= Solution
Use $\mathbb F_{32}=\mathbb F_2[\omega]$ with $\omega^5=\omega^2+1$. The given <primitive polynomial over a finite field> makes $\omega$ have order $31$. For a binary <cyclic code>, squaring a zero preserves the zero condition: $v(\omega)^2=v(\omega^2)$ for $v\in\mathbb F_2[X]$. Thus the <binary cyclotomic cosets modulo an odd integer> governing the four specified zeros are
$$
\boxed{C_1=C_2=C_4=\{1,2,4,8,16\},\qquad C_3=\{3,6,12,24,17\}\pmod{31}.}
$$
The two displayed orbits are disjoint, each has size five, and together contain $1,2,3,4$. Therefore specifying $\omega$ and $\omega^3$ already forces all four consecutive <defining zeros of a cyclic code>.
The <minimal polynomial of an algebraic element> of $\omega$ is $m_1(X)=X^5+X^2+1$. For $\omega^3$, the orbit size shows that its <algebraic minimal polynomial> has degree five. The candidate $m_3(X)=X^5+X^4+X^3+X^2+1$ vanishes there, since
$$
m_3(\omega^3)=\omega^{15}+\omega^{12}+\omega^9+\omega^6+1=0.
$$
For example, the five-bit representations of these terms are $11111,01110,11010,01010,00001$, whose sum is zero in $\mathbb F_2^5$. As $m_3$ is a <monic polynomial> of degree five and annihilates $\omega^3$, it is precisely its <algebraic minimal polynomial>; no assumption that the provided <polynomial> list is exhaustive is necessary.
The narrow-sense primitive <BCH code> is
$$
\mathcal X=\{v\in\mathbb F_2[X],\ \deg v<31:\ v(\omega)=v(\omega^3)=0\}.
$$
Both <algebraic minimal polynomials> divide each such $v$, and they are <coprime polynomials>. Their product is a divisor of $X^{31}-1$. Hence the <generator polynomial> is their product. Direct multiplication with <field characteristic> two gives
$$
\boxed{g(X)=m_1(X)m_3(X)=X^{10}+X^9+X^8+X^6+X^5+X^3+1,\qquad\dim\mathcal X=21.}
$$
The <designed distance> is five, because $g$ vanishes at the four consecutive powers $\omega,\omega^2,\omega^3,\omega^4$.
Here is an explicit <BCH bound> proof for these zeros. If a nonzero <codeword> had <support of a vector> of size $\nu\leq4$, with distinct positions $i_1,\ldots,i_\nu$ and nonzero coefficients $b_1,\ldots,b_\nu$, put $\lambda_r=\omega^{i_r}$. Evaluating at the first $\nu$ defining powers gives
$$
\sum_{r=1}^\nu b_r\lambda_r^j=0\qquad(1\leq j\leq\nu).
$$
The coefficient <matrix> has <determinant> $\bigl(\prod_r\lambda_r\bigr)\prod_{r<s}(\lambda_s-\lambda_r)$, a nonzero <Vandermonde determinant> because the $\lambda_r$ are distinct and nonzero. It therefore forces every $b_r=0$, a contradiction. Thus the <minimum Hamming distance> is at least five.
For the matching upper bound, multiplication gives the actual <codeword>
$$
(1+X^2+X^3)g(X)=1+X^2+X^7+X^8+X^{13},
$$
which has <Hamming weight> five and degree below $31$. Hence
$$
\boxed{\mathcal X\text{ has parameters }[31,21,5]_2.}
$$
This proves the actual distance without needing the optional theorem. Its numerical hypothesis is also satisfied: with $s=5$ and the error parameter $l=2$, $2^{sl}=2^{10}=1024$, whereas $\sum_{i=0}^{3}\binom{31}{i}=1+31+465+4495=4992$.
Now calculate the <root-evaluation syndrome for a cyclic code> from the received <polynomial>. Reduction modulo $m_1(X)$ gives $u(X)\equiv X^3$, so $S_1=u(\omega)=\omega^3$. Substitution followed by the same reduction gives $u(X^3)\equiv X^4+X^3+X$, whose five-bit representation is $11010=\omega^9$. Thus
$$
\boxed{S_1=\omega^3,\qquad S_3=u(\omega^3)=\omega^9=S_1^3.}
$$
The <single-error test from two binary BCH syndromes> identifies coordinate $3$, since a one-bit error $X^j$ has <syndromes> $\omega^j,\omega^{3j}$. Flip that coordinate to obtain
$$
\boxed{c_{\mathrm{correct}}(X)=u(X)+X^3=X^{12}+X^{11}+X^9+X^7+X^6+X^3+X^2+1.}
$$
Both defining evaluations vanish: $c_{\mathrm{correct}}(\omega)=\omega^3+\omega^3=0$ and $c_{\mathrm{correct}}(\omega^3)=\omega^9+\omega^9=0$. More directly,
$$
c_{\mathrm{correct}}(X)=(1+X^2)g(X),
$$
so it is certainly a <codeword> of $\mathcal X$. It is at <Hamming distance> one from $u$. Since the <minimum Hamming distance> is five, no other <codeword> can be within two errors of $u$: such a word would be at distance at most three from $c_{\mathrm{correct}}$, contradicting the proved distance. Thus this is the unique <bounded-distance decoding> under the <linear code>'s two-error guarantee.
\b[The decoded <polynomial> printed in the original PDF is incorrect.] It instead adds $X^{10}$ to $u$, giving a <polynomial> $c_{\mathrm{printed}}=u+X^{10}$. Its defining evaluations are
$$
c_{\mathrm{printed}}(\omega)=\omega^3+\omega^{10}=\omega^{25}\ne0,\qquad
c_{\mathrm{printed}}(\omega^3)=\omega^9+\omega^{30}=\omega^3\ne0.
$$
Therefore it is not a <codeword>, and the request to verify that it belongs to $\mathcal X$ is a false premise. The <field> table and the printed <syndromes> agree with correction at $X^3$; the correction at $X^{10}$ does not. The boxed corrected <polynomial> resolves this source error and supplies the intended decoding and membership proof.
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