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

Sapewa

Sapewa

Answered question

2021-12-30

Let F be a field. Suppose that a polynomial p(x)=a0+a1x++anxn is reducible in F[x]. Prove that the polynomial q(x)==xnp(1x)=an+an1x+a0xn is also reducible.

Answer & Explanation

Paineow

Paineow

Beginner2021-12-31Added 30 answers

Step 1
Definition: Let (F, +, ) be a field and let fF[x]. 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 n=1
P(x)=a0+a1x is reducible in field a0a1F(x) and q(x)=x(a0+a1x)=a0x+q1 is also reducible in field as a0a1F(x)
Now let for n=k is true
P(x)=a0+a1x+akxk is reducible in fidd
So any polynomial of degree dess than k with a0a1
akF(x) is reducible
q(x)=a0xk+a1xk1+ak is reducible in field
Now for n=k+1
P(x)=a0+a1x+akxk+ak+1xk+1 in reducible
a0ak+1f(x) with polynomial of degree less or equal to k is reducible
Step 3
Now k=1
q(x)=ak+1+qkx+q0xk+1
q(x)=ak+1+x(qk+a0xk)
Since ak+a0xk is reducible in f(x) and k+1 degree polinomial with qk+1f(x) can also reducible
q(x) is reducible.

lenkiklisg7

lenkiklisg7

Beginner2022-01-01Added 29 answers

Step 1
Suppose p(x)=a0+a1x++anxnF[x] is not irreducible, i.e., there is q, rF[x] such that deg(q), deg(r)>0 and p=qr
We know that deg(p)=deg(q)+deg(r), so we may assume WLG that
1) q(x)=b0+b1x++bmxm
2) r(x)=c0+c1x++cnmxnm
Step 2
Then
a0xn+a1xn1++an=
xnp(1x)=
xnq(1x)r(1x)=
xmq(1x)xnmr(1x)=
(b0xm+b1xm1++bm)(c0xnm+c1xnm++cnm)
Therefore a0xn+a1xn1++an is not irreducible.
The implication in the opposite direction is analogous.
karton

karton

Expert2022-01-09Added 613 answers

Substitute x1z. We get
a0+a11z++an(1z)n
=1zn(a0zn+a1zn1++an)=1znq(z)
p(z)is irreducible if q(z) is irreducible.

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