Fields and Galois theory

a. Let $m=2$. By Eisenstein criterion for $p=2$, since the polynomial is monic, $2$ divides every other coefficient, and $2^2 = 4$ does not divide the constant term, the polynomial is irreducible over $\mathbb{Q}$.

b. Since $\alpha$ is a root of a polynomial $f(x)$ of degree $3$, and the splitting field has degree $6$, then the polynomial does not completely factor in $\mathbb{Q}(\alpha)$ (which is a degree $3$ extension), so we can simply apply Euclidean division and get \[ f(x) = (x-\alpha)(x^2 +(\alpha-4)x + \alpha^2 - 4\alpha). \] In fact, multiplying we have \[ (x-\alpha)(x^2 +(\alpha-4)x + \alpha^2 - 4\alpha) = x^3 - \alpha x^2 + (\alpha-4) x^2 - \alpha(\alpha-4) x + \alpha^2 x - \alpha^3 - 4 \alpha x - 4 \alpha^2 \\ = x^3 - \alpha x^2 + \alpha x^2 -4 x^2 - \alpha^2 x + 4 \alpha x + \alpha^2 x - \alpha^3 - 4 \alpha x + 4 \alpha^2 \\ = x^3 -4 x^2 - \alpha^3 + 4 \alpha^2 \\ x^3 - 4x^2 - 6 \]
since $\alpha$ is a root of $x^3 - 4x^2 - 6$, so $\alpha^3 - 4\alpha^2 = 6$.

c. We have that $\alpha^3 = 4\alpha^2 + 6$, so \[ \alpha^5 = \alpha^2 \cdot \alpha^3 \\ = \alpha^2 (4 \alpha^2 + 6) \\ = 4 \alpha^4 + 6\alpha^2 \\ = 4\alpha(4\alpha^2 + 6) + 6 \alpha^2 \\ = 16 \alpha^3 + 6 \alpha^2 + 24 \alpha \\ = 16(4\alpha^2 + 6) + 6 \alpha^2 + 24 \alpha \\ = 70 \alpha^2 + 24 \alpha + 96. \]