[proofplan] We show $\operatorname{Gal}(p/\mathbb{Q}) \cong S_5$ and then invoke the fact that $S_5$ is not solvable. The identification proceeds in three stages. First, we apply [Eisenstein's Criterion](/theorems/859) at the prime $2$ to establish that $p(t) = t^5 - 4t + 2$ is irreducible over $\mathbb{Q}$. Second, we perform a calculus analysis of $p'(t) = 5t^4 - 4$ to show that $p$ has exactly three real roots and two non-real roots (a complex conjugate pair). These two facts verify the hypotheses of the [Recognition Criterion for $S_p$](/theorems/1291): an irreducible polynomial of prime degree $5$ over $\mathbb{Q}$ with exactly $5 - 2 = 3$ real roots has Galois group $S_5$. Since $S_5$ is not solvable (it contains the simple non-abelian group $A_5$), the polynomial $t^5 - 4t + 2$ is not solvable by radicals. [/proofplan] [step:Apply Eisenstein's Criterion at $p = 2$ to show $t^5 - 4t + 2$ is irreducible over $\mathbb{Q}$] Write $p(t) = t^5 + 0 \cdot t^4 + 0 \cdot t^3 + 0 \cdot t^2 + (-4)t + 2$, so the coefficients are $a_5 = 1$, $a_4 = 0$, $a_3 = 0$, $a_2 = 0$, $a_1 = -4$, $a_0 = 2$. We verify the three conditions of [Eisenstein's Criterion](/theorems/859) with the prime $q = 2$: 1. **$q \nmid a_5$:** The leading coefficient is $a_5 = 1$, and $2 \nmid 1$. 2. **$q \mid a_i$ for all $0 \le i \le 4$:** We have $2 \mid 0$, $2 \mid 0$, $2 \mid 0$, $2 \mid (-4)$, and $2 \mid 2$. 3. **$q^2 \nmid a_0$:** The constant term is $a_0 = 2$, and $4 \nmid 2$. All three conditions are satisfied. By [Eisenstein's Criterion](/theorems/859), $p(t) = t^5 - 4t + 2$ is irreducible over $\mathbb{Q}$. [guided] Eisenstein's Criterion provides a sufficient condition for a polynomial in $\mathbb{Z}[t]$ to be irreducible over $\mathbb{Q}$. Given a prime $q$, the criterion requires: (i) $q$ does not divide the leading coefficient, (ii) $q$ divides every other coefficient, and (iii) $q^2$ does not divide the constant term. For $p(t) = t^5 - 4t + 2$, the prime $q = 2$ is the natural candidate because the non-leading coefficients are $0, 0, 0, -4, 2$, all divisible by $2$. The leading coefficient $1$ is not divisible by $2$, and the constant term $2$ is not divisible by $2^2 = 4$. So all three Eisenstein conditions are satisfied. Why does Eisenstein's Criterion work? The proof (see [Eisenstein's Criterion](/theorems/859)) proceeds by contradiction: assuming a nontrivial factorisation $p(t) = g(t)h(t)$ in $\mathbb{Z}[t]$ (which suffices by Gauss's Lemma), one reduces modulo $q$. The conditions force $\bar{g}(t) = t^a$ and $\bar{h}(t) = t^b$ in $\mathbb{F}_q[t]$ (where $a + b = 5$), so $q$ divides the constant terms of both $g$ and $h$, giving $q^2 \mid g(0)h(0) = p(0) = 2$, i.e., $4 \mid 2$, a contradiction. Note that irreducibility of $p(t)$ also establishes separability, since $\operatorname{char}(\mathbb{Q}) = 0$ and every irreducible polynomial over a field of characteristic zero is separable (its formal derivative is nonzero and has strictly smaller degree, so $\gcd(p, p') = 1$). [/guided] [/step] [step:Analyse the critical points of $p(t)$ to determine the number of real roots] The derivative of $p(t) = t^5 - 4t + 2$ is \begin{align*} p'(t) = 5t^4 - 4. \end{align*} Setting $p'(t) = 0$ gives $t^4 = 4/5$, i.e., $t^2 = \sqrt{4/5} = 2/\sqrt{5}$. Since $t^4 = 4/5 > 0$, the real solutions are $t = \pm (4/5)^{1/4}$. There are exactly two real critical points: \begin{align*} t_- &:= -(4/5)^{1/4} < 0, \\ t_+ &:= (4/5)^{1/4} > 0. \end{align*} Since $p''(t) = 20t^3$, we have $p''(t_-) = 20 t_-^3 < 0$ (so $t_-$ is a local maximum) and $p''(t_+) = 20 t_+^3 > 0$ (so $t_+$ is a local minimum). We evaluate the sign of $p$ at the critical points. At the local maximum $t_-$: \begin{align*} p(t_-) &= t_-^5 - 4t_- + 2 = t_-(t_-^4 - 4) + 2 = t_-(4/5 - 4) + 2 = t_- \cdot (-16/5) + 2. \end{align*} Since $t_- < 0$, the product $t_- \cdot (-16/5) > 0$, so $p(t_-) > 2 > 0$. At the local minimum $t_+$: \begin{align*} p(t_+) &= t_+^5 - 4t_+ + 2 = t_+(t_+^4 - 4) + 2 = t_+ \cdot (-16/5) + 2. \end{align*} Since $t_+ = (4/5)^{1/4} > 0$, we have $t_+ \cdot (-16/5) < 0$. We bound this more precisely: $(4/5)^{1/4} > (0.8)^{1/4}$. Since $0.8 > 0.6^4/(0.6^4) \ldots$ we compute directly. We have $(4/5)^{1/4} = (0.8)^{0.25}$. Since $0.9^4 = 0.6561 < 0.8$ and $1^4 = 1 > 0.8$, we get $0.9 < t_+ < 1$. Then \begin{align*} p(t_+) &= t_+ \cdot (-16/5) + 2 < 1 \cdot (-16/5) + 2 = -16/5 + 2 = -6/5 < 0. \end{align*} Therefore $p(t_-) > 0$ and $p(t_+) < 0$. [guided] The goal of this step is to determine the exact number of real roots of $p(t)$. We use basic calculus: find the critical points, determine whether they are maxima or minima, and evaluate the sign of $p$ at each. **Finding critical points.** Setting $p'(t) = 5t^4 - 4 = 0$ yields $t^4 = 4/5$. Since $t^4$ is an even function, the real solutions come in a $\pm$ pair: $t = \pm (4/5)^{1/4}$. There are no other real critical points (the equation $t^4 = 4/5$ has exactly two real solutions because $t \mapsto t^4$ is strictly decreasing on $(-\infty, 0)$ and strictly increasing on $(0, \infty)$). **Classifying critical points.** The second derivative $p''(t) = 20t^3$ is negative at $t_- < 0$ and positive at $t_+ > 0$, so $t_-$ is a local maximum and $t_+$ is a local minimum. Since these are the only critical points of $p$ (a degree-$5$ polynomial with positive leading coefficient), $p$ is increasing on $(-\infty, t_-)$, decreasing on $(t_-, t_+)$, and increasing on $(t_+, \infty)$. **Sign analysis at critical points.** At the local maximum: $p(t_-) = t_-(t_-^4 - 4) + 2 = t_- \cdot (-16/5) + 2$. Since $t_- < 0$, the term $t_-(-16/5) = -16t_-/5 > 0$, so $p(t_-) > 2 > 0$. At the local minimum: $p(t_+) = t_+ \cdot (-16/5) + 2$. Since $t_+ > 0$, the term $-16t_+/5 < 0$. We need to check that the negative contribution outweighs $+2$. Since $0.9^4 = 0.6561 < 0.8 = 4/5$ and $1^4 = 1 > 4/5$, we have $t_+ \in (0.9, 1)$. Using the lower bound $t_+ > 0.9$: \begin{align*} p(t_+) < -\frac{16 \cdot 0.9}{5} + 2 = -\frac{14.4}{5} + 2 = -2.88 + 2 = -0.88 < 0. \end{align*} (In fact, even the crude bound $t_+ < 1$ gives $p(t_+) < -16/5 + 2 = -6/5 < 0$.) So $p(t_+) < 0$. Why do we care about the signs at the critical points? Because $p$ is a degree-$5$ polynomial with positive leading coefficient, $p(t) \to -\infty$ as $t \to -\infty$ and $p(t) \to +\infty$ as $t \to +\infty$. Combined with the shape of $p$ (increasing, then decreasing, then increasing) and the signs at the critical values, the number of real roots is determined by the Intermediate Value Theorem in the next step. [/guided] [/step] [step:Conclude that $p$ has exactly three real roots and two non-real roots] Since $\deg p = 5$ and the leading coefficient is positive, we have $p(t) \to -\infty$ as $t \to -\infty$ and $p(t) \to +\infty$ as $t \to +\infty$. The function $p$ is continuous, has exactly two critical points $t_- < t_+$, and satisfies: - $p$ is strictly increasing on $(-\infty, t_-)$, from $-\infty$ to $p(t_-) > 0$. - $p$ is strictly decreasing on $(t_-, t_+)$, from $p(t_-) > 0$ to $p(t_+) < 0$. - $p$ is strictly increasing on $(t_+, \infty)$, from $p(t_+) < 0$ to $+\infty$. By the Intermediate Value Theorem, $p$ has exactly one real root in each of the three intervals where it changes sign: 1. One root in $(-\infty, t_-)$, where $p$ rises from $-\infty$ to $p(t_-) > 0$. 