Question

Prove that ${\mathbf{x}}^{\mathbf{4}}\mathbf{-}\mathbf{16}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{4}$ is irreducible in $\mathbf{Q}\left[x\right]$.

Short answer

${\mathrm{x}}^4-16{\mathrm{x}}^2+4$ is irreducible.

Step-by-step solution

Definition of irreducible polynomial.

Irreducible polynomial is a non constant polynomial that cannot be factored into the product of two non constant polynomial. The property of irreducible depends on the field or the ring to which the coefficients to belong.

Showing that x4-16x2+4 is irreducible.

Let$f\left(x\right)={\mathrm{x}}^4-16{\mathrm{x}}^2+4$

Let if possible f(x) is reducible in Q[x].

Gauss lemma mentioned below,

If a polynomial with integer coefficient is reducible over Q , then it is reducible over Z.

Therefore it must factor as a product of two quadratic polynomial in Z(x).

Since the coefficient of $x^3$is zero, this factorization must be of the form.

${\mathrm{x}}^4-16{\mathrm{x}}^2+4=\left(x^2+ax+b\right)\left(x^2-ax+c\right)$

Where

$bc=4,a\left(c-b\right)=0,b+c+16=a^2$

If a=0then $b+c+16=0$ . Which is impossible b,cfor because bc=4 .

c-b=0

c=b, bc=4

If

b=c

=2

So,$a^2=20,a^2=12$

Both are impossible as a is an interger. Therefore $f\left(x\right)$ is irreducible in $Q\left[x\right]$.