Polynomial Division and the Factor Theorem · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

Polynomial Division and the Factor Theorem

Mathematics · Algebra · ages 16-17
Name ______________________   Date ____________

  1. (x + 3) is a factor of a cubic f(x). What does that tell you?

    • f(3) = 0
    • f(-3) = 0
    • f(0) = 3
    • f(-3) = 3

  2. x³ + 5x² - 2x + 8 is divided by (x + 1). What is the first term of the quotient?

    • x³
    • x
    • x²
    • 1

  3. Let f(x) = x³ + 2x² - 5x - 6. What is f(2)?

    Answer: ______________

  4. For f(x) = x³ - 4x² + x + 6 you find that f(2) = 0. Which expression is a factor of f(x)?

    • (x - 2)
    • (x + 2)
    • (x - 6)
    • (2x - 1)

  5. Lena says (x + 2) is a factor of x³ - 2x² - 5x + 6. Is Lena right?

    Circle one:   True   False

  6. Given that (x - 1) is a factor of x³ - 2x² - 5x + 6, what is the largest root of the cubic?

    Answer: ______________

  7. Divide x³ + x² - 10x + 8 by (x - 2). What is the quotient?

    • x² - 3x - 4
    • x² + 3x + 4
    • x² + 3x - 4
    • x² - 10x + 8

  8. When f(x) is divided by (x - 5) the remainder is 12. What is f(5)?

    • 0
    • 12
    • 5
    • -12

  9. Which of these is a factor of f(x) = x³ - 2x² - 11x + 12?

    • (x - 3)
    • (x + 1)
    • (x + 4)
    • (x + 3)

  10. Sam writes x³ + 2x² - 9x - 18 = (x + 2)(x² - 9) and says the factorisation is finished. Where is the slip?

    • x² - 9 still factorises, into (x - 3)(x + 3)
    • the first bracket should have been (x - 2)
    • a cubic can only ever have two brackets
    • he should have divided by (x + 3) instead
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Answer key

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

Polynomial Division and the Factor Theorem W1-mt_haL2UkGa5C-s1

  1. f(-3) = 0 · The bracket (x + 3) is zero when x = -3, so -3 is the root and f(-3) = 0.
  2. x² · The first step divides the leading term of the polynomial by the leading term of the divisor: x³ divided by x is x².
  3. 0 · f(2) = 8 + 8 - 10 - 6 = 0, so (x - 2) is a factor.
  4. (x - 2) · A root of 2 gives the factor (x - 2), because that bracket is zero exactly when x is 2.
  5. True · f(-2) = -8 - 8 + 10 + 6 = 0, so (x + 2) is indeed a factor.
  6. 3 · The cubic factorises as (x - 1)(x - 3)(x + 2), so the roots are 1, 3, -2 and the largest is 3.
  7. x² + 3x - 4 · x²(x - 2) leaves 3x², then 3x(x - 2) leaves -4x, and -4(x - 2) clears the rest. The quotient is x² + 3x - 4.
  8. 12 · The remainder theorem works both ways: the remainder on dividing by (x - 5) is f(5), so f(5) = 12.
  9. (x + 3) · Testing the root each bracket names: f(3) = -12, f(-1) = 20, f(-4) = -40, and f(-3) = -27 - 18 + 33 + 12 = 0. Only -3 works.
  10. x² - 9 still factorises, into (x - 3)(x + 3) · The division is right, but the quotient still factorises. A pair multiplying to -9 and adding to 0 is -3 and 3, so x² - 9 is (x - 3)(x + 3) and the cubic has three brackets.
Worksheet · LightMySky