2. One root in $(t_-, t_+)$, where $p$ falls from $p(t_-) > 0$ to $p(t_+) < 0$. 3. One root in $(t_+, \infty)$, where $p$ rises from $p(t_+) < 0$ to $+\infty$. On each of these intervals $p$ is strictly monotone, so each contains exactly one real root. Since $p$ has degree $5$, there are $5$ roots in $\mathbb{C}$ (counted with multiplicity). We have identified $3$ real roots, so the remaining $2$ roots are non-real. Since $p$ has real coefficients, non-real roots occur in conjugate pairs, confirming that the $2$ non-real roots form a single conjugate pair. [guided] This step assembles the information from the critical-point analysis into a root count. The key structural fact is that a degree-$5$ polynomial with positive leading coefficient has the end behaviour $p(t) \to -\infty$ as $t \to -\infty$ and $p(t) \to +\infty$ as $t \to +\infty$. Between these extremes, $p$ has one local maximum at $t_-$ and one local minimum at $t_+$. The sign pattern is: \begin{align*} p(-\infty) &= -\infty, & p(t_-) &> 0, & p(t_+) &< 0, & p(+\infty) &= +\infty. \end{align*} Each sign change forces at least one real root (by the Intermediate Value Theorem), and the strict monotonicity on each subinterval forces at most one root per interval. So there are exactly three real roots. Since $p$ has degree $5$, the Fundamental Theorem of Algebra gives $5$ roots in $\mathbb{C}$ (counted with multiplicity). We showed in the first step that $p$ is irreducible over $\mathbb{Q}$, hence separable (characteristic zero), so all $5$ roots are distinct. Three of them are real, so $5 - 3 = 2$ are non-real. Since $p \in \mathbb{R}[t]$, non-real roots come in conjugate pairs, and indeed $2$ roots form exactly one conjugate pair. Could the polynomial have fewer than three real roots? Only if $p(t_-) \le 0$ (the local maximum does not rise above the axis) or $p(t_+) \ge 0$ (the local minimum does not dip below the axis). In either case, the number of sign changes would decrease, reducing the number of real roots to $1$. Our explicit computation ruled this out. [/guided] [/step] [step:Apply the Recognition Criterion for $S_p$ to identify $\operatorname{Gal}(p/\mathbb{Q}) \cong S_5$] We have established: 1. $p(t) = t^5 - 4t + 2 \in \mathbb{Q}[t]$ is irreducible over $\mathbb{Q}$ (by [Eisenstein's Criterion](/theorems/859) at the prime $2$). 2. $\deg p = 5$, which is prime. 3. $p$ has exactly $3$ real roots and $2$ non-real roots (i.e., exactly $5 - 2 = 3$ real roots). These are precisely the hypotheses of the [Recognition Criterion for $S_p$](/theorems/1291): if $f \in \mathbb{Q}[t]$ is an irreducible polynomial of prime degree $p$ with exactly $p - 2$ real roots, then $\operatorname{Gal}(f/\mathbb{Q}) \cong S_p$. Applying the criterion with $p = 5$, we conclude \begin{align*} \operatorname{Gal}(t^5 - 4t + 2 / \mathbb{Q}) \cong S_5. \end{align*} [guided] The [Recognition Criterion for $S_p$](/theorems/1291) synthesises two pieces of information about an irreducible polynomial $f$ of prime degree $p$ over $\mathbb{Q}$: - **Irreducibility** ensures the Galois group acts transitively on the $p$ roots, so (via the orbit-stabiliser theorem and Cauchy's theorem) $\operatorname{Gal}(f/\mathbb{Q})$ contains an element of order $p$. Since $p$ is prime, such an element must be a $p$-cycle in $S_p$. - **Exactly $p - 2$ real roots** means there is one conjugate pair of non-real roots. Complex conjugation restricts to a $\mathbb{Q}$-automorphism of the