$ we have $$p(x)q(x)=a_n b_mx^{m+n}+\sum_{k=0}^{n+m-1}c_kx^k$$ for some  constants $c_k \;(k=0,...,n+m-1)$
Applying the result of the first paragraph to the polynomial $p(x)q(x),$ we see that $p(x)q(x)$ is not constantly $0.$
A: 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.
A: 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.
