Remainder Theorem: Formula, Rules, Tricks and Examples

Remainder Theorem gives the remainder obtained when a polynomial is divided by a linear expression. For division by x − a, substitute x = a in the polynomial. This page covers the theorem formula, division by ax + b, factor-related rules, shortcut methods and solved numerical examples.

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What Is the Remainder Theorem?

If a polynomial P(x) is divided by x − a, the remainder is P(a). Thus, the remainder can be found by direct substitution without performing polynomial division.

For a polynomial P(x), division by x − a can be written as P(x) = (x − a)Q(x) + P(a), where Q(x) is the quotient. Since the remainder has degree less than the divisor and x − a has degree 1, the remainder is a constant. For example, when P(x) = x² + 3x + 5 is divided by x − 2, the remainder is P(2) = 4 + 6 + 5 = 15.

Remainder Theorem Formula & Tricks

Important Formulas

Division by x − a
Remainder = P(a)

Substitute a for x in the polynomial when the divisor is x − a.

Division by x + a
Remainder = P(−a)

Rewrite x + a as x − (−a), then substitute −a in the polynomial.

Division by ax + b
Remainder = P(−b/a), a ≠ 0

Set the linear divisor equal to zero and substitute its root into P(x).

Polynomial division identity
P(x) = D(x)Q(x) + R(x), where degree of R(x) < degree of D(x)

The remainder always has a lower degree than the divisor.

Quick Tricks

Change the sign of the constant term

For a divisor x − a, use a directly. For x + a, use −a. The substitution value is the root of the divisor.

Example: For P(x) = x³ + 2x + 1 divided by x + 2, use x = −2: P(−2) = −8 − 4 + 1 = −11.
Use the divisor's root for any linear divisor

For a divisor ax + b, solve ax + b = 0. The remainder is P(−b/a), avoiding long division.

Example: For P(x) = 2x² + x + 3 divided by 2x − 1, use x = 1/2: P(1/2) = 1/2 + 1/2 + 3 = 4.
Check divisibility through substitution

A polynomial is exactly divisible by x − a if P(a) = 0. This is the factor theorem consequence of the Remainder Theorem.

Example: P(x) = x² − 5x + 6 has P(2) = 4 − 10 + 6 = 0, so x − 2 is a factor.
Group terms before substituting

When powers are large or the polynomial has many terms, combine like terms or calculate powers of the substitution value first.

Example: For P(x) = x⁴ − 3x² + 2 and divisor x − 2, P(2) = 16 − 12 + 2 = 6.

Remainder Theorem Concepts

Remainder for a divisor x − a

The remainder on dividing P(x) by x − a is P(a).

The quotient-remainder form is P(x) = (x − a)Q(x) + r. Putting x = a makes the quotient term zero, so P(a) = r. Therefore, only the value of the polynomial at a is needed.

Example: For P(x) = 3x³ − 2x + 7 and divisor x − 1, the remainder is P(1) = 3 − 2 + 7 = 8.

Divisors of the form x + a

For division by x + a, substitute x = −a in the polynomial.

Because x + a = x − (−a), its zero is −a. This sign change is the most common source of error in direct-substitution questions.

Example: If P(x) = x² − 4x + 1 and the divisor is x + 3, the remainder is P(−3) = 9 + 12 + 1 = 22.

Divisor of the form ax + b

For a linear divisor ax + b, where a ≠ 0, the remainder is P(−b/a).

The root of ax + b is −b/a. Since the divisor has degree one, the remainder is a constant, and substituting this root into P(x) gives that constant.

Example: For P(x) = x² + 4x + 1 divided by 2x + 4, use x = −2. The remainder is P(−2) = 4 − 8 + 1 = −3.

Connection with the Factor Theorem

If P(a) = 0, then x − a is a factor of P(x).

The Remainder Theorem states that the remainder on division by x − a is P(a). Hence, P(a) = 0 means the remainder is zero, so the divisor divides the polynomial exactly.

Example: For P(x) = x³ − 4x² + x + 6, P(2) = 8 − 16 + 2 + 6 = 0. Therefore, x − 2 is a factor.

Remainder after division by a product of factors

The direct Remainder Theorem gives a constant remainder only for a linear divisor; division by a higher-degree divisor can produce a polynomial remainder.

If the divisor has degree n, the remainder must have degree less than n. For example, division by (x − 1)(x − 2), a quadratic divisor, gives a remainder of the form ax + b. Its values can be found using P(1) and P(2).

Example: If the remainder is R(x) = ax + b, then R(1) = P(1) and R(2) = P(2). These two equations determine a and b.

Remainder Theorem Video Lessons

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Remainder Tricks: Fermat and Euler

Learn how Fermat’s pattern and Euler’s approach simplify remainder calculations for powers, including the key conditions and steps needed to apply these methods accurately.

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Practice Remainder Theorem Questions

Practise published questions related to this topic.

1Least Common Multiple of A and B is 15. Least Common Multiple of C and D is 25. What is the Least Common Multiple of A, B, C and D?→ 2What is the largest 2 digit number which is exactly divisible by both 9 and 3?→ 3The number 8,33,525 is NOT divisible by:→ 4If √210.25 = 14.5, then what is the value of √0.00021025?→ 5Three numbers are in the ratio 9 : 7 : 5, and their LCM is 2205. Their HCF is:→ 6What is the Highest Common Factor of 24 × 53 and 26 × 52?→ 7If x = √144 ÷ √36 × √49 , then what is the value of x² ?→ 8Find the HCF of 148, 172, and 198.→ 9Which is the largest number that divides 38, 50, 74 and 98 to leave the same remainder in each case?→ 10What is the smallest number which when increased by 3 is divisible by 27, 35, 25, and 21?→

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Quick Revision Notes

Remainder Theorem Quick Revision

Use the divisor's zero as the substitution value and apply the degree rule for the remainder.

  • For division by x − a, remainder = P(a).
  • For division by x + a, remainder = P(−a).
  • For division by ax + b, remainder = P(−b/a), with a ≠ 0.
  • A linear divisor gives a constant remainder.
  • A divisor of degree n gives a remainder of degree less than n.
  • P(a) = 0 if and only if x − a is a factor of P(x).
  • For a higher-degree divisor, determine the remainder polynomial using values at the divisor's roots when applicable.

Remainder Theorem FAQs

What is the Remainder Theorem formula?

If P(x) is divided by x − a, the remainder is P(a). For example, the remainder when x² + 2x + 3 is divided by x − 1 is P(1) = 1 + 2 + 3 = 6.

What value of x should be used for the divisor x + 5?

Use x = −5 because x + 5 = x − (−5). The remainder is P(−5).

How do you find the remainder when the divisor is 3x − 6?

Set 3x − 6 = 0, giving x = 2. Therefore, the remainder is P(2).

When is a polynomial exactly divisible by x − a?

It is exactly divisible when P(a) = 0. In that case, the remainder is zero and x − a is a factor of P(x).

What is the remainder when x³ − 2x² + 4 is divided by x − 2?

Substitute x = 2: P(2) = 8 − 8 + 4 = 4. Therefore, the remainder is 4.

Can the Remainder Theorem be applied directly to a quadratic divisor?

Not as a constant-remainder formula. A quadratic divisor produces a remainder of degree less than 2, generally of the form ax + b.

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