splitting field that transposes the two non-real roots and fixes the $p - 2$ real roots, giving a transposition in $\operatorname{Gal}(f/\mathbb{Q})$. A subgroup of $S_p$ ($p$ prime) containing both a $p$-cycle and a transposition must equal $S_p$: conjugation by powers of the $p$-cycle shifts the transposition through all positions, generating all adjacent transpositions, and these generate $S_p$. We verified all hypotheses: $p(t)$ is irreducible (Eisenstein at $2$), has prime degree $5$, and has exactly $3 = 5 - 2$ real roots. Therefore $\operatorname{Gal}(p/\mathbb{Q}) \cong S_5$. [/guided] [/step] [step:Conclude that $t^5 - 4t + 2$ is not solvable by radicals] The group $S_5$ is not solvable. To see this, recall that the alternating group $A_5$ is the unique normal subgroup of $S_5$ (apart from $\{e\}$ and $S_5$ itself). A group is solvable if and only if its composition factors are all cyclic of prime order. The composition series of $S_5$ is \begin{align*} \{e\} \trianglelefteq A_5 \trianglelefteq S_5, \end{align*} with composition factors $A_5 / \{e\} \cong A_5$ and $S_5 / A_5 \cong \mathbb{Z}/2\mathbb{Z}$. The factor $\mathbb{Z}/2\mathbb{Z}$ is abelian, but $A_5$ is a non-abelian simple group (it has order $60$ and no normal subgroups other than $\{e\}$ and $A_5$). In particular, $A_5$ is not cyclic of prime order, so $S_5$ is not solvable. By the Galois-theoretic criterion for solvability by radicals, a polynomial $f \in \mathbb{Q}[t]$ is solvable by radicals over $\mathbb{Q}$ only if $\operatorname{Gal}(f/\mathbb{Q})$ is a solvable group. Since $\operatorname{Gal}(t^5 - 4t + 2 / \mathbb{Q}) \cong S_5$ and $S_5$ is not solvable, the polynomial $t^5 - 4t + 2$ is not solvable by radicals over $\mathbb{Q}$. [guided] This final step connects the Galois group computation to the original question about solvability by radicals. The bridge is one of the central theorems of Galois theory: a polynomial $f \in \mathbb{Q}[t]$ is solvable by radicals if and only if its Galois group $\operatorname{Gal}(f/\mathbb{Q})$ is a solvable group. **Why is $S_5$ not solvable?** A group $G$ is solvable if there exists a chain of subgroups \begin{align*} \{e\} = G_0 \trianglelefteq G_1 \trianglelefteq \cdots \trianglelefteq G_r = G \end{align*} with each quotient $G_{i+1}/G_i$ abelian. For $S_5$, the only normal subgroups are $\{e\}$, $A_5$, and $S_5$ itself. Any subnormal series must therefore pass through $A_5$, and the quotient $A_5/\{e\} \cong A_5$ must appear as one of the factors. But $A_5$ is not abelian: it is a non-abelian group of order $60$. In fact, $A_5$ is *simple* (it has no nontrivial proper normal subgroups), so $A_5$ cannot be broken down further into abelian pieces. This is the obstruction to solvability. Why does this fail for $S_4$ but not $S_5$? The group $A_4$ is *not* simple: it has a normal subgroup, the Klein four-group $V_4 = \{e, (12)(34), (13)(24), (14)(23)\}$, giving the composition series $\{e\} \trianglelefteq V_4 \trianglelefteq A_4 \trianglelefteq S_4$ with abelian factors $V_4 \cong (\mathbb{Z}/2\mathbb{Z})^2$, $A_4/V_4 \cong \mathbb{Z}/3\mathbb{Z}$, and $S_4/A_4 \cong \mathbb{Z}/2\mathbb{Z}$. So $S_4$ is solvable, and every degree-$4$ polynomial is solvable by radicals (the quartic formula). The passage from degree $4$ to degree $5$ is precisely where simplicity of $A_n$ (for $n \ge 5$) prevents solvability. Since $\operatorname{Gal}(t^5 - 4t + 2/\mathbb{Q}) \cong S_5$ is not solvable, the Galois-theoretic criterion yields the conclusion: $t^5 - 4t + 2$ is not solvable by radicals over $\mathbb{Q}$. [/guided] [/step]