## Problem 65 on GR9768

Forum for the GRE subject test in mathematics.
faster
Posts: 1
Joined: Mon Mar 30, 2009 9:48 pm

### Problem 65 on GR9768

Let p(x) be the polynomial x^3+ax^2+bx+c where a,b,c are real constants. If p(-3)=p(2)=0 and p'(-3)<0 which of the following is a possible value of?
a)-27
b)-18
c)-6
d)-3
e)-1/2

The Correct Answer is a): -27.
I dont understand how we are given enough to solve this. Any ideas?

lime
Posts: 129
Joined: Tue Dec 04, 2007 2:11 am
First of all, note that this polynomial is of odd degree and hence
p(+inf) = +inf and p(-inf) = -inf.

Also, since two real roots exist, the third one must also be real.

Finally, since graph is decreasing in point x=-3, the last root must be in interval (-inf, -3).

Then the negative root product = c < -((-3)*(-3)*2) = -18. Only A satisfies this requirement.

origin415
Posts: 61
Joined: Fri Oct 23, 2009 11:42 pm

### Re: Problem 65 on GR9768

I still don't understand this answer: using the rational roots theorem, all the roots must be a divisor of c over a divisor of 1, so both 3 and -2 should divide c, Seeing this I put -18 without much other thought, as -27 I thought should be impossible.

As further evidence, if p(-3) = p(2) = 0, then p(x) = (x + 3)(x - 2)(x - r) where r is the third root. Expanding this gives
p(x) = x^3 + (1-r)x^2 - (r + 6)x + 6r
So c again divisible by 6.

Edit: Nevermind, I looked again at the hypotheses of the rational roots theorem, and it only holds for integer coefficients, a and b could be any real. r = -9/2 produces c = -27 and satisfies all of the other stipulations.

mag487
Posts: 13
Joined: Thu Aug 06, 2009 3:59 am

### Re: Problem 65 on GR9768

p(x) = (x - 2)(x + 3)(x - r) = x^3 + (1 - r)x^2 - (6 + r)x + 6r.
So a = 1 - r, b = -(6 + r), c = 6r.

Also, p'(x) = 3x^2 + 2(1 - r)x - (6 + r). So p'(-3) < 0 implies 27 - 6(1 - r) - (6 + r) = 15 + 5r < 0 implies r < -3 implies c = 6r < -18. So -27 is the only possible answer.