Question

Which of these polynomials are irreducible in $\mathrm{\mathbb{Q}}\left[x\right]$:

(b)${\mathit{x}}^{\mathbf{4}}\mathbf{+}\mathit{x}\mathbf{+}\mathbf{1}$0

Short answer

The given polynomial is irreducible in$\mathrm{\mathbb{Q}}\left[x\right]$ .

Step-by-step solution

Step 1:To factorization of the polynomial as a product of two quadratics

Let us see thedefinitions of Reducible polynomial and Irreducible polynomial,

A nonconstant polynomial $f\left(x\right)\in F\left[x\right]$, where F is a field, is said to be irreducible if its only divisors are its associates and the nonzero constant polynomials(units).

A nonconstant polynomial that is not irreducible is called reducible polynomial.

Let $f\left(x\right)=x^4+x+1$be the given polynomial.

We claim that $f\left(x\right)$is irreducible in .$\mathrm{\mathbb{Q}}\left[x\right]$

By contradiction, let if possible, $f\left(x\right)$is reducible, then it can be factored as the product of two nonconstant polynomials in $\mathrm{\mathbb{Q}}\left[x\right]$.

If either of these factors have degree 1, then polynomial $f\left(x\right)$has a root in$\mathrm{\mathbb{Q}}\left[x\right]$ .

But by the Rational Root Test,$f\left(x\right)$ has no roots in$\mathrm{\mathbb{Q}}\left[x\right]$ .

The only possibilities are $\pm 1$, and neither is a root.

Thus, by Theorem 4.2, $f\left(x\right)$if is reducible, then it can be factored as the product of two quadratics.

Therefore, the given polynomial can be factored as follows:

$\begin{array}{rcl}f\left(x\right)& =& x^4+x+1=\left(x^2+ax+b\right)\left(x^2+cx+d\right)\\ & \Rightarrow & x^4+0x^3+0x^2+x+1=x^4+c{x}^3+x^{2}d+a{x}^3+ac{x}^2+adx+b{x}^2+bcx+bd\\ & \Rightarrow & x^4+0x^3+0x^2+x+1=x^4+\left(a+c\right)x^3+\left(d+ac+b\right)x^2+\left(ad+bc\right)x+bd \cdots \cdots \left(A\right)\\ & & \end{array}$

To show fxis irreducible in ℚx

From $\left(A\right)$in step 1, equal polynomials have equal coefficients, therefore, on comparing coefficients, we get,

$$\begin{gathered} a+c=0 \cdots \cdots \left(1\right), \\ ac+b+d=0 \cdots \cdots \left(2\right), \\ ad+bc=1 \cdots \cdots \left(3\right), \\ bd=1 \cdots \cdots \left(4\right) \end{gathered}$$

Then from (1), $c=-a$.

From (4) , $b=d=\pm 1$.

From(2) , $a^2=\pm 2\Rightarrow a=\pm \sqrt{2}\notin$.

Hence, $f\left(x\right)$is irreducible in $\mathrm{\mathbb{Q}}\left[x\right]$.