[Math] “If x – a is a factor of polynomial P(x), then a is a factor of the constant term of the polynomial.” – Confused with proof

polynomials

I have recently started learning about polynomials. I've been able to grasp polynomial long division algorithm and the remainder and factor theorems and also a few other common-sense theorems about polynomials. There's just one property of polynomials I don't quite understand the proof of.

The property: "If x – a is a factor of polynomial P(x), then a is a factor of the constant term of the polynomial."

There are 2 proofs that I've seen so far that prove this theorem. The first proof I understand and makes complete sense to me. In my view, I think proof 1 is easier to understand.

Proof 1:

Proof 2:

The second proof is the one I don't understand. More specifically, the part that I don't understand is how:

Can someone please carefully explain how those two expressions are equal to each other? I just don't see how those expressions are equal. I can't find any common factors that have been taken out or what logic has been used to rewrite the expression in that way.

Best Answer

I presume you're talking about polynomials with integer coefficients: if the coefficients can be real numbers, the question makes no sense because any nonzero real number can divide any other real number.

Then the point is that $p(x) = q(x) r(x)$ implies $p(0) = q(0) r(0)$, where $p(0)$, $q(0)$ and $r(0)$ are the constant terms of $p, q$ and $r$.

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[Math] Using resultants to check if multivariate polynomials have a common factor – is the proof correct

Your proof seems to work, but it looks incomplete because you need to use more explicitly the fact that the ring of polynomials is an UFD.

The condition that the resultant $\mathrm{Res}(f,g)=0$ means that there are some polynomials $p,q$ such that $\deg p<\deg f$, $\deg q<\deg g$ and $fq=gp$ (so $p/q=f/g$). In a UFD this is equivalent to having a common factor of $f$ and $g$.

To complete your approach you must replace the coefficient ring by its quotient field. You need the unique factorization at the end when you simplify the coefficients to be ring elements again.

For zero divisor free rings, without unique factorization, it will not work. For example, consider the polynomials $x^2-4y^2$ and $x^2+4xy+4y^2$. In some rings, like $(\mathbb{R}[y])[x]$, they have the common factor $x+2y$. In other rings, for example in $(\mathbb{Z}[4y,4y^2])[x]$, they don't. However, the resultants will be the same zero polynomial in all rings. So this method does not work if the coefficients are from the ring $\mathbb{Z}[4y,4y^2]$.

Another possible approach is building a single (but huge) resultant. If $f,g\in\mathbb{R}[x_1,\dots,x_n]$ are nonzero polynomials then take all polynomials of the form $x_1^{a_1}\cdots x_n^{a_n}f$ with $a_1+\cdots+a_n<\deg g$ and $x_1^{b_1}\cdots x_n^{b_n}g$ with $b_1+\cdots+b_n<\deg f$. These polynomials are linearly dependent if and only if there are some polynomials $p,q$ with $\deg p<\deg f$, $\deg q<\deg g$ and $p/q=f/g$.

Sorry, I cannot recommend you books other than I can find by internet search... :-(

[Math] SAGE: Is it possible to extract the irreducible factor of a polynomial for the purpose of constructing a Number Field

I'm not sure what you mean by "the" irreducible factor, since some polynomials will have many irreducible non-linear (I guess you mean this) factors over $\mathbb{Q}$. But if you know exactly what it looks like, you could use this syntax.

sage: F = (x^2+1)*(x^2+2)
sage: F.factor_list().index((x^2+2,1))
0

Don't forget that the symbolic ring factors over the integers; you'll need to use polynomials rings to get what you want, probably. Naturally, you are right that the for i in list: do something loop syntax is what you want to go through the loop, although "list comprehensions" will end up being nicer in the long run.

Also, if you don't know the exact factor, then you could still search for the form of it by doing something

<inside your loop>
if the_factor_im_looking_at[0].degree(x) == 2:
    do something with it...

I'm giving less detail because you asked for that. Good luck! A little bit of basic programming will go a LONG way for you.

Best Answer

Related Solutions

[Math] Using resultants to check if multivariate polynomials have a common factor – is the proof correct

[Math] SAGE: Is it possible to extract the irreducible factor of a polynomial for the purpose of constructing a Number Field

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