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How would I prove that if two polynomials (over reals) have a product of zero, then one of those polynomials must be zero?

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closed as off-topic by Najib Idrissi, Pedro Tamaroff Feb 25 at 20:56

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What did you tried to solve what you asked? – choco_addicted Feb 25 at 10:06

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.

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Of course, this is called the degree of the non-zero polynomial. Your proof more or less says that if both factors are non-zero, they have degrees $\ge 0$, and then the product polynomial can be seen to have the degree sum as its degree, and hence the product polynomial is non-zero as well. – Jeppe Stig Nielsen Feb 25 at 16:18
    
Yep. This is the problem with answering any question about something so basic: you have to ask yourself what well-established concepts and theorems are OK to use. Note I explained "high order term"; had I used degree I would have felt an urge to explain that. I once answered a question invoking the open mapping theorem for holomorphic functions when I could have used Liouville's theorem, and got a comment saying that the answer was esoteric. – ForgotALot Feb 25 at 17:20
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This is one of those “proofs by contradiction” that’s really a proof by contraposition. Better to just show directly that $p \neq 0 \wedge q \neq 0 \implies pq \neq 0$. – Lynn Feb 25 at 17:43
    
This answer assumes that the only polynomial corresponding to the constant function 0 is the constant 0 polynomial. How can you be sure that some combination of lower order terms could not "cancel out" the highest order term? I would probably argue this with limits, but these are not available at this lever. – Steven Gubkin Feb 25 at 18:32
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As regards how to prove that "some combination of lower order terms could not 'cancel out' the highest order term": you are suggesting in effect that there might be a polynomial with nonzero coefficients (the higher order term + the cancelling lower order terms) which is nonetheless zero for all real $x$. The most elementary proof that there isn't seems to me with an inequality showing that for large $x$ the high order term is larger in modulus than all others together. As always with these basic questions it is hard to know what well-known stuff can be taken for granted. – ForgotALot Feb 25 at 18:57

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.

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This is good, but relies on the fact that a polynomial has only finitely many roots, which may not be obvious. – Steven Gubkin Feb 25 at 18:28
    
@StevenGubkin Sure, though one can show that easily just using induction, or appeal to the fundament theorem of algebra. – Macavity Feb 25 at 18:34

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.

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I don't understand how this would help prove my case. – user317627 Feb 25 at 6:14
    
If $p,q$ are not zero they have, say, $m$ and $n$ roots. Then $pq$ has a maximum of $m+n$ roots and can't be the zero polynomial. – David Feb 25 at 6:17
    
Is there a rigorous proof for this? – user317627 Feb 25 at 6:24
    
This does not work over finite fields, since your argument forgets the nature of polynomials. You only work with functions. – MooS Feb 25 at 6:33
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@MooS I didn't forget the nature of polynomials, but I also didn't forget that the OP specifically asked about real polynomials :) – David Feb 25 at 6:55

$\bullet \;$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 $0.$

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This is unnecessarily complicated. – Pedro Tamaroff Feb 25 at 20:56

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.

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Just because one of the $a_ib_j$ terms is not zero, it doesn't follow that $p(x)q(x)$ is not zero, because $a_ib_j$ isn't the only coefficient of $x^{i+j}$ and there could be cancellation. – Ben Millwood Feb 25 at 15:50
    
You are absolutely right. I'll edit and make the argument for the last coefficient. – Cheese Feb 25 at 16:46
    
What do you mean when you say "Take the contraposition and you're done."? – user317627 Feb 25 at 19:47
    
I proved that if you start with a non zero $p$ AND a non zero $q$ then it implies that $p\cdot q$ must be non zero. The contraposition of a logical proposition $(A\land B) \Rightarrow C$ is simply $\neg C \Rightarrow (\neg A \lor \neg B$). So we get that if $p\cdot q$ is zero then either $p$ or $q$ must be zero. – Cheese Feb 25 at 20:38

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.

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This needs to be proved. It depends on the fact that the product of the two leading terms is not zero. This is not always true in general rings. – Matt Samuel Feb 25 at 7:18
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The degree is $n+m$, not $nm$. – Erick Wong Feb 25 at 7:21
    
@MattSamuel I didn't claim to prove every detail, but the OP isn't working over an arbitrary ring and that fact for the real line is obvious enough that I felt justified omitting it. The OP should fill in the details if they wish. – JLA Feb 25 at 17:23

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.

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That looks more like a test than proof. – pjs36 Feb 25 at 20:45
    
Yeah, saw that, expanded the proof to validate the source product also. The question was about proving the assumption that one or both terms will resolve to zero when the product is zero. – Zorro Borealis Feb 25 at 20:56

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