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Remainder Theorem

Class 9Polynomials

The Remainder Theorem connects polynomial division with polynomial evaluation. It provides a quick way to find the remainder when a polynomial is divided by a linear divisor, without performing long division.


Instead of going through the lengthy process of polynomial long division, you simply substitute a value into the polynomial to get the remainder.


The Remainder Theorem is closely linked to the Factor Theorem and serves as a stepping stone to synthetic division, polynomial roots, and the fundamental theorem of algebra.

What is Remainder Theorem?

Definition: If a polynomial p(x) is divided by a linear polynomial (x − a), then the remainder is p(a).


To find the remainder, substitute x = a into p(x). The resulting value is the remainder.


Mathematical form:

p(x) = (x − a) × q(x) + r


Where:

  • q(x) = quotient polynomial
  • r = remainder (a constant, since the divisor has degree 1)

Proof:

  1. Put x = a in the equation: p(a) = (a − a) × q(a) + r
  2. p(a) = 0 × q(a) + r
  3. p(a) = r

Extensions:

  • Dividing by (x + a): remainder = p(−a), since x + a = x − (−a)
  • Dividing by (ax − b): remainder = p(b/a), since the zero of (ax − b) is x = b/a
  • Dividing by (ax + b): remainder = p(−b/a)

Important: The Remainder Theorem gives only the remainder, not the quotient. For the quotient, use long division or synthetic division.

Remainder Theorem Formula

Key Formulas:


1. Remainder Theorem (basic form):

When p(x) is divided by (x − a), remainder = p(a)


2. Division algorithm for polynomials:

  • p(x) = (x − a) × q(x) + p(a)

3. When dividing by (x + a):

  • Remainder = p(−a)

4. When dividing by (ax − b):

  • Remainder = p(b/a)

5. When dividing by (ax + b):

  • Remainder = p(−b/a)

6. Connection to Factor Theorem:

  • If p(a) = 0 (remainder is zero), then (x − a) is a factor of p(x).

7. Finding unknown coefficients:

  • If the remainder when p(x) is divided by (x − a) is R, then p(a) = R. Solve for the unknown.

Derivation and Proof

Proof of the Remainder Theorem:


Step 1: State the Division Algorithm

  • When p(x) is divided by d(x), we get quotient q(x) and remainder r(x) such that:
  • p(x) = d(x) × q(x) + r(x)
  • where degree of r(x) < degree of d(x), or r(x) = 0.

Step 2: Apply to division by (x − a)

  • Let d(x) = (x − a), which has degree 1.
  • The remainder r(x) must have degree less than 1, so r(x) is a constant r.
  • p(x) = (x − a) × q(x) + r ... (equation 1)

Step 3: Substitute x = a

  1. p(a) = (a − a) × q(a) + r
  2. p(a) = 0 × q(a) + r
  3. p(a) = r

Step 4: Conclusion

  • The remainder r = p(a). This proves the Remainder Theorem.

Geometric understanding: When x = a, the factor (x − a) becomes zero, eliminating the quotient term. What remains is the constant remainder.


Extension to (ax − b):

  1. The zero of (ax − b) is x = b/a.
  2. p(x) = (ax − b) × q(x) + r
  3. Put x = b/a: p(b/a) = (a × b/a − b) × q(b/a) + r = 0 + r
  4. So remainder = p(b/a).

Types and Properties

Types of Remainder Theorem Problems:


  • Type 1: Finding the remainder directly — Given p(x) and divisor (x − a), compute p(a). No division needed.
  • Type 2: Division by (x + a) — Rewrite as (x − (−a)) and find remainder as p(−a). Remember to use the negative value.
  • Type 3: Division by (ax − b) or (ax + b) — Find the zero of the divisor. For (ax − b), zero = b/a. For (ax + b), zero = −b/a.
  • Type 4: Finding unknown coefficients — If the remainder is given, set up the equation p(a) = remainder and solve for the unknown.
  • Type 5: Checking if a linear expression is a factor — If p(a) = 0, the divisor is a factor. This connects to the Factor Theorem.
  • Type 6: Finding remainders for multiple divisors — Evaluate the polynomial at different values for different divisors.
  • Type 7: Verification problems — Verify a remainder by both the Remainder Theorem and polynomial long division.

Solved Examples

Example 1: Example 1: Finding remainder when dividing by (x − 2)

Problem: Find the remainder when p(x) = x³ − 3x² + 4x − 12 is divided by (x − 2).


Solution:

By the Remainder Theorem: remainder = p(2)

  • p(2) = (2)³ − 3(2)² + 4(2) − 12
  • = 8 − 12 + 8 − 12
  • = −8

Answer: The remainder is −8.

Example 2: Example 2: Finding remainder when dividing by (x + 3)

Problem: Find the remainder when p(x) = 2x³ + x² − 5x + 1 is divided by (x + 3).


