Polynomials: irreducibility $\iff$ no zeros in F.

Question

Given is the polynomial $f \in F[x]$, with $deg(f)=3$. I have to prove, that f is irreducible iff $f$ has no zeros in $F$.

"$\Rightarrow$": let's prove the contrapositive: "if $f$ has zeros in $F$, then $f$ is not irreducible."

If $f$ has zeros in $F$, it means there's at least one zero and we can represent $f$ as product of two polynomials: $f = (ax + b) * q(x)$, where $deg(q)=2$. But for a polynomial to be irreducible, the only way to factor it into two polynomials is in a way that one of them is of degree $0$, i.e. it's not irreducible.

There are gaps in the prove which follows from some blind spots in my understanding of the material. Please, help.

QUESTIONS:

1) If a polynomial of degree 3 has a zero point, can I say, we can represent it in one of the following ways?

$f=(ax+b)q(x)\\ f= f = (ax+b)(cx+d)g(x)\\ f = (ax+b)(cx+d)(gx+h)$

What about polynomials of type $ax^3 + b$? I can also say that the zero point is $\sqrt[3]{-\frac ba}$. Which one should I use?

2) I saw people writing $(x - a)$ when they factor a polynomial or talk about zero points. If which cases I should use $(x-a)$ and $(ax+b)$ notations? Should I have used $(x-a_1)$ notaion in 1)

3) In my proof I do nothing with the fact that $deg(f) = 3$. I guess, this fact is given for some reason. Could you tell me please, what the reason is?

Sure, some people will find these question stupid and the answers straightforward, but I need your help to put every piece of information about polynomials on right places in my mind.

Answer 1

1) If $f(a)=0$, then all you know is that $f(x)=(\square x+\square)g(x)$ for some $g(x)$. You do not know from this if $g(x)$ can be factored further, so your other expressions are not generally valid. You also need to be able to justify how you know that $f(x)$ has a linear factor from the existence of a root.

It is true though that $f(x)$ will factor in one of two ways: a linear factor times a quadratic irreducible factor, or three linear factors. (Note that the second two factorizations in your list are actually the same, because $\deg f=3$.) You want to stick with the first factorization for your proof, because (a) it suffices and (b) it comes directly out of the application of the factor theorem.

2) The linear factor $x-a$ is technically of the form $\square x+\square$, but writing $x-a$ emphasizes the fact that you are using the factor theorem in your proof, and that you are using a zero (namely $a$) in order to construct the linear factor. For these reasons, writing $x-a$ is definitely better.

3) You have only done one half of the proof. You showed that if $f$ has a zero then it is reducible, and this fact actually does not depend on $\deg f=3$, as it's true for all polynomials. Now you need to show the other direction, that if $f$ is reducible, then it has a zero. Here you will use $\deg f=3$: if this is the case when $f$ is reducible, then what are the possible degrees of $f$'s factors?

Note that the originally problem asked you to show that $f$ is irreducible iff it has no zeros, and this is equivalent to showing $f$ is reducible iff it has a zero. This is because $\rm A\Leftrightarrow B$ and $\rm\neg A\Leftrightarrow\neg B$ are logically equivalent propositions (split the iff into two implications then use contraposition).

Answer 2

If $f(x)$ is a polynomial and $g(x)$ is any nonzero polynomial, you can always write $f(x)=q(x)g(x)+r(x)$ with polynomials $q(x)$, $r(x)$, $\deg r<\deg g$ (polynomial division with remainder). Especially, if you let $g(x)=x-a$ be a linear polynomial,you obtain $f(x)=q(x)\cdot(x-a)+r$ where $r$ is constant. If $a$ happens to be a root of $f$, observe that from $f(a)=q(a)\cdot 0+r$ you obtain $r=0$, i.e. $f(x)=q(x)(x-a)$. So if $\deg f>1$ and $f(a)=0$ for some $a\in F$, $f$ cannot be irreducible.

The property that $\deg f=3$ is needed for the other direction of the proof (For example $x^4-5x^2+6$ is reducible in $\mathbb Q[x]$ but has no root in $\mathbb Q$). Assume $f$ is reducible, say $f(x)=g(x)h(x)$ for some nonconstant polynomials $g,h$. What can you say about the degrees of $g$ and $h$?