Remainder Theorem

• Divide polynomials and find quotient and remainder.
• State and apply the Remainder Theorem: p(x) ÷ (x − a) has remainder p(a).
• Use remainder theorem to evaluate remainders without long division.
• Connect remainder zero to factor (x − a).
• Solve NCERT Ex. 2.3 problems on remainders.

1. Polynomial division — Dividend = divisor × quotient + remainder; deg(remainder) < deg(divisor).
2. Remainder Theorem — When p(x) is divided by (x − a), remainder = p(a).
3. Example — p(x) = x² − 3x + 2; remainder on ÷ (x − 1): p(1) = 0.
4. Linear divisor — Theorem applies directly when divisor is (x − a); for (ax − b) use x = b/a.
5. Applications — Find k if x − k divides p(x) exactly (remainder 0).
6. Long division check — Remainder theorem gives quick verification.
7. Degree insight — Remainder upon division by linear factor is a constant.
8. NCERT Example 2.9 — Find remainder when x³ + 3x² + 3x + 1 divided by x + 1.
9. Synthetic division — Shortcut equivalent for linear divisors (extension).
10. Exam tip — Substitute x = a into p(x); do not confuse with p(−a) unless divisor is (x + a).

NCERT: Remainder when p(x) = x³ − 2x² + x + 1 is divided by x − 2

Method — Remainder Theorem with a = 2:
Step 1 — p(2) = (2)³ − 2(2)² + 2 + 1
Step 2 — = 8 − 8 + 2 + 1 = 3
Remainder = 3 (without performing long division)
Check: long division also gives remainder 3

• Using p(−a) when divisor is (x − a) (sign error).
• For (x + 1), use a = −1, not a = 1.
• Forgetting all powers when substituting into cubic/quartic.
• Confusing remainder with quotient.
• Applying theorem when divisor degree > 1 without adjustment.

• State the Remainder Theorem.
• Find remainder: p(x) = x² + 5x + 6 divided by x − 1.
• If p(x) = x³ − kx + 4 has remainder 10 on ÷ (x − 2), find k.
• Remainder when x³ + 1 divided by x + 1?
• Why is remainder upon ÷ (x − a) always a constant?

Open the Practice tab for graded questions on Remainder Theorem.