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

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 $h$? 
