Finding number of roots by observation

Question

So here is a question : Given $ p(x) = x^4 + 4x^3 -4x-13 $ Then find number and position of real roots of $p(x)$ .

So far, using Descartes rule of signs, I can say it has at most one positive root or at most three negative roots.

The possibility of having all four roots imaginary is ruled out since product of roots is negative.

So either all the roots are real 3 negative and one positive or two roots are real one being positive and other being negative.

Now I can't decide which of the two possibility is correct. And about position of roots I am completely clueless. I can't find the tag for homework and exercises. Please edit and add the tag.

Answer 1

From Descartes, there is exactly one positive root, and at least one negative root (alternately observe $p(0) < 0$ and $\lim_{x\to \pm \infty} p(x) > 0$). The question remains whether the remaining two roots are negative or complex.

For this, note $p(x) = (x^3-4)(x+4)+3$, and for negative $x$, a sign change can come only from the $(x+4)$ term which is positive close to $0$ and negative far from it, so it can happen only once. Hence there can be only one negative root.

Of course a similar argument could have been used to conclude there is only one positive root, as the sign change can come from only $(x^3-4)$ in that regime.

P.S. the above gives the positions of the root wrt the origin. One can of course test signs across intervals around $x=-4$ and $x=\sqrt[3]4$ to narrow down the position.

Answer 2

You have upper bounds from Descartes. On the other hand $p(0)<0$ and $p(x)>0$ as $x\to \pm\infty$, shows that there is at least one positive and at least one negative root (this step could replace your argument that there cannot be four non-real roots). We still do not know if there are three or just one negative root, and to decide this, we may try new methods:

We have the derivative $p'(x)=4x^3+12x^2-4=4\cdot(x^3+3x^2-1)$. At points where $p'(x)=0$, we have $$ p(x)=p(x)-\frac x4p'(x)=x^3-3x-13$$ and you can show that this negative for $x<0$. Smarter, and following a comment by Macavity, at points where $p'(x)=0$, we have $$p(x)=p(x)-\frac {x+1}4p'(x)=-3x^2-3x-12 =-3(x-\tfrac12)^2-11\tfrac14\le -11\tfrac14<0.$$ If $p$ had three negative roots, it would have a local maximum between them (or at a multiple root) with non-negative value.