Question

Show that each polynomial is irreducible in, as in Example 3.

(a) ${\mathbf{x}}^{\mathbf{4}}\mathbf{+}\mathbf{2}{\mathbf{x}}^{\mathbf{3}}\mathbf{+}\mathbf{x}\mathbf{+}\mathbf{1}$

(b)${\mathbf{x}}^{\mathbf{4}}\mathbf{-}\mathbf{2}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{8}\mathbf{x}\mathbf{+}\mathbf{1}$

Short answer

b) It is proved $x^4-2x^2+8x+1$is irreducible in $\mathrm{\mathbb{Q}}\left[x\right]$.

Step-by-step solution

Statement of Theorem 4.23

Theorem 4.23states consider$f\left(x\right)$with integer coefficients. Therefore,$f\left(x\right)$factors as a product of polynomials of degrees $m$and $n$in $\mathrm{\mathbb{Q}}\left[x\right]$and only if $f\left(x\right)$factors as a product of polynomials with degrees $m$and $n$in $\mathrm{\mathbb{Z}}\left[x\right]$.

Show that the polynomial is irreducible in

b)

When $x^4-2x^2+8x+1$contains a linear factor, it shall have a certain rational root, and the only possible root is $\pm 1$according to the rational root test. The fact that neither is a root can be easily verified.

As a result, when this polynomial contains a factorization, it will be into quadratic polynomials. Assume that there are such factors $x^2+a_{1}x+a_0$ and$x^2+b_{1}x+b_0$ with integer coefficients according to theorem 4.23. Then,

$$\begin{gathered} \left\{a_1+b_1=0 \\ {a}_0+b_0+a_1{b}_1=-2 \\ {a}_0{b}_1+a_1{b}_0=8 \\ {a}_0{b}_0=1\right. \end{gathered}$$

It can be deduced$a_0=b_0=1$ or$a_0=b_0=-1$ from the previous equation.

There is $b_1=-a_1$from the first. The third equation yields that role="math" localid="1649755400589" $0=8$, which is a contradiction.

It concludes $x^4-2x^2+8x+1$is irreducible.

Hence, it is proved $x^4-2x^2+8x+1$is irreducible in $\mathrm{\mathbb{Q}}\left[x\right]$.