Solution:

x + 3 = x − (−3), so remainder = p(−3)

  • p(−3) = 2(−3)³ + (−3)² − 5(−3) + 1
  • = 2(−27) + 9 + 15 + 1
  • = −54 + 9 + 15 + 1
  • = −29

Answer: The remainder is −29.

Example 3: Example 3: Finding remainder when dividing by (2x − 1)

Problem: Find the remainder when p(x) = 4x³ − 12x² + 11x − 3 is divided by (2x − 1).


Solution:

The zero of (2x − 1) is x = 1/2.

Remainder = p(1/2)

  • p(1/2) = 4(1/2)³ − 12(1/2)² + 11(1/2) − 3
  • = 4(1/8) − 12(1/4) + 11/2 − 3
  • = 1/2 − 3 + 11/2 − 3
  • = (1 + 11)/2 − 6
  • = 6 − 6 = 0

Answer: The remainder is 0, which means (2x − 1) is a factor of p(x).

Example 4: Example 4: Finding an unknown coefficient using the Remainder Theorem

Problem: If p(x) = x³ + 2x² − kx + 5 leaves a remainder of 11 when divided by (x − 1), find k.


Solution:

By the Remainder Theorem: p(1) = 11

  1. p(1) = (1)³ + 2(1)² − k(1) + 5 = 1 + 2 − k + 5 = 8 − k
  2. Set 8 − k = 11
  3. −k = 3
  4. k = −3

Answer: k = −3

Example 5: Example 5: Verifying the Remainder Theorem by long division

Problem: Find the remainder when p(x) = x² + 5x + 6 is divided by (x − 1) using both methods.


Solution:

Method 1: Remainder Theorem

  • Remainder = p(1) = 1 + 5 + 6 = 12

Method 2: Long Division

  • x² + 5x + 6 = (x − 1)(x + 6) + 12
  • Check: (x − 1)(x + 6) + 12 = x² + 6x − x − 6 + 12 = x² + 5x + 6 ✓

Answer: Both methods give a remainder of 12.

Example 6: Example 6: Remainder when dividing by (3x + 2)

Problem: Find the remainder when p(x) = 9x³ + 3x² − 5x + 7 is divided by (3x + 2).


Solution:

The zero of (3x + 2) is x = −2/3.

Remainder = p(−2/3)

  • p(−2/3) = 9(−2/3)³ + 3(−2/3)² − 5(−2/3) + 7
  • = 9(−8/27) + 3(4/9) + 10/3 + 7
  • = −8/3 + 4/3 + 10/3 + 7
  • = (−8 + 4 + 10)/3 + 7
  • = 6/3 + 7 = 2 + 7 = 9

Answer: The remainder is 9.

Example 7: Example 7: Finding the value of 'a' when the polynomial is exactly divisible

Problem: Find the value of a if (x − 2) is a factor of p(x) = x³ − ax² + 6x − 8.


Solution:

If (x − 2) is a factor, then p(2) = 0.

  1. p(2) = (2)³ − a(2)² + 6(2) − 8
  2. = 8 − 4a + 12 − 8
  3. = 12 − 4a
  4. Set 12 − 4a = 0
  5. 4a = 12 ⇒ a = 3

Answer: a = 3

Example 8: Example 8: Multiple remainders for the same polynomial

Problem: For p(x) = x⁴ − 2x³ + x² − 2x + 1, find the remainder when divided by (a) (x − 1), (b) (x + 1), (c) (x − 2).


Solution:

  • (a) p(1) = 1 − 2 + 1 − 2 + 1 = −1
  • (b) p(−1) = 1 + 2 + 1 + 2 + 1 = 7
  • (c) p(2) = 16 − 16 + 4 − 4 + 1 = 1

Answer: The remainders are −1, 7, and 1 respectively.

Example 9: Example 9: Finding two unknowns using the Remainder Theorem

Problem: The polynomial p(x) = x³ + ax² + bx − 6 gives a remainder of 0 when divided by (x − 1) and −4 when divided by (x + 1). Find a and b.


Solution:

Condition 1: p(1) = 0

  • 1 + a + b − 6 = 0 ⇒ a + b = 5 ... (i)

Condition 2: p(−1) = −4

  • −1 + a − b − 6 = −4 ⇒ a − b = 3 ... (ii)

Solving:

  1. Adding (i) and (ii): 2a = 8 ⇒ a = 4
  2. Substituting a = 4 into (i): 4 + b = 5 ⇒ b = 1

Answer: a = 4 and b = 1.

Example 10: Example 10: Applying the Remainder Theorem to a quartic polynomial

Problem: Find the remainder when p(x) = 3x⁴ + 2x³ − x² + 7x − 4 is divided by (x + 2).


Solution:

x + 2 = x − (−2), so remainder = p(−2).

  • p(−2) = 3(−2)⁴ + 2(−2)³ − (−2)² + 7(−2) − 4
  • = 3(16) + 2(−8) − 4 − 14 − 4
  • = 48 − 16 − 4 − 14 − 4
  • = 10

Answer: The remainder is 10.

