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TMUA Polynomials and the Factor Theorem

A polynomial is a sum of non-negative whole-number powers of xx. The specification covers the factor and remainder theorems, which find unknown coefficients and roots quickly.

Paper 1 and Paper 2 topic. Specification codes: MM1.6. Last checked 12 November 2026.

Remainder theorem

When p(x)p(x) is divided by (x−a)(x-a), the remainder is p(a)p(a). When divided by (x−a)(x−b)(x-a)(x-b), the remainder has the form cx+dcx+d.

Factor theorem

(x−a)(x-a) is a factor of p(x)p(x) exactly when p(a)=0p(a)=0. Use this to find a missing coefficient, then divide to find the other factors.

Algebraic division

Divide term by term, or compare coefficients after writing p(x)=(x−a)q(x)+rp(x)=(x-a)q(x)+r.

Roots, graphs and coefficients

A repeated factor such as (x−2)2(x-2)^2 means the graph touches the axis at x=2x=2 instead of crossing it. A cubic with positive leading coefficient runs from bottom left to top right, so it always has at least one real root.

If two polynomials are equal for all xx, match the coefficients of each power of xx. This is often the quickest way to find unknown constants.

Division has the form p(x)=(x−a)q(x)+rp(x)=(x-a)q(x)+r. Setting x=ax=a kills the quotient, which is why the remainder theorem works.

Worked examples

Example 1

Find the remainder when x3−2x2+5x−1x^3-2x^2+5x-1 is divided by x−2x-2.

Show the worked solution

Remainder =p(2)=8−8+10−1=9=p(2)=8-8+10-1=9.

Answer: 9

Example 2

(x+2)(x+2) is a factor of x3+ax2−5x+6x^3+ax^2-5x+6. Find aa and all the roots.

Show the worked solution

p(−2)=−8+4a+10+6=0p(-2)=-8+4a+10+6=0, so a=−2a=-2. Then x3−2x2−5x+6=(x+2)(x2−4x+3)=(x+2)(x−1)(x−3)x^3-2x^2-5x+6=(x+2)(x^2-4x+3)=(x+2)(x-1)(x-3).

Answer: a=−2a=-2; roots −2, 1, 3-2,\,1,\,3

Example 3

A polynomial p(x)p(x) leaves remainder 4 on division by x−1x-1 and remainder 2 on division by x+1x+1. Find the remainder on division by x2−1x^2-1.

Show the worked solution

Write the remainder as ax+bax+b. Then p(1)=a+b=4p(1)=a+b=4 and p(−1)=−a+b=2p(-1)=-a+b=2, so b=3b=3 and a=1a=1.

Answer: x+3x+3

Common mistakes

  • Evaluating p(−a)p(-a) instead of p(a)p(a) for a divisor x−ax-a.
  • Forgetting that dividing by a quadratic leaves a remainder of degree up to 1.
  • Stopping after finding one factor when the question asks for all roots.
  • Sign slips when substituting negative values into odd powers.

Test yourself on this topic with our free practice questions.

Quick answers

What is the factor theorem?

If p(a)=0p(a)=0 then (x−a)(x-a) is a factor of p(x)p(x), and the converse also holds.

How do you find the remainder without long division?

Substitute the root of the divisor into the polynomial. For x−ax-a the remainder is p(a)p(a).

How many roots can a cubic have?

A cubic has at most three real roots, and at least one.

Practise it under timed conditions

A free timed sample paper with a worked solution for every question. No sign-up. For two full papers and a weak-spot drill, see Full Access.

Free sample paper

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