Question

By finding quadratic factors show that $\mathit{Q}\mathbf{(}\sqrt{\mathbf{2}}\mathbf{,}\sqrt{\mathbf{3}}\mathbf{)}$ is a splititing field of ${\mathit{x}}^{\mathbf{4}}\mathbf{+}\mathbf{2}{\mathit{x}}^{\mathbf{3}}\mathbf{-}\mathbf{8}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{6}\mathit{x}\mathbf{-}\mathbf{1}$over$\mathit{Q}$ .

Short answer

It is proved that $Q(\sqrt{2},\sqrt{3})$ is splititing field of$x^4+2x^3-8x^2-6x-1$.

Step-by-step solution

Definition of splitting field

A splititing field of a polynomial with coefficient in a smallest field over which the polynomial split or split into linear factors.

Showing thatQ(2,3)   is splitting field of  x4+2x3−8x2−6x−1

$Q(-2\pm \sqrt{3},1\pm \sqrt{2})=Q(\sqrt{2},\sqrt{3})$Let$f\left(x\right)=x^4+2x^3-8x^2-6x-1$

Now find the root of the polynomial$f\left(x\right)=x^4+2x^3-8x^2-6x-1$

$\begin{array}{c}{x}^4+2x^3-8x^2-6x-1=0\\ ((x-2)x-1)((x+4)x+1)=0\\ (x^2-2x-1)(x^2+4x+1)=0\end{array}$

Simplify further,

$\begin{array}{c}(x^2+4x+1)=0,(x^2-2x-1)=0\\ x=-2\pm \sqrt{3},x=1\pm \sqrt{2}\end{array}$

Thus the splititing field of$f\left(x\right)$is$Q(-2\pm \sqrt{3},1\pm \sqrt{2})$. As

Therefore the splititing field of is$Q(\sqrt{2},\sqrt{3})$