Polynomials path · A09

Before this lesson: Remainder and factor theorems, Prime numbers, Modular arithmetic and congruences, How to write a proof

Your goal: Prove irreducibility with a valid criterion over a specified field.

Start with the idea and one example. Take a break before the written task if you need it. The questions are original teaching exercises, not official past-paper questions.

Irreducible polynomials: the key idea

Reducibility depends on the coefficient field. The polynomial x²−2 is irreducible over the rationals but factors over the reals. A degree-two or degree-three polynomial over a field is reducible exactly when it has a root in that field; this shortcut fails in degree four. For integer coefficients, the rational-root theorem restricts a reduced root u/v: u divides the constant coefficient and v divides the leading coefficient. Eisenstein’s criterion gives irreducibility over the rationals when one prime divides every coefficient except the leading one, does not divide that leading coefficient, and its square does not divide the constant. Reduction modulo a prime can also prove irreducibility when the degree is preserved. These criteria require their hypotheses; a failed test proves nothing.

A worked example

Prove x³−2 is irreducible over the rationals.

A reducible cubic over the rationals would have a rational root. Since it is monic, any rational root is an integer dividing 2, hence one of ±1,±2. None has cube 2. Therefore there is no rational root, and the cubic is irreducible. This degree-three argument cannot simply be used for a quartic.

Your turn: change one thing

Why does “x⁴+3x²+2 has no rational root” not prove irreducibility?

Try this on paper before opening the explanation.

Compare your reasoning

It factors as (x²+1)(x²+2), though neither quadratic has a rational root. A quartic can split into two quadratics without a linear factor.

Pause and check

A trap to avoid: No rational root does not imply irreducibility for degree four or higher.

Practise and adjust the level

Foundation checks the language; Core applies the method; Stretch asks you to choose or justify an idea. These are levels within this lesson. A session selects six of the nine questions; unused questions allow the level to change. Advanced theory still needs written practice.

Interactive practice loads here. You can also use the complete question set below.

Write a complete argument

Prove that a reducible cubic over a field has a root in that field.

Planning hint

List the assumptions and the exact conclusion. Identify the definition or theorem in this lesson that connects them. Explain why its conditions hold before using it.

Read the full solution after your attempt

A nontrivial factorisation splits degree 3 into positive degrees, necessarily 1 and 2 (or three linear factors). Thus a linear factor ax+b occurs with a≠0. Its root −b/a belongs to the field and is a root of the cubic. Conversely, a root gives a linear factor by the factor theorem.

My proof notebook

Write on paper, or keep a draft here. Compare your reasoning with the solution only after a real attempt. The checklist is your own review, not an automatic mark.

All nine practice questions

Prefer paper or have JavaScript switched off? The complete question set, hints and solutions are here. Interactive practice uses these same questions in an order chosen from your answers.

1. Over which field does x²−2 factor as (x−2)(x+2)(x-\sqrt{2})(x+\sqrt{2})?

Foundation

  1. The rationals using rational coefficients
  2. The integers using integer coefficients
  3. No field
  4. The real numbers
Hint

The proposed factors contain 2\sqrt{2}.

Answer and reasoning

The real numbers. Their coefficients are real, but 2\sqrt{2} is not rational.

2. For a quadratic over the rationals, irreducibility is equivalent to what?

Foundation

  1. Having no rational root
  2. Having no integer coefficient
  3. Being monic
  4. Having a positive constant
Hint

A nontrivial split must use linear factors.

Answer and reasoning

Having no rational root. A rational root produces a rational linear factor, and a split quadratic has such a root.

3. Which list contains every possible rational root of x³−2?

Core

  1. All real cube roots
  2. ±1, ±2
  3. 0,1,2
  4. ±1/2 only
Hint

Apply the rational-root theorem to a monic polynomial.

Answer and reasoning

±1, ±2. A rational root must be an integer dividing the constant −2.

4. Which prime proves x⁵+10x+5 irreducible by Eisenstein?

Core

  1. 7
  2. 5
  3. 2
  4. 3
Hint

Check all nonleading coefficients and the square condition.

Answer and reasoning

5. 5 divides 10,5 and the zero intermediate coefficients, does not divide 1, and 25 does not divide 5.

5. Which example refutes the degree-four no-rational-root shortcut?

Stretch

  1. x³−1
  2. (x²+1)(x²+2)
  3. x−1
  4. x²−1
Hint

Seek a product of quadratics with no rational roots.

Answer and reasoning

(x²+1)(x²+2). This quartic is reducible by construction, but neither factor vanishes at a rational number.

6. Eisenstein fails for your polynomial with p=2. What follows?

Stretch

  1. The polynomial is irreducible
  2. The degree is even
  3. That test did not decide irreducibility
  4. The polynomial is reducible
Hint

A sufficient criterion need not be necessary.

Answer and reasoning

That test did not decide irreducibility. Failure of its hypotheses supplies no conclusion; another prime or method may work.

7. Irreducibility is defined relative to what?

Foundation

  1. Only the degree
  2. The drawing of the graph
  3. Only positive inputs
  4. A specified coefficient field or ring
Hint

Factorisations depend on allowed coefficients.

Answer and reasoning

A specified coefficient field or ring. x²−2 is irreducible over ℚ but factors over ℝ.

8. Is x³−2 irreducible over ℚ?

Core

  1. No, every cubic factors over ℚ
  2. Only if 2 is negative
  3. Yes
  4. No, because it has a real root
Hint

Use Eisenstein at 2 or the rational-root test.

Answer and reasoning

Yes. For degree three, absence of rational roots proves irreducibility; ±1,±2 are not roots.

9. Does having no rational roots prove every quartic irreducible over ℚ?

Stretch

  1. Yes always
  2. Only for monic quartics
  3. Only for even quartics
  4. No
Hint

A quartic can split into two quadratics.

Answer and reasoning

No. For example (x²+1)(x²+2) has no rational roots but is reducible.

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