Polynomial Rings, Irreducibility and Unique Factorisation · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.
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Dividing polynomials and factoring uniquely

Mathematics · Abstract Algebra · ages 22-24
Name ______________________   Date ____________

  1. Divide x^2 + 7x + 10 by x + 2. The quotient is ax + b. What is a + b?

    Answer: ______________

  2. For f(x) = x^3 - 2x^2 + 3x - 6, what does division by x - 2 give as its remainder?

    • f(2) = 0; x - 2 is a factor
    • f(2) = 4; x - 2 is not a factor
    • f(0) = -6; x - 2 is not a factor

  3. Divide x^2 + 5x + 3 by x + 1. Which result is correct?

    • Quotient x + 6; remainder 3
    • Quotient x + 4; remainder -1
    • Quotient x + 4; remainder 0

  4. What is the remainder when x^2 - 6x + 11 is divided by x?

    Answer: ______________

  5. Divide x^3 + 2x^2 + x + 4 by x^2 + 1. What constant remainder remains?

    Answer: ______________

  6. In Q[x], the ideal generated by x^2 - 1 and x^2 + 2x + 1 has a monic generator g. What is degree(g)?

    Answer: ______________

  7. While dividing x^2 + 2x + 5 by x - 1, a student subtracts x(x - 1) and gets x + 5. What should that subtraction give?

    • x + 5, because the two x terms cancel
    • 3x + 5, because subtracting -x adds x
    • 2x + 5, because only x^2 is subtracted
    • 3x - 5, because subtraction changes the constant

  8. Which correctly compares x^3 - 5 over Q and over R?

    • It factors over Q but is irreducible over R
    • It is irreducible over both coefficient fields
    • It has the rational root 5 and a real linear factor
    • It is irreducible over Q but has a real linear factor

  9. In a nonzero ideal I of F[x], choose nonzero g of least degree. Why must division of any f in I by g leave remainder 0?

    • A nonzero remainder would contradict minimality
    • Every remainder is a nonzero constant polynomial
    • The remainder lies outside I by the division theorem
    • A field contains only constant polynomials

  10. For f(x) = x^3 + 14x + 7, which prime p satisfies Eisenstein's criterion?

    Answer: ______________

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Answer key

For grown-ups. Fold this page away before handing over the rest.

Dividing polynomials and factoring uniquely W1-mt_m2G9PIHj6y-s1

  1. 6 · Division gives x + 5, so a + b = 1 + 5 = 6.
  2. f(2) = 0; x - 2 is a factor · The Remainder Theorem gives f(2) = 8 - 8 + 6 - 6 = 0.
  3. Quotient x + 4; remainder -1 · Successive subtraction leaves 4x + 3, then -1.
  4. 11 · The divisor is x - 0, so the remainder is f(0) = 11.
  5. 2 · The quotient is x + 2, whose product with the divisor leaves remainder 2.
  6. 1 · Their gcd is x + 1, so the generated ideal has a degree-1 monic generator.
  7. 3x + 5, because subtracting -x adds x · Subtract x^2 - x from x^2 + 2x + 5, producing 3x + 5.
  8. It is irreducible over Q but has a real linear factor · Eisenstein with prime 5 proves rational irreducibility; the real cube root of 5 gives a real linear factor.
  9. A nonzero remainder would contradict minimality · The remainder f - qg belongs to I and, if nonzero, has smaller degree than g.
  10. 7 · Seven divides 0, 14 and 7, does not divide the leading coefficient 1, and its square does not divide 7.
Worksheet · LightMySky