Exercise on polynomials

1. Factorize the following polynomials over ℂ, ℝ, ℚ, ℤ₂

   x⁵ + 2x⁴ −5x³ − 10x² + 6x + 12
   x⁵ + 2x⁴ −5x³ − 2x² − 2x + 12
   x⁵ + 3x⁴ −x³ − 3x² − 2x −6

2. For which values of a ∈ ℤ the polynomial 3x³ + 20ax² + 50a²x + 60 is irreducible over ℂ, ℝ, ℚ.

3. Check whether x⁵ + 7x⁴ +2x³ + 6x² −x + 8 is irreducible over ℚ.

4. Show that x³ + 4x² + 6x + 2 is irreducible in ℚ[x].

5. Show that x⁴ + 2x³ + 2x2 − x +1 is irreducible in ℚ[x].

6. Factorize over ℤ₅ following polynomials

   x⁵ +x⁴ +x³ + x+ 1
   x⁴ + 2x + 3
   x⁶ + 4x + 1
   x⁴ − 1
   x⁴ +1

Solutions

• A root is clearly −2, so a factor is (x+2). Dividing the polynomial by this factor we obtain x⁴ −5x² +6. The substitution y = x² yields

  y² −5y +6

  which has roots 3 and 2. So the factorization over ℚ is (x+2)(x²−2)(x²−3)

  Over ℂ and ℝ, (x+2)(x+√2)(x–√2)(x+√3)(x–√3).

  In ℤ₂: (x+2)(x²− 2)(x² − 3) ⟶ (x)(x²)(x² + 1), since 2 = 0 mod 2 and −3 = 1 mod 2, so

  x³ (x² +1), and as x² +1 factors as: (x+1)(x+1) we obtain x³ (x+1)² ■.

• −2 is clearly a root so a factor is (x+2). Dividing by this factor x⁴ − x² − 2 is obtained. The substitution y = x² yields

  y² −y −2

  which has roots 2 and −1. So the factorization over ℚ and over ℝ is (x + 2)(x² − 2)(x² + 1)

  Over ℂ (x+2)(x+√2)(x–√2)(x+√i)(x–√i).

  In ℤ₂ since 2 = 0 mod 2

  x³ (x² + 1) and as x² + 1 factors as: (x+1)(x+1) we obtain x³ (x+1)² ■.

• A root is clearly −3, so a factor is (x + 3). Dividing the polynomial by this factor we obtain x⁴ −x² −2. The substitution y = x² yields

  y² −y −2

  which has roots 2 and −1. So the factorization over ℚ and over ℝ is (x + 3)(x² − 2)(x² + 1)

  Over ℝ, (x + 3)(x + √2)(x – √2)(x² + 1).

  Over ℂ since x² = −1, you just factorize the x²+1 into (x − i)(x + i)(x + 3)(x + √2)(x – √2)(x² + 1). ■

1. Every cubic polynomial is reducible over both ℂ and ℝ. With regards to ℚ, it is possible to apply Eisenstein, p = 5 and notice that it is irreducilbe over ℚ, whatever a is chosen. ■

2. If we reduce modulo 2, we get x⁵ + x⁴ + x ∈ ℤ/2ℤ[x], which is NOT irreducible. So we try with p = 3 and get x⁵ + x⁴ +2x³ +2x +2. So if it factorazies it does so as cubic ⋅ quadratic, both irreducible ((they have no roots). After some calculation is found that it's not possible, so f is irreducible in ℚ[x]. ■

3. We could use the rational roots theorem to show that that f has no root in ℚ. However, reducing modulo 3 is much easier. Then f̄ = x³ + x² + 2, which clearly has no root in ℤ₂[x]. ■

4. Reduction modulo 2 gives f̄ = x⁴ + x + 1, which clearly has no root in ℤ₂. It could factor into two quadratics. These must be irreducible (they have no roots) so they must be both equal to x² + x + 1. But (x² + x + 1)² = x⁴ + x² + 1 ≠ f̄, Hence f̄ is irreducible in ℤ/2ℤ[x], ando f is irreducible in ℚ[x]. ■

5. • We test the five elements. A root is 1 which is congruent to −4, so a factor is (x + 4); another root is 3 which is congruent to −2 so a factor is (x + 2) and another is, dividing x⁵ +x⁴ + x³ + x +1 by (x + 4)⋅ (x + 2) = x² + 2x + 4x +8 ≡ x² +6x +8 we find x³ +3x +2.

   • A root is 3, confruent to −2 so a factor is (x + 2); dividing x⁴ + 2x +3 by (x + 2) we find x³ −2x² +4x −6. If the cubic factor further splits, then it must have a linear factor itself; 3 is a root and the factor (x +2) is found. Dividing the cubic polynomial by x +2 we find x² + x + 2 is irreducible. So x⁴ + 2x +3 (x² + x +2) (x + 2)².

   • x⁴ − 1 = (x + 2)(x + 3) (x + 1) (x + 4);

   • x⁴ + 1 = (x² + a) (x² + ab) = x⁴ +bx² +ax² +ab = x⁴ (b+a)x² + ab. So it follows that a = 2 and b = 3 and a factorization is x⁴ +1 = (x² + 2) (x² + 3).