Discussion
Let \(p(x) = 9x^{5} + 10x^{3} + 5x + 15\) and \(q(x) = x^{3} - x^{2} - x - 2\) be two polynomials in \(\mathbb{Q}[x]\). Then, over \(\mathbb{Q}\),
Answer: C
\(p(x)\) is irreducible but \(q(x)\) is reducible.
\(p(x) = 9x^{5} + 10x^{3} + 5x + 15\) satisfies Eisenstein's criterion at the prime 5: 5 divides 10, 0 (the other coefficients), 5 and 15; it does not divide the leading coefficient 9; and \(5^{2} = 25\) does not divide the constant 15. So \(p\) is irreducible over \(\mathbb{Q}\).
\(q(2) = 8 - 4 - 2 - 2 = 0\), so \(x - 2\) is a factor: \(q(x) = (x - 2)(x^{2} + x + 1)\), which is reducible.
That rules out the options calling \(q\) irreducible or \(p\) reducible.
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