Let P(x) = x³ + a*x² + b*x + c where a, b, c are real numbers. Given P(-3) = 0, P(2) = 0, and P'(-3)

Question

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

Accepted Answer

-27. P(x) = (x+3)(x-2)(x-r). c = P(0) = 3*(-2)*(-r) = 6r. P'(-3) = (-3-2)*(-3-r) = 5*(3+r). For P'(-3)<0: r<-3, so c=6r<-18. Only c=-27 gives r=-4.5<-3. c=-18 gives r=-3 (boundary, P'(-3)=0, not <0). c=-6 and -3 give r>-3 (invalid).

Let P(x) = x³ + ax² + bx + c where a, b, c are real numbers. Given P(-3) = 0, P(2) = 0, and P'(-3) < 0, which of the following is a possible value of c?

Solution

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