The polynomial x^3 + 2x^2 − 5x + k is exactly divisible by x + 1 for all real x. Using the Remainder Theorem or factor theorem, what is the value of the constant term k that makes this divisibility possible?

Introduction / Context:This question focuses on polynomial division and the use of the Remainder Theorem. When a polynomial is divisible by a linear factor like x + 1, the remainder must be zero when x is replaced by the value that makes the factor equal to zero. Here, we use that idea to determine the constant k in the given cubic polynomial.

Given Data / Assumptions:

• The polynomial is f(x) = x^3 + 2x^2 − 5x + k.
• The polynomial is exactly divisible by x + 1.
• Exact divisibility means the remainder is zero when dividing by x + 1.
• We assume real coefficients and real arithmetic.

Concept / Approach:The Remainder Theorem states that if a polynomial f(x) is divided by (x − a), the remainder is f(a). Here, the factor is x + 1, which can be written as x − (−1). Therefore, the remainder when f(x) is divided by x + 1 is f(−1). Exact divisibility implies that this remainder is zero. So we set f(−1) = 0 and solve the resulting equation for k.

Step-by-Step Solution:Write the polynomial as f(x) = x^3 + 2x^2 − 5x + k.For division by x + 1, use a = −1. Compute f(−1): f(−1) = (−1)^3 + 2(−1)^2 − 5(−1) + k.Calculate each term: (−1)^3 = −1, 2(−1)^2 = 2, and −5(−1) = 5.So f(−1) = −1 + 2 + 5 + k = 6 + k.Exact divisibility requires f(−1) = 0, so 6 + k = 0. Hence k = −6.

Verification / Alternative check:We can verify by substituting k = −6 back into the polynomial and performing synthetic division or direct substitution. With k = −6, f(−1) becomes −1 + 2 + 5 − 6 = 0, confirming the remainder is zero. Alternatively, synthetic division of x^3 + 2x^2 − 5x − 6 by x + 1 will yield a quotient with no remainder, which confirms that x + 1 is indeed a factor when k = −6.

Why Other Options Are Wrong:

• Values like −1, 0, 2, or 6 produce non zero remainders when substituted into f(−1), so the polynomial is not exactly divisible by x + 1 for those choices.
• Only k = −6 makes the remainder vanish, satisfying the divisibility condition.

Common Pitfalls:

• Confusing the sign when substituting x = −1 into the polynomial.
• Forgetting that the divisor x + 1 corresponds to evaluating the polynomial at x = −1, not at x = 1.
• Making arithmetic errors while combining terms, which could lead to an incorrect value of k.

Final Answer:-6