Let F be a field. Suppose that a polynomial p(x)=a_{0}+a_{1}x+\cdots+a_{n}x^{n}

Step 1
Definition: Let ${\displaystyle (F,+,\cdot )}$ be a field and let ${\displaystyle f\in F\left[x\right]}$. Then f is said to be Irreducible over F if f cannot be factored into a product of polynomials all of which having lower degree than f. If f is not irreducible over F then we say that f is Reducible over F.
Step 2
We will prove with the help of principal of mathematical equation.
For ${\displaystyle n=1}$
${\displaystyle P\left(x\right)=a_0+a_{1}x}$ is reducible in field ${\displaystyle \Rightarrow a_0{a}_1\in F\left(x\right)}$ and ${\displaystyle q\left(x\right)=x(a_0+\frac{a_1}{x})=a_{0}x+q_1}$ is also reducible in field as ${\displaystyle a_0{a}_1\in F\left(x\right)}$
Now let for ${\displaystyle n=k}$ is true
${\displaystyle P\left(x\right)=a_0+a_{1}x+\cdots a_k{x}^k}$ is reducible in fidd
So any polynomial of degree dess than k with ${\displaystyle a_0{a}_1}$
${\displaystyle a_k\in F\left(x\right)}$ is reducible
${\displaystyle \Rightarrow q\left(x\right)=a_0{x}^k+a_1{x}^{k-1}\cdots +a_k}$ is reducible in field
Now for ${\displaystyle n=k+1}$
${\displaystyle P\left(x\right)=a_0+a_{1}x+\cdots a_k{x}^k+a_{k+1}{x}^{k+1}}$ in reducible
${\displaystyle \Rightarrow a_0\cdot a_{k+1}\in f\left(x\right)}$ with polynomial of degree less or equal to k is reducible
Step 3
Now ${\displaystyle k=1}$
${\displaystyle q\left(x\right)=a_{k+1}+q_{k}x+\cdots q_0{x}^{k+1}}$
${\displaystyle q\left(x\right)=a_{k+1}+x(q_k+\cdots a_0{x}^k)}$
Since ${\displaystyle a_k+\cdots a_0{x}^k}$ is reducible in f(x) and k+1 degree polinomial with $q_{k+1}\in f(x)$ can also reducible
${\displaystyle \Rightarrow q\left(x\right)}$ is reducible.

Step 1
Suppose ${\displaystyle p\left(x\right)=a_0+a_{1}x+\cdots +a_n{x}^n\in F\left[x\right]}$ is not irreducible, i.e., there is ${\displaystyle q,r\in F\left[x\right]}$ such that ${\displaystyle deg\left(q\right),deg\left(r\right)>0}$ and ${\displaystyle p=qr}$
We know that ${\displaystyle deg\left(p\right)=deg\left(q\right)+deg\left(r\right)}$, so we may assume WLG that
1) ${\displaystyle q\left(x\right)=b_0+b_{1}x+\cdots +b_m{x}^m}$
2) ${\displaystyle r\left(x\right)=c_0+c_{1}x+\cdots +c_{n-m}{x}^{n-m}}$
Step 2
Then
${\displaystyle a_0{x}^n+a_1{x}^{n-1}+\cdots +a_n=}$
${\displaystyle x^{n}p\left(\frac{1}{x}\right)=}$
${\displaystyle x^{n}q\left(\frac{1}{x}\right)r\left(\frac{1}{x}\right)=}$
${\displaystyle x^{m}q\left(\frac{1}{x}\right)x^{n-m}r\left(\frac{1}{x}\right)=}$
${\displaystyle (b_0{x}^m+b_1{x}^{m-1}+\cdots +b_m)(c_0{x}^{n-m}+c_1{x}^{n-m}+\cdots +c_{n-m})}$
Therefore ${\displaystyle a_0{x}^n+a_1{x}^{n-1}+\cdots +a_n}$ is not irreducible.
The implication in the opposite direction is analogous.

Substitute $x\to \frac{1}{z}$. We get
$a_0+a_1\frac{1}{z}+\cdots +a_n(\frac{1}{z}{)}^n$
$=\frac{1}{z^n}(a_0{z}^n+a_1{z}^{n-1}+\cdots +a_n)=\frac{1}{z^n}q(z)$
$p(z)$is irreducible if $q(z)$ is irreducible.