Discussion
The number of linear factors of the polynomial \(X^{16} + 1\) in R[X] is :
Answer: A
\(X^{16} + 1\) has 0 linear factors in R[X].
A linear factor \(X - c\) with \(c\) real would mean \(c^{16} + 1 = 0\). But \(c^{16} \ge 0\) for every real \(c\), so \(c^{16} + 1 \ge 1 \gt 0\). The polynomial has no real roots and hence no real linear factors.
Its 16 complex roots, the 32nd roots of unity that are not 16th roots of unity, come in conjugate pairs, so over R it splits into 8 irreducible quadratics and nothing else. That rules out 1, 2 and 4.
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