Polynomials • Topic 3 of 4

Remainder Theorem

The remainder theorem states that when a polynomial P(x) is divided by (x − a), the remainder is simply P(a) — no long division needed. To divide by a linear factor of the form (bx − c), evaluate P at x = c/b. The theorem is the workhorse for "find the remainder" questions in both algebra and number theory, because a large number written in a base can be read as a polynomial in that base. For instance, the remainder when a number is divided by 9 equals the digit sum reduced mod 9, which is exactly P(1) mod 9 with x as the base. In CAT, watch for two layers: a direct evaluation, or a "find the remainder, then use it" set-up where the remainder feeds a second equation. When two remainders are given for two divisors, you usually get two equations P(a₁) = r₁ and P(a₂) = r₂ to solve simultaneously.

✅ Solved examples

1. Find the remainder when x³ − 4x² + 5x − 7 is divided by (x − 2).
Remainder = P(2) = 8 − 16 + 10 − 7 = −5.
2. Find the remainder when 2x³ + 3x² − x + 4 is divided by (x + 1).
Divisor x + 1 ⇒ a = −1. P(−1) = −2 + 3 + 1 + 4 = 6.
3. Find the remainder when 4x³ − 3x + 2 is divided by (2x − 1).
2x − 1 = 0 ⇒ x = 1/2. P(1/2) = 4(1/8) − 3(1/2) + 2 = 0.5 − 1.5 + 2 = 1.
4. P(x) leaves remainder 5 when divided by (x − 1) and 7 when divided by (x − 3). If P(x) = ax² + bx + 2, find a and b.
P(1) = a + b + 2 = 5 ⇒ a + b = 3. P(3) = 9a + 3b + 2 = 7 ⇒ 9a + 3b = 5. Solving: from a + b = 3, b = 3 − a; 9a + 9 − 3a = 5 ⇒ 6a = −4 ⇒ a = −2/3, b = 11/3.

✏️ Practice — try these, take hints as needed

1. Remainder when x³ + 2x² − 3x + 1 is divided by (x − 1)?
Use P(1).
1 + 2 − 3 + 1.
Add the terms.
1
2. Remainder when x⁴ − 2x² + 3 is divided by (x + 2)?
a = −2.
16 − 8 + 3.
Powers of −2 are positive when even.
11
3. Remainder when 3x³ − 2x² + x − 5 is divided by (3x − 1)?
Set 3x − 1 = 0 ⇒ x = 1/3.
Evaluate P(1/3).
3/27 − 2/9 + 1/3 − 5.
−5 (P(1/3) = 1/9 − 2/9 + 3/9 − 45/9 = −43/9)
4. If x³ − ax + 6 leaves remainder 2 on division by (x − 2), find a.
Set P(2) = 2.
8 − 2a + 6 = 2.
14 − 2a = 2.
a = 6
5. P(x) = x³ + bx + c leaves remainders 4 and 12 on division by (x − 1) and (x − 3). Find b and c.
P(1) = 1 + b + c = 4.
P(3) = 27 + 3b + c = 12.
Subtract the two equations.
b = −8, c = 11

📝 Topic test — 8 questions

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