Possible values of c in the polynomial $x^3+ax^2+bx+c$.

Question

I'm studying for a GRE and this is a question from an old test:

Let $p(x)$ be the polynomial $x^3+ax^2+bx+c$, 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 $c$?

Of the options given (-27, -18, -6, -3, -0.5), I have already solved the problem to see that -27 is the only possible value given, but my solution is lengthy, to say the least, and this is the GRE, so ideally problems should be solvable in approximately 2.5 minutes. I was wondering if anyone could think of faster ways to see how to solve this problem that avoids the copious amount of arithmetic I sort through to find the answer?

Here's my solution:

$$p(-3)=-27+9a-3b+c=0=8+4a+2b+c=p(2)$$ and so we may see that $a=b+7$ after several steps of time consuming arithmetic. Then, $$p'(x)=3x^2+2(7+b)x+b$$ $$\Rightarrow p'(-3)=27-6(7+b)+b<0 $$ and so we get that $b>-3$ after several more steps of arithmetic. Now, since $p$ has two real roots in $\mathbb{R}$, it must split completely in $\mathbb{R}$, and so $$p(x)=(x+3)(x-2)(x-\alpha)$$ for some $\alpha \in \mathbb{R}$. Expanding this expression we get, $$x^3+x^2(1-\alpha)+x(-\alpha-6)+6\alpha $$ $$\Rightarrow a=1-\alpha\\ b=-\alpha-6\\c=6\alpha $$ And so, finally, since $b>-3 \Rightarrow \alpha<-3 \Rightarrow c<-18$. And so we're done.

This isn't a particularly challenging problem, in my opinion, but the solution is lengthy even without me including my actual arithmetic. Just writing it out takes longer that 3 minutes, let alone the time built in to actually thinking about the problem/ figuring out where to go next.

So, I feel like there must be a simpler way to think about this problem and approach finding a solution, no?

Answer 1

Since $p$ has positive leading coefficient, the fact that $p(-3) = 0$ but $p'(-3) < 0$ implies that $-3$ must be the middle root of $3$ real roots (as the derivatives of $p$ at the first and third roots are positive). Since $-3 < 2$, the remaining root $\alpha$ satisfies $\alpha < -3$---which is the same conclusion that you derived by substituting values into the general monic polynomial. Then, as you write, Vieta's formula for the constant term gives that $c = 6 \alpha < 6 (-3) = -18$.

Answer 2

The value of the information that $ \ p(-3) \ = \ 0 \ $ and $ \ p'(-3) \ < \ 0 \ $ is to tell us that $ \ x \ = \ -3 \ $ is not a "double zero" ("touching intercept"), so this cubic polynomial must have three distinct zeroes, the third of which you called $ \ \alpha \ \ . \ $

It is not actually all that helpful otherwise to work with $ \ p(-3) \ = \ p(2) \ = \ 0 \ \ . \ $ Applying the work you've done, the important result you had (which is what the Viete relations* give you) is $ \ a \ = \ - [ \ (-3) + 2 + \alpha \ ] \ = \ 1 - \alpha \ \ , \ \ b \ = \ (-3)·2 + (-3)·\alpha + 2·\alpha \ = \ -\alpha - 6 \ \ , $ $ c \ = \ - [ \ (-3)·2·\alpha \ ] \ = \ 6\alpha \ \ . \ $

$ ^{*} $ you will find these useful (and time-saving) to know for the GRE or if you get involved with math contests; they can also just be generally "handy" at times

Inserting these into the expression for the derivative function at $ \ x \ = \ -3 \ $ then produces $$ p'(-3) \ \ = \ \ 27 \ - \ 6a \ + \ b \ \ = \ \ 27 \ - \ 6·(1 - \alpha) \ + \ (-\alpha - 6) \ \ = \ \ 15 \ + \ 5·\alpha \ \ < \ \ 0 \ \ . $$

From these, as you found, the third root is constrained to be $ \ \alpha \ < \ -3 $ $ \Rightarrow \ c \ < \ 6·\alpha \ = \ -18 \ \ , \ $ leaving only one correct option for the answer. This by-passes working with the other coefficients of $ \ p(x) \ \ , \ $ which we don't really have any need to know.