The remainder theorem

The remainder theorem says that when a polynomial f(x) is divided by (x − a), the remainder is f(a). So to find a remainder you do not need to divide at all: just substitute x = a into the polynomial. If the remainder is 0, then (x − a) is a factor, which is the factor theorem.

The remainder theorem

• Dividing by (x − 3)? The remainder is f(3).
• Dividing by (x + 2)? Write it as (x − (−2)), so the remainder is f(−2).
• Dividing by (2x − 1)? x = 1/2 makes the divisor zero, so substitute 1/2: the remainder is f(1/2).

In general, dividing by (ax − b) leaves remainder f(b/a): substitute the value of x that makes the divisor zero.

Why the remainder theorem works

Dividing f(x) by (x − a) gives a quotient q(x) and a remainder r. Because the divisor is linear, the remainder is just a number. So:

This is true for every x, so put x = a. The bracket (x − a) becomes 0, which wipes out the whole first term:

So the remainder r equals f(a). That is the whole proof.

The factor theorem

If f(a) = 0, the remainder is 0, so (x − a) divides f(x) exactly: it is a factor. And if (x − a) is a factor, then f(a) = 0. AQA A-level Maths names the factor theorem and algebraic division by linear expressions; Cambridge International AS & A Level (Papers 2 and 3) names both the factor theorem and the remainder theorem, including divisors such as (ax + b) and calculating remainders.

Worked examples

Examples 1 to 3 use f(x) = x³ − 4x² + 2x + 5.

Example 1: dividing by (x − 3)

• Remainder = f(3) = 27 − 36 + 6 + 5 = 2.
• Check by division: f(x) = (x − 3)(x² − x − 1) + 2. Expanding the right-hand side gives x³ − 4x² + 2x + 5 back, with the 2 left over.

Example 2: dividing by (x + 2)

• Remainder = f(−2) = −8 − 16 − 4 + 5 = −23. Take care with the signs: (−2)³ = −8 and −4 × (−2)² = −16.

Example 3: dividing by (2x − 1)

• Substitute x = 1/2, the value that makes 2x − 1 zero: f(1/2) = 1/8 − 1 + 1 + 5 = 41/8, or 5.125.

Check remainders like these with the Polynomial Division Calculator, which divides by long or synthetic division and shows the remainder.

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Example 4: one unknown coefficient

When 2x³ + kx² − 5x + 4 is divided by (x − 2), the remainder is 14. Find k.

1. By the remainder theorem, f(2) = 14.
2. f(2) = 2(8) + k(4) − 5(2) + 4 = 16 + 4k − 10 + 4 = 4k + 10.
3. 4k + 10 = 14, so k = 1.

Example 5: two unknowns

f(x) = x³ + ax² + bx − 12. (x − 3) is a factor of f(x), and dividing f(x) by (x − 1) leaves a remainder of −12. Find a and b.

1. Factor theorem: f(3) = 0, so 27 + 9a + 3b − 12 = 0, which simplifies to 3a + b = −5.
2. Remainder theorem: f(1) = −12, so 1 + a + b − 12 = −12, which simplifies to a + b = −1.
3. Subtract the second equation from the first: 2a = −4, so a = −2 and b = 1.
4. Check: f(x) = x³ − 2x² + x − 12, and f(3) = 27 − 18 + 3 − 12 = 0.

Two conditions give two simultaneous equations. Practise solving a pair in this question from the Algebra course.

Example 6: factorising a cubic

Show that (x − 2) is a factor of g(x) = x³ − 2x² − 9x + 18, and factorise g(x) fully.

1. g(2) = 8 − 8 − 18 + 18 = 0, so (x − 2) is a factor.
2. Divide, or compare coefficients: g(x) = (x − 2)(x² − 9).
3. x² − 9 is a difference of two squares: g(x) = (x − 2)(x − 3)(x + 3).

Dividing by a quadratic

The remainder theorem is about linear divisors. When you divide by a quadratic, the remainder can be linear, px + q, so you need two equations. The same idea still works: write f(x) = (divisor) × q(x) + px + q and substitute the values of x that make the divisor zero.

Example 7: remainder on dividing by (x − 1)(x − 2)

Find the remainder when f(x) = x⁴ + x + 1 is divided by (x − 1)(x − 2).

1. Write f(x) = (x − 1)(x − 2)q(x) + px + q.
2. x = 1: f(1) = 3, so p + q = 3.
3. x = 2: f(2) = 16 + 2 + 1 = 19, so 2p + q = 19.
4. Subtract: p = 16, then q = −13. The remainder is 16x − 13.

Common mistakes

• Using the wrong sign. For (x + 2), substitute x = −2, not x = 2.
• Forgetting to divide for (ax − b). For (3x − 2), substitute x = 2/3, not x = 2.
• Sign slips with negative powers. (−2)³ = −8 but (−2)² = +4; use brackets on the calculator.
• Thinking the remainder theorem gives the quotient. It only gives the remainder; use division for the quotient.
• Missing terms. If a power of x is missing, its coefficient is 0; it matters in division, though not when you substitute.

Practise with a short timed paper

Polynomials, the factor theorem and the remainder theorem come up in the daily A-level Maths: Algebra & Functions paper, a short timed paper of 8 questions with a new one every day. For a fast way to divide, see the synthetic division guide, and for the basics, polynomials.

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