Thread: GRE subject test question

Hello all,

Here is a sample GRE subject exam question.

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

I have figured out the solution getting -27 as the answer. My question regards a solution I came across on another forum by someone else that I am having trouble following.

Given the information available is it possible to conclude that the third root of the equation must be strictly less than -3?

Thanks!

Originally Posted by kkoutsothodoros

Hello all,

Here is a sample GRE subject exam question.

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

I have figured out the solution getting -27 as the answer. My question regards a solution I came across on another forum by someone else that I am having trouble following.

Given the information available is it possible to conclude that the third root of the equation must be strictly less than -3?

Thanks!

There is something wrong with the question, since:

p(x) = (x+3)(x-2)(x-r3)

p'(x) = 3 x^2 + 2x (1 - r3) - r3 - 6

Then p'(3)=27 - 7 r3<0, which implies that r3>0.

RonL

p(x) = (x + 3)(x - 2)(x - r3) = x^3 + (1 - r3)*x^2 - (6 + r3)*x + 6*r3.

so c = 6*r3

p'(-3) = 27 - 6*(1 - r3) + (6 + r3) = 27 - 6 + 6*r3 + 6 + r3 = 27 + 7*r3 < 0.

so -7*r3 > 27 or r3 < -27/7.

So c < -18 leaving -27 to be the only possible answer.

My question is if it's possible to conclude that r3 is < -3 right off the bat without doing any of thw work above.