Polynomials with the same remainder (Recovery applied) If the polynomials ax^3 + 4x^2 − 3x − 4 and x^3 − 4x + a leave the same remainder when divided by (x + 3), find the value of a.

Introduction / Context:Two polynomials leave the same remainder upon division by a linear factor if and only if they take the same value at the corresponding root. Recovery-First: the original stem with (x − 3) yields a non-integer a not in options; replacing by (x + 3) corrects the standard intent and aligns with the answer set.

Given Data / Assumptions:

• P1(x) = a x^3 + 4x^2 − 3x − 4.
• P2(x) = x^3 − 4x + a.
• Same remainder on division by (x + 3) ⇒ evaluate both at x = −3.

Concept / Approach:Remainder Theorem: remainder of division by (x − r) equals P(r). Equal remainders ⇒ P1(r) = P2(r).

Step-by-Step Solution:Compute P1(−3) = a(−27) + 4·9 − 3(−3) − 4 = −27a + 36 + 9 − 4 = −27a + 41.Compute P2(−3) = (−27) − 4(−3) + a = −27 + 12 + a = −15 + a.Set equal: −27a + 41 = −15 + a ⇒ −28a = −56 ⇒ a = 2.

Verification / Alternative check:At x = −3, both polynomials evaluate to −13, confirming the equal remainder.

Why Other Options Are Wrong:−1, 3, 4, 0 do not satisfy the equality at x = −3.

Common Pitfalls:Using x = 3 (from the uncorrected stem) yields a non-listed fractional a; always tie the evaluation point to the divisor's root.

Final Answer:2