Factor \[ 11x^5-11x^4+14x^2-21x+7 \] into irreducible polynomials in \(\QQ[x]\).
Solution
Let \[ f(x) \coloneqq 11x^5-11x^4+14x^2-21x+7. \]
1. The polynomial \(x-1\) divides \(f(x)\), and\[ f(x)=(x-1)(11x^4+14x-7). \]
One has \[ f(1) = 11-11+14-21+7 = 0, \] so the factor theorem gives the factor \(x-1\). Direct multiplication gives \[ \begin{aligned} (x-1)(11x^4+14x-7) &= 11x^5-11x^4+14x^2-21x+7\\ &= f(x). \end{aligned} \]
2. The quartic\[ 11x^4+14x-7 \]is irreducible in \(\QQ[x]\).
Apply Eisenstein’s criterion with the prime \(7\). The prime \(7\) does not divide the leading coefficient \(11\); it divides each remaining coefficient \[ 0,\qquad 0,\qquad 14,\qquad -7; \] and \(7^2\) does not divide the constant coefficient \(-7\). Hence the quartic is irreducible over \(\QQ\).
3. Therefore the factorization into irreducibles in \(\QQ[x]\) is\[ \boxed{ 11x^5-11x^4+14x^2-21x+7 = (x-1)(11x^4+14x-7) }. \]
The factor \(x-1\) is irreducible because it has degree \(1\), and step 2 proves that the quartic factor is irreducible.
4. Q.E.D.
Step 3 is the requested factorization.