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Let P(x) = x³ + a*x² + b*x + c where a, b, c are real numbers. Given that P(-3) = P(2) = 0 and P'(-3) < 0. Which of the following is a possible value of c?

1. -27
2. -18
3. -6
4. -3

Correct answer: -27

Solution

With roots -3, 2, and r: P(x) = (x+3)(x-2)(x-r). c = P(0) = 6r. P'(-3) = 5(3+r) < 0 => r < -3 => c = 6r < -18. Among options, c = -27 gives r = -4.5 < -3 (valid). c = -18 gives r = -3 (boundary, P'(-3)=0, not strictly < 0). c = -6 => r=-1 (fails). c = -3 => r=-0.5 (fails). Only c = -27 satisfies all conditions.

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