Divide polynomials and verify quotient and remainder
Align missing powers, divide by a linear or quadratic expression and rebuild the original polynomial as a check.
Exam scope and assessment notes
Pure Mathematics 2 (Paper 2, AS only) and Pure Mathematics 3 (Paper 3, A Level) · Shared Algebra foundations, Mathematics 9709, 2026–2027. Paper 1 knowledge is assumed. These modulus and polynomial requirements occur in both sections 2.1 and 3.1. Paper 3 additionally requires partial fractions and rational-power binomial expansions; follow its separate lessons for those methods. The AS route using Papers 1 + 2 cannot be carried forward to A Level. Both sections use linear modulus graphs; nonlinear y = |f(x)| and y = f(|x|) graphs are excluded. Polynomial dividends have degree at most 4 and divisors are linear or quadratic. Written practice is self-checked.
Before you begin
- Expand and collect polynomial expressions, including missing powers.
- Divide leading terms and subtract signed coefficients accurately.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which aligned form matches x⁴ − 2x³ + 3x − 2 for long division?
What you will learn
- Divide a polynomial of degree at most 4 by a linear or quadratic polynomial.
- Identify a lower-degree remainder and verify f = divisor × quotient + remainder.
Repeat leading-term division, multiplication and subtraction
Arrange the dividend and divisor in descending powers, inserting zero coefficients for missing terms. Divide the leading remaining term by the divisor's leading term to find the next quotient term, multiply the whole divisor by that term, then subtract the whole product. Repeat until the remainder has degree less than the divisor. A linear divisor leaves a constant remainder; a quadratic divisor can leave a linear remainder.
Divide x⁴ − 2x³ + 0x² + 3x − 2 by x² + 1. The leading quotient term x² removes x⁴ + x², leaving −2x³ − x² + 3x − 2. The term −2x removes −2x³ − 2x, leaving −x² + 5x − 2. The term −1 removes −x² − 1, leaving 5x − 1. Thus the quotient is x² − 2x − 1 and the remainder is 5x − 1.
The rebuilt identity checks every sign and coefficient
The result must satisfy f(x) = D(x)Q(x) + R(x) as a polynomial identity. For the quartic example, (x² + 1)(x² − 2x − 1) + (5x − 1) expands to x⁴ − 2x³ + 3x − 2. This check is valid even at roots of D because it involves multiplication, not division. A zero remainder means D is an exact factor; a remainder of too high a degree means division has not finished.
A non-monic divisor may produce fractional quotient coefficients. For example dividing 2x⁴ − x³ + 3x² + 4 by 2x² + 1 gives Q = x² − x/2 + 1 and R = x/2 + 3. Keep these coefficients exact. When writing f/D = Q + R/D, exclude inputs where D = 0; cancellation of a zero-remainder factor does not restore an input that was undefined in the original rational expression.
Pause and explain
For division by x² + 1, a working remainder still contains 2x². What should happen?
Worked example
Divide x⁴ − 2x³ + 3x − 2 by x² + 1, giving the quotient, remainder and a rebuilt check.
Show the worked solution
- Insert the missing 0x² term. The first quotient term x² leaves −2x³ − x² + 3x − 2 after subtraction.
- The term −2x leaves −x² + 5x − 2; the term −1 then leaves 5x − 1.
- Rebuild: (x² + 1)(x² − 2x − 1) + (5x − 1) = x⁴ − 2x³ + 3x − 2.
Answer Quotient x² − 2x − 1; remainder 5x − 1.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Divide x³ + 2x² − 5x + 1 by x − 1 and verify the result.
Give me a hint
Find one quotient term at a time, then expand divisor × quotient + remainder.
Compare my reasoning
- The first quotient term x² leaves 3x² − 5x + 1; the next term 3x leaves −2x + 1.
- The final quotient term −2 leaves remainder −1, giving Q = x² + 3x − 2.
- (x − 1)(x² + 3x − 2) − 1 expands to x³ + 2x² − 5x + 1.
Quotient x² + 3x − 2; remainder −1.
Look for these in your work
- Subtract each full product accurately.
- Reconstruct every coefficient of the original polynomial.
Divide 2x⁴ − x³ + 3x² + 4 by 2x² + 1, keeping exact coefficients.
Give me a hint
Insert a zero x coefficient and allow a fractional quotient term when the leading division requires it.
Compare my reasoning
- The term x² leaves −x³ + 2x² + 0x + 4.
- The term −x/2 leaves 2x² + x/2 + 4; the term 1 leaves x/2 + 3.
- The result is Q = x² − x/2 + 1, R = x/2 + 3; rebuilding with 2x² + 1 recovers the dividend.
Quotient x² − x/2 + 1; remainder x/2 + 3.
Look for these in your work
- Retain the divisor's leading coefficient 2.
- Keep fractional coefficients and a lower-degree remainder.
Simplify R(x) = (x³ − 4x² + x + 6)/(x − 2), state its original domain exclusion and evaluate R(3).
Give me a hint
A zero-remainder polynomial identity can simplify the rule while the original denominator still restricts the domain.
Compare my reasoning
- Division gives x³ − 4x² + x + 6 = (x − 2)(x² − 2x − 3) with zero remainder.
- Hence R(x) = x² − 2x − 3 for x ≠ 2; the original fraction is undefined at 2 even though the simplified polynomial is defined there.
- At x = 3, R(3) = 9 − 6 − 3 = 0, an allowed input.
R(x) = x² − 2x − 3 for x ≠ 2; R(3) = 0.
Look for these in your work
- Separate a polynomial identity from division at a zero denominator.
- Retain the original exclusion after cancellation.
Common mistakes
- Include a zero column for every missing power.
- Subtract the entire divisor product, preserving its signs.
- A remainder has degree less than the divisor, and a cancelled rational expression keeps its original exclusions.
What can you explain now?
What degree can a non-zero remainder have when dividing by a quadratic?
Compare with the explanation
At most 1: linear or constant.
The remainder degree must be less than 2; a remaining quadratic term requires another division step.
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Make it stick
Explain the repeatable division steps, the remainder-degree rule and the reconstructed identity. State why cancellation does not restore an excluded rational input.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.