Polynomials path · A02

Before this lesson: Polynomial functions, Algebraic identities and factorisation

Your goal: Write and check P = DQ + R with the correct remainder degree.

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.

Polynomial Long Division: the key idea

Polynomial division writes P(x)=D(x)Q(x)+R(x), where D is nonzero and either R=0 or deg R<deg D. Work from the leading term: choose the next quotient term to cancel the current highest power, subtract the whole product, and continue. Missing powers can be written with zero coefficients to keep columns aligned. Division is performed over a specified coefficient field, usually the reals or rationals here. Always multiply back to check the identity. A quadratic divisor can leave a linear remainder; a linear divisor leaves a constant.

A worked example

Divide x³−1 by x−1.

The successive quotient terms are x², x and 1. Multiplying back gives (x−1)(x²+x+1)=x³−1, so the quotient is x²+x+1 and the remainder is 0. This identity is valid even at x=1, although the corresponding cancelled fraction is not defined there.

Your turn: change one thing

Write x²+1 as a multiple of x−2 plus a remainder.

Try this on paper before opening the explanation.

Compare your reasoning

x²+1=(x−2)(x+2)+5. The quotient is x+2 and the constant remainder is 5.

Pause and check

A trap to avoid: Forgetting deg R < deg D or failing to multiply back.

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

Why are the quotient and remainder unique over the reals?

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

Suppose P=DQ₁+R₁=DQ₂+R₂ with both remainder degrees less than deg D. Then D(Q₁−Q₂)=R₂−R₁. If Q₁−Q₂ were nonzero, the left side would have degree at least deg D, while the right side has smaller degree or is zero. This is impossible. Thus Q₁=Q₂ and then R₁=R₂.

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. What is the largest possible degree of a nonzero remainder after division by a cubic?

Foundation

  1. 2
  2. 0
  3. 3
  4. 4
Hint

The remainder degree is strictly smaller.

Answer and reasoning

2. A cubic has degree 3, so a remainder has degree at most 2.

2. Which identity checks quotient Q and remainder R?

Foundation

  1. P=Q/R
  2. P=DQ+R
  3. P=D+Q+R
  4. P=DR+Q always
Hint

Recall the division identity.

Answer and reasoning

P=DQ+R. Dividend equals divisor times quotient plus remainder.

3. What is the remainder when x²+1 is divided by x−2?

Core

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

Evaluate the division identity at x=2.

Answer and reasoning

5. The multiple of x−2 vanishes, leaving 2²+1=5.

4. Divide x²−9 by x−3. What is the quotient?

Core

  1. x−3
  2. x²+3
  3. 3x
  4. x+3
Hint

Use a difference of squares.

Answer and reasoning

x+3. x²−9=(x−3)(x+3), so the quotient is x+3.

5. A remainder after division by x²+1 is written ax²+bx+c. What must be true?

Stretch

  1. a=0
  2. b=0
  3. c=0
  4. a=1
Hint

Compare degrees.

Answer and reasoning

a=0. A nonzero x² term would make the remainder degree 2, which is not smaller than the divisor’s degree.

6. Why can two valid remainders for the same division not differ?

Stretch

  1. All remainders are positive
  2. Every polynomial has one root
  3. The leading coefficient is always 1
  4. Their difference would be a multiple of D of smaller degree
Hint

Subtract the two division identities.

Answer and reasoning

Their difference would be a multiple of D of smaller degree. A nonzero multiple of D cannot have degree below deg D. The difference must therefore be zero.

7. In P=DQ+R, the divisor D must be what?

Foundation

  1. Always linear
  2. Always monic
  3. A nonzero polynomial
  4. The zero polynomial
Hint

Polynomial division cannot use a zero divisor.

Answer and reasoning

A nonzero polynomial. The algorithm requires D≠0.

8. Divide x²+1 by x−1: what is the remainder?

Core

  1. 1
  2. −1
  3. 2
  4. 0
Hint

Evaluate at x=1.

Answer and reasoning

2. P(1)=2; indeed x²+1=(x−1)(x+1)+2.

9. Why must deg R<deg D in the usual division result?

Stretch

  1. Otherwise another leading term can be removed
  2. So R must always be zero
  3. So Q is linear
  4. Because degrees are negative
Hint

The remainder is what is left after reduction stops.

Answer and reasoning

Otherwise another leading term can be removed. A remainder at least as large in degree as D is not fully reduced.

Choose your next step

Continue to Remainder and factor theorems. If this felt difficult, return to a prerequisite above. Every lesson stays open.

Open my revision list →

Original teaching material · IMOolympiad.com. Send a specific correction through our contact page. Learning progress is optional and stays in this browser.