Real-World Applications

Applications of the Remainder Theorem:


  • Quick polynomial evaluation: Provides a fast, structured way to evaluate polynomials at specific values.
  • Factorisation: By testing values and finding which give remainder zero, we identify factors of the polynomial (trial-and-error method).
  • Checking divisibility: To check if a linear expression divides a polynomial exactly, evaluate the polynomial at the zero of the divisor.
  • Finding unknown coefficients: Given conditions about remainders or factors, set up equations using p(a) = R and solve for unknowns.
  • Synthetic division (Horner's method): The Remainder Theorem underlies this efficient algorithm for polynomial evaluation used in computer science.
  • Error-correcting codes: Polynomial remainder calculations are fundamental to cyclic redundancy check (CRC) codes in data transmission.
  • Interpolation: Related to polynomial interpolation methods used in data science and engineering.

Key Points to Remember

  • When p(x) is divided by (x − a), the remainder is p(a).
  • To find the remainder, substitute the zero of the divisor into the polynomial.
  • Dividing by (x + a): remainder = p(−a).
  • Dividing by (ax − b): remainder = p(b/a).
  • The remainder when dividing by a linear polynomial is always a constant.
  • If the remainder is 0, the divisor is a factor — this leads to the Factor Theorem.
  • The theorem can find unknown coefficients when the remainder is given.
  • Division algorithm: p(x) = divisor × quotient + remainder.
  • The theorem gives only the remainder, not the quotient.
  • Works only for linear divisors (degree 1).

Practice Problems

  1. Find the remainder when p(x) = x³ + 3x² + 3x + 1 is divided by (x + 1).
  2. Find the remainder when p(x) = 4x³ − 3x² + 2x − 1 is divided by (x − 3).
  3. If the remainder when x⁴ − 2x³ + kx² − 3x + 2 is divided by (x − 1) is 5, find the value of k.
  4. Find the remainder when p(x) = 2x³ + 3x² − 5x + 7 is divided by (2x + 1).
  5. The polynomial ax³ + bx² + x − 6 has (x + 2) as a factor and leaves a remainder 4 when divided by (x − 2). Find a and b.
  6. Find the remainder when x⁵¹ + 51 is divided by (x + 1).
  7. Without actual division, prove that 2x⁴ − 6x³ + 3x² + 3x − 2 is divisible by x² − 3x + 2. (Hint: Factorise x² − 3x + 2 first.)
  8. If p(x) = x³ − 6x² + 11x + k and p(1) + p(2) = 0, find the value of k.

Frequently Asked Questions

Q1. What is the Remainder Theorem in simple words?

When you divide a polynomial p(x) by (x − a), the remainder is p(a) — the value you get by substituting x = a. For example, dividing x² + 3x + 2 by (x − 1): remainder = p(1) = 1 + 3 + 2 = 6.

Q2. How do you use the Remainder Theorem to find the remainder?

Step 1: Identify the divisor and note its zero (for (x − a), use a; for (x + a), use −a; for (ax − b), use b/a). Step 2: Substitute this value into p(x). Step 3: The result is the remainder.

Q3. What is the relationship between the Remainder Theorem and Factor Theorem?

The Factor Theorem is a special case of the Remainder Theorem. When the remainder p(a) = 0, then (x − a) is a factor of p(x). Conversely, if (x − a) is a factor, then p(a) = 0.

Q4. Can the Remainder Theorem be used for divisors of degree 2 or higher?

No. The standard Remainder Theorem applies only to linear divisors (degree 1). For higher-degree divisors, the remainder is a polynomial (not a constant), and you need polynomial long division.

Q5. What is the remainder when a polynomial is divided by (x + a)?

The remainder is p(−a), since x + a = x − (−a). For example, dividing x³ + 1 by (x + 1): remainder = p(−1) = (−1)³ + 1 = 0.

Q6. How do you find the remainder when dividing by (2x − 3)?

Find the zero of the divisor: 2x − 3 = 0 gives x = 3/2. The remainder is p(3/2). Substitute x = 3/2 into the polynomial and evaluate.

Q7. Is the Remainder Theorem part of the CBSE Class 9 syllabus?

Yes. It is part of Chapter 2: Polynomials. Students must understand the theorem, apply it to find remainders, check factors, and solve problems with unknown coefficients.

Q8. Can the Remainder Theorem give a negative remainder?

Yes. Unlike integer division, polynomial remainders can be positive, negative, or zero. For example, p(x) = x² − 5x + 3 divided by (x − 1): remainder = p(1) = 1 − 5 + 3 = −1.

Q9. How is the Remainder Theorem different from polynomial long division?

Long division gives both the quotient and remainder and works for any divisor. The Remainder Theorem gives only the remainder and works only for linear divisors. However, it is much faster — just one substitution.

Q10. What happens if you apply the Remainder Theorem and get 0?

A remainder of 0 means (x − a) divides p(x) exactly — it is a factor. This result is the Factor Theorem, used to factorise polynomials.

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