Another problem

[Leviathan]

Let p(x) be the polynomial x^3 + a*x^2 + b*x + c, where a, b, and c are real constants. if p(-3) = p(2) = 0 and p'(3) < 0, which of the following is a possible value of c?
(A) -27
(B) -18
(C) -6
(D) -3
(E) -1/2

Can anyone solve this please? Thanks!

[Dragonfinity]

We are given the real polynomial p(x) = x^3 + a*x^2 + b*x + c has -2 and 3 for roots. Since 2 of its roots are real (and p is of odd degree), the third root R must also be real. Hence,

p(x) = (x + 3)(x - 2)(x - R) = (x^2 + x - 6)(x - R) = x^3 + (1 - R)*x^2 - (6 + R)*x + 6*R

Thus, c = 6*R. We can further compute

p'(x) = 3*x^2 + 2*(1 - R)*x - (6 + R)
=> 0 > p'(3) = 27 + 6*(1 - R) - (6 + R) = 27 + 6 - 6*R - 6 - R = 27 - 7*R
=> 7*R > 27
=> R > 27/7 > 0
=> c = 6*R > 0

Hence, none of the values given is a possible value for c.

[juphuy]

dragonfinity, your answer was great. But the given question was wrong. It should be p'(-3) < 0. Following your way, the answer is (a): c = -27.

[Leviathan]

Originally Posted by Dragonfinity

We are given the real polynomial p(x) = x^3 + a*x^2 + b*x + c has -2 and 3 for roots. Since 2 of its roots are real (and p is of odd degree), the third root R must also be real. Hence,

p(x) = (x + 3)(x - 2)(x - R) = (x^2 + x - 6)(x - R) = x^3 + (1 - R)*x^2 - (6 + R)*x + 6*R

Thus, c = 6*R. We can further compute

p'(x) = 3*x^2 + 2*(1 - R)*x - (6 + R)
=> 0 > p'(3) = 27 + 6*(1 - R) - (6 + R) = 27 + 6 - 6*R - 6 - R = 27 - 7*R
=> 7*R > 27
=> R > 27/7 > 0
=> c = 6*R > 0

Hence, none of the values given is a possible value for c.

Dragonfinity, thanks for the response. Your method was right. One thing though. The roots of the function are 2 and -3 (not 3), hence
p'(-3) < 0 <==> R < -3 <==> 6*R < -18 <==> c < -18
So we can conclude that the answer is (A) -27

Thanks again.

[Dragonfinity]

Originally Posted by Leviathan

Let p(x) be the polynomial x^3 + a*x^2 + b*x + c, where a, b, and c are real constants. if p(-3) = p(2) = 0 and p'(3) < 0, which of the following is a possible value of c?
(A) -27
(B) -18
(C) -6
(D) -3
(E) -1/2

Can anyone solve this please? Thanks!

Note that my answer was based on the original problem as written above. Apparently, there was a typo in that problem.

[MAM]

When I first time saw this problem I tried to do the same, but the problem is much easier. If you just try to sketch a graph of p(x)=0 (using the facts, that p(2)=p(-3)=0 and p'(-3)<0), you can see, that the third root has to be strictly less than -3, but c is a negative product of all three roots. So, we have c<-2*(-3)*(-3) or c<-18. It takes 30 seconds to figure out this!!!

[Pensive]

MAM,You are a genius!