Minimal polynomial of an algebraic number expressed in terms of another algebraic number I am working on the following exercise: Let $\alpha \in \mathbb{C}$ be a root of the polynomial $f(X) = X^4 - 3X - 5$.


*

*Prove that $f$ is irreducible in $\mathbb{Q}[X]$.

*Find the minimal polynomial of $2\alpha - 3$ over $\mathbb{Q}$.

*Find the minimal polynomial of $\alpha^2$ over $\mathbb{Q}$.


Here are my thoughts:
I am okay with question one ($f$ is irreducible of $\mathbb{Z}_2$ and hence over $\mathbb{Q}$) but struggling with the rest of the exercise. I have found this relevant question but fail to apply Gerry’s answer to this example. Could someone give me a hint?
 A: 3.
$(\alpha^2)^2-5=3\alpha$ implies that $((\alpha^2)^2-5)^2=9\alpha^2$.$\alpha^2$ is a root of $(X^2-5)^2-9X=X^4-10X^2-9X+25$ this implies that $g(\alpha^2)=0$ where $g(X)=X^4-10X^2-9X+25$.
Remark that $\alpha={{(\alpha^2)^2-5}\over 3}$ we deduce that $\alpha\in Q(\alpha^2)$ and $Q(\alpha^2)=Q(\alpha)$. This implies that the degree of $[Q(\alpha^2):Q]=4$ and $g$ is the minimal polynomial of $\alpha^2$.
A: 2) Let $\beta = 2 \alpha - 3$.  As user1952009 points out, for suitably chosen $a,b$, you can show that $f(aX+b)$ has $\beta$ as a root.  As a further hint, note that $\alpha = \frac{\beta+3}{2}$.
3) Here's a solution in a similar vein to Gerry Myerson's linked answer.  As a vector space over $\mathbb{Q}$, $\mathbb{Q}(\alpha)$ has $1, \alpha, \alpha^2, \alpha^3$ as a basis.  We compute the matrix of the multiplication map
\begin{align*}
\lambda_{\alpha^2}: \mathbb{Q}(\alpha) &\to \mathbb{Q}(\alpha)\\
x &\mapsto \alpha^2 x
\end{align*}
with respect to this basis, which is
$$
A =
\left(\begin{array}{rrrr}
0 & 0 & 5 & 0 \\
0 & 0 & 3 & 5 \\
1 & 0 & 0 & 3 \\
0 & 1 & 0 & 0
\end{array}\right) \, .
$$
(The fancy name for this is the regular representation.) One can show that $\alpha^2$ is a root of the characteristic polynomial of $A$.  (This is a general fact and there is nothing special about considering $\alpha^2$; see Dummit and Foote, $\S13.2$, Exercise $20$, p. $531$.)  You'll still have to show that this polynomial is irreducible, but actually the same trick you used for 1) should work.
A: 

*put $x=2\alpha-3 \Rightarrow x+3=2\alpha\Rightarrow (x+3)^4=16\alpha^4 \Rightarrow \frac {(x+3)^4}{16}=3\alpha+5; \alpha^4=3\alpha+5 \Rightarrow \frac {(x+3)^4}{16}=3(\frac{x+3}{2}) +5 \Rightarrow \frac {(x+3)^4}{16}-3(\frac{x+3}{2}) -5=0 $ then  $ f(x)= \frac {x^4}{16}+\frac{3x^3}{4}+\frac{27x^2}{8}+\frac{21x}{4}-\frac{71}{16}=0 $ is minimal polynomial of $2\alpha-3$

