Remainder Theorem
Find the remainder of a polynomial division in seconds by substituting one value. Learn the idea, memorise the rules, solve 10 worked MCQs, then take a timed 20-question test.
- Bank, SSC & RRB exams
- Placement aptitude
- Beginner to advanced
Learning objectives
State the theorem
Explain what the remainder theorem says and why it works.
Find remainders fast
Use f(a) and f(−b/a) instead of long division.
Link to factors
Apply the factor theorem and solve for unknown constants.
Handle harder cases
Quadratic divisors, large powers and remainders of numbers.
Subtopics covered
- Polynomial division basics
- Remainder theorem
- Divisor ax + b
- Factor theorem
- Finding unknown constants
- Two divisors, one condition
- Quadratic divisors
- Large powers
- Remainders of numbers
- Negative remainders
- Fermat’s little theorem
- Number System models
Concept explanation
Step 1: The statement
If a polynomial f(x) is divided by (x − a), the remainder is f(a). The remainder is always a constant, because the divisor has degree 1.
Remainder on dividing f(x) by (x − a) = f(a)
Step 2: Why it works
Division gives f(x) = (x − a)·q(x) + r, where r is a constant. Put x = a. The bracket (a − a) becomes 0, so the whole first term vanishes and f(a) = r.
Step 3: Divisor of the form ax + b
Set the divisor to zero and solve. For (ax + b), the root is x = −ba, so the remainder is f(−ba).
Example: for the divisor (2x − 1), use x = 12.
Step 4: Factor theorem (the special case r = 0)
If f(a) = 0, then (x − a) is a factor of f(x), and a is a root of f(x) = 0. The reverse is also true.
Step 5: Quadratic divisor with two roots
The remainder has degree at most 1, so write it as ux + v. Put each root of the divisor into f(x) and solve the two equations for u and v.
Example 1: direct substitution
Find the remainder when f(x) = x3 − 4x2 + 2x + 7 is divided by (x − 3).
- Root of the divisor: x = 3.
- f(3) = 27 − 36 + 6 + 7 = 4.
Remainder = 4.
Example 2: divisor 3x − 1 and a factor check
Is (3x − 1) a factor of 3x2 + 5x − 2?
- Root: x = 13.
- f(13) = 3×19 + 5×13 − 2 = 13 + 53 − 2 = 0.
The remainder is 0, so (3x − 1) is a factor. Check: (3x − 1)(x + 2) = 3x2 + 5x − 2.
Example 3: finding an unknown constant
The polynomial x3 − kx2 + 5x − 6 leaves remainder 8 when divided by (x − 2). Find k.
- f(2) = 8 − 4k + 10 − 6 = 12 − 4k.
- Set 12 − 4k = 8, so 4k = 4.
k = 1.
Example 4: quadratic divisor
Find the remainder when x3 + 2 is divided by (x2 − 1).
- Roots of the divisor: x = 1 and x = −1. Let the remainder be ux + v.
- f(1) = 3, so u + v = 3. f(−1) = 1, so −u + v = 1.
- Add: 2v = 4, so v = 2 and u = 1.
Remainder = x + 2.
Remainder theorem in the Number System
The same idea works for ordinary numbers. A number can be replaced by its remainder (or by a negative remainder) before you add, multiply or raise to a power. This is the fastest way to solve "remainder when an is divided by m" questions.
Step 1: Remainder arithmetic
Let a leave remainder r1 and b leave remainder r2 on division by m. Then:
- (a + b) leaves the same remainder as (r1 + r2).
- (a × b) leaves the same remainder as (r1 × r2).
- an leaves the same remainder as r1n.
After each calculation, reduce the result by m again.
Step 2: Use negative remainders
If a number is 1 less than a multiple of m, treat its remainder as −1. Then (−1)n = 1 for even n and −1 for odd n. If the final remainder is negative, add m to it.
64 = 65 − 1, so 64 leaves remainder −1 on division by 13
Step 3: Link with the polynomial theorem
Put a number in place of x. Since xn − an is divisible by (x − a), and xn + an is divisible by (x + a) when n is odd, you get results such as: 525 + 1 is divisible by 5 + 1 = 6.
Step 4: Fermat's little theorem
If p is prime and a is not a multiple of p, then ap−1 leaves remainder 1 on division by p. Example: 212 leaves remainder 1 on division by 13.
Model 1: 236 divided by 13
- 26 = 64 = 65 − 1, so 26 leaves remainder −1.
- 236 = (26)6, which leaves the same remainder as (−1)6 = 1.
Remainder = 1. Check: 36 = 3×12, and 212 leaves remainder 1 by Fermat.
Model 2: a product
Find the remainder when 17 × 23 × 29 is divided by 15.
- 17 leaves 2, 23 leaves 8, 29 leaves 14 (or −1).
- 2 × 8 = 16, which leaves 1. Then 1 × (−1) = −1.
- Add 15 to the negative remainder: −1 + 15 = 14.
Remainder = 14. Check: 17 × 23 × 29 = 11339 = 15 × 755 + 14.
Model 3: a cycle of powers
Find the remainder when 2100 is divided by 7.
- 23 = 8 leaves remainder 1.
- 100 = 3 × 33 + 1, so 2100 = (23)33 × 2, which leaves 1 × 2.
Remainder = 2.
Model 4: a larger divisor
Find the remainder when 7100 is divided by 50.
- 72 = 49 = 50 − 1, so 72 leaves −1.
- 7100 = (72)50, which leaves (−1)50 = 1.
Remainder = 1.
Model 5: a sum with a factorial
Find the remainder when 1! + 2! + 3! + … + 50! is divided by 12.
- 4! = 24 is divisible by 12, so every term from 4! onwards is divisible by 12.
- Only 1! + 2! + 3! = 1 + 2 + 6 = 9 remains.
Remainder = 9.
Formula sheet and quick revision
Remainder theorem
Divisor (x − a): remainder = f(a)
General linear divisor
Divisor (ax + b): remainder = f(−ba)
Factor theorem
f(a) = 0 ⇔ (x − a) is a factor of f(x)
Powers
xn ÷ (x − a) leaves an. xn ÷ (x + a) leaves (−a)n.
Special values
Sum of coefficients = f(1). Alternating sum = f(−1). Constant term = f(0).
Remainder arithmetic
Sums, products and powers can use remainders: reduce by m after each step.
Negative remainder
If a = km − 1, then an leaves (−1)n. Add m if the result is negative.
Fermat's little theorem
p prime, a not a multiple of p: ap−1 leaves remainder 1 on division by p.
Divisibility of powers
an − bn is divisible by (a − b). an + bn is divisible by (a + b) for odd n.
Quadratic divisor
Remainder = ux + v. Use both roots to get two equations.
Quick revision notes
- The remainder of a division by a degree-1 divisor is a constant. For a degree-2 divisor it is at most degree 1.
- Always write the divisor as zero first, then solve for x.
- (−1) raised to an even power is 1, and to an odd power is −1.
- Multiplying the divisor by a constant does not change the remainder: dividing by (2x + 4) and by (x + 2) leaves the same remainder.
10 solved MCQs
Timed practice test
| Questions | 20, picked at random from a pool of 32 |
|---|---|
| Time allowed | 20 minutes |
| Marks | 1 mark per correct answer |
| Negative marking | None |
| Tools | Mark for review and a jump-to-question panel |
| Level | Mixed: basic, intermediate and advanced |
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