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

Let F be a field. Suppose that a polynomial $$\displaystyle{p}{\left({x}\right)}={a}_{{{0}}}+{a}_{{{1}}}{x}+\cdots+{a}_{{{n}}}{x}^{{{n}}}$$ is reducible in $$\displaystyle{F}{\left[{x}\right]}$$. Prove that the polynomial $$\displaystyle{q}{\left({x}\right)}=={x}^{{{n}}}{p}{\left({\frac{{{1}}}{{{x}}}}\right)}={a}_{{{n}}}+{a}_{{{n}-{1}}}{x}+\cdots{a}_{{{0}}}{x}^{{{n}}}$$ is also reducible.

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Paineow

Step 1
Definition: Let $$\displaystyle{\left({F},\ +,\ \cdot\right)}$$ 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 $$\displaystyle{q}{\left({x}\right)}={x}{\left({a}_{{{0}}}+{\frac{{{a}_{{{1}}}}}{{{x}}}}\right)}={a}_{{{0}}}{x}+{q}_{{{1}}}$$ 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 $$\displaystyle{k}={1}$$
$$\displaystyle{q}{\left({x}\right)}={a}_{{{k}+{1}}}+{q}_{{{k}}}{x}+\cdots{q}_{{{0}}}{x}^{{{k}+{1}}}$$
$$\displaystyle{q}{\left({x}\right)}={a}_{{{k}+{1}}}+{x}{\left({q}_{{{k}}}+\cdots{a}_{{{0}}}{x}^{{{k}}}\right)}$$
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.

###### Not exactly what you’re looking for?
lenkiklisg7
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 $$\displaystyle{d}{e}{g{{\left({q}\right)}}},\ {d}{e}{g{{\left({r}\right)}}}{>}{0}$$ and $$\displaystyle{p}={q}{r}$$
We know that $$\displaystyle{d}{e}{g{{\left({p}\right)}}}={d}{e}{g{{\left({q}\right)}}}+{d}{e}{g{{\left({r}\right)}}}$$, 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
$$\displaystyle{a}_{{{0}}}{x}^{{{n}}}+{a}_{{{1}}}{x}^{{{n}-{1}}}+\cdots+{a}_{{{n}}}=$$
$$\displaystyle{x}^{{{n}}}{p}{\left({\frac{{{1}}}{{{x}}}}\right)}=$$
$$\displaystyle{x}^{{{n}}}{q}{\left({\frac{{{1}}}{{{x}}}}\right)}{r}{\left({\frac{{{1}}}{{{x}}}}\right)}=$$
$$\displaystyle{x}^{{{m}}}{q}{\left({\frac{{{1}}}{{{x}}}}\right)}{x}^{{{n}-{m}}}{r}{\left({\frac{{{1}}}{{{x}}}}\right)}=$$
$$\displaystyle{\left({b}_{{{0}}}{x}^{{{m}}}+{b}_{{{1}}}{x}^{{{m}-{1}}}+\cdots+{b}_{{{m}}}\right)}{\left({c}_{{{0}}}{x}^{{{n}-{m}}}+{c}_{{{1}}}{x}^{{{n}-{m}}}+\cdots+{c}_{{{n}-{m}}}\right)}$$
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.
karton

Substitute $$x\rightarrow\frac{1}{z}$$. 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.