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# So I know the answer is -27,

but I'm not sure why. I know it has to have a third real zero since one complex number would imply it's conjugate as a fourth zero, which is impossible. I also know that the product of the roots = c, so c = 6 times the third root. But how do I know if the third root is positive or negative? Since all the answer choices are negative, I know the third root is negative, but I don't know why. And even then, I'm not sure how to go about finding what values of c are "possible". I don't see why it couldn't theoretically have a root in between it's two known ones. Any help is appreciated! I guess I don't know much about the behavior of cubic functions, or how the derivative is relevant here. 65. Let p(x) be the polynomial x3 + ax2 + bx + c, where a, b, and c are real constants. If
p(—-3) = p(2) = 0 and p'(—3) &lt; 0, which of the following is a possible value of c '2‘ (A) -27
(B) —18
(C) -6
(D) -3 1 (E) —5

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