Question

Problem 4. Consider f(x) = x⁵+ x⁴ + 2x³ + 3x² + 4x + 5 ∈ Q[x] and our goal is to determine if f is irreducible over Q. We compute f(1), f(−1), f(5), f(−5) directly and see that none of them is zero. By the Rational Roots Theorem, f has no root in Q. So if f is reducible over Q, it cannot be factored into the product of a linear polynomial and a quartic polynomial (i.e. polynomial of degree 4). By the useful consequence of Gauss’ Lemma (i.e. Theorem 6 in Chapter 4.2), if f is reducible, it has factorization in Z[x]. So the only possibility is: f(x) = (x² + ax + b)(x³ + cx² + dx + e) with a, b, c, d, e ∈ Z. Comparing the coefficients of x ⁴ , x ³ ,... of both sides, we have the system of equations: (∗) c + a = 1, d + ac + b = 2, e + ad + bc = 3, ae + bd = 4, be = 5.

Now the last equation be = 5 yields 4 cases: Case 1: b = 5 and e = 1. The remaining equations become:

(1) c + a = 1 .(2) d + ac + 5 = 2 (3) 1 + ad + 5c = 3   (4) a + 5d = 4. The first and last equation allow us to get a and c in terms of d, namely a = 4 − 5d and c = 1 − a = 1 − (4 − 5d) = −3 + 5d. Then we “plug in” the second equation to get: d + (4 − 5d)(−3 + 5d) + 5 = 2. Simplify this and we get the quadratic equation: 25x² − 36x + 9 = 0. We get 2 roots (18 ± 3 √ 11)/25 neither of which is an integer (actually neither is a rational number since √ 11 $\notin$ ( Q). Note that even if we had an integer solution here, we would still need to find a and c, make sure they are integers, then check that the third equation holds (we have not touched the third equation before). Anyway, this case does not yield any integer solutions a, b, c, d, e of the original system (∗). Case 2: b = 1 and e = 5. We don’t have any integer solutions in this case either. Case 3: b = −1 and e = −5. No solution either. Here is the question for you. Case 4: b = −5 and e = −1. Provide all the details as in Case 1, check if you can find any integer solutions of the system (∗), then conclude whether f is irreducible over Q or not.

Answer 1

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