The polynomial x² + a x + b leaves the same remainder when divided by (x − 1) and by (x + 1). What are the values of a and b respectively?

Introduction / Context:
This algebra question involves remainders of a polynomial when divided by linear factors. It uses the Remainder Theorem, which states that the remainder when a polynomial P(x) is divided by (x − c) is simply P(c). Here, we know that the remainders for two different divisors, (x − 1) and (x + 1), are equal, and we must deduce conditions on the coefficients a and b.

Given Data / Assumptions:

- The polynomial is P(x) = x² + a x + b.

- When P(x) is divided by (x − 1), the remainder is P(1).

- When P(x) is divided by (x + 1), the remainder is P(−1).

- These two remainders are equal.

- We must find a and b that satisfy this condition.

Concept / Approach:
By the Remainder Theorem, set P(1) equal to P(−1). This gives an equation in a and b. Solve this equation to find the relationship between a and b. If only a is fixed and b remains unrestricted, then b can be any integer. We then match this with the most appropriate option.

Step-by-Step Solution:
Step 1: Compute P(1). P(1) = 1² + a * 1 + b = 1 + a + b. Step 2: Compute P(−1). P(−1) = (−1)² + a * (−1) + b = 1 − a + b. Step 3: Use the condition that the remainders are the same. P(1) = P(−1). So 1 + a + b = 1 − a + b. Step 4: Simplify this equation. 1 + a + b = 1 − a + b implies a = −a. So 2a = 0 and hence a = 0. Step 5: Observe that b cancels out and is not restricted by this equation. Therefore, b can be any integer.

Verification / Alternative check:
Take any specific integer value for b, such as b = 3, and with a = 0, form P(x) = x² + 3. Then: P(1) = 1 + 3 = 4. P(−1) = 1 + 3 = 4. The remainders are equal. If we choose a different b, the equality still holds, showing that only a must be zero and b remains arbitrary.

Why Other Options Are Wrong:
- 4 and 0: This fixes a = 4, but we have shown a must be 0.

- 0 and 3: This is just one special case; the problem allows any integer b, not only 3.

- 3 and 0: Again this forces a = 3, contradicting a = 0.

Common Pitfalls:
A common error is to misapply the Remainder Theorem, or to forget that both remainders must be equal, leading to incorrect equations. Some students mistakenly assume that both a and b must be uniquely determined, not recognising that one coefficient can remain free. Always follow the algebra carefully to see whether each parameter is fully determined.

Final Answer:
The correct condition is that a = 0 and b can be any integer, corresponding to 0 and any integer.