# What theorem is being invoked in this solution to deciding if this polynomial is solvable by radicals?

I was reading these solutions online http://campus.lakeforest.edu/trevino/Spring2019/Math331/Homework7Solutions.pdf for practice for an upcoming exam I have . But in question $$2$$ part $$(a)$$ the author uses a trick I'm not familiar with , he takes the derivative of the polynomial $$x^5-12x^2+2$$ , notes that the roots of said derivative are $$0$$ and $$\sqrt{\tfrac{24}{5}}$$ and then from this deduces that there are at most 3 real roots to our original polynomial .

What theorem is the author invoking here ?

• Looks like Rolle's Theorem. – Torsten Schoeneberg Jul 14 at 6:12
• a lot of it is calculus: Rolle's theorem with MVT and IVT with FTA, conjugate pair theorem – user29418 Jul 14 at 6:14
• @user29418 I know MVT stands for mean value theorem , what does IVT and FTA stand for though – excalibirr Jul 14 at 6:17
• @user29418 I think IVT stands for intermediate value theorem ,not sure about FTA still though – excalibirr Jul 14 at 6:21
• FTA is almost surely the Fundamental Theorem of Algebra – Rylee Lyman Jul 14 at 6:35

## 1 Answer

Just to expand on my comment, and to see that no other theorem (certainly not IVT or the Fundamental Theorem) need to be used here:

The standard version of Rolle's Theorem asserts that if a real-valued function is continuous on the interval $$[a,b]$$ and differentiable inside that interval, and $$f(a) = f(b)$$, then there is at least one $$c \in (a,b)$$ with $$f'(c)=0$$.

Now, since any polynomial $$p$$ is differentiable and hence continuous everywhere, Rolle's Theorem asserts that between any two of its zeroes (taking the role of $$a$$ and $$b$$ above), its derivative $$p'$$ must also have a zero. This easily generalises to:

If $$p$$ has $$n$$ different zeroes, then $$p'$$ has at least $$n-1$$ different zeroes

(namely, at least one zero of $$p'$$ between the first and second zero of $$p$$, then at least one between the second and third, ..., and at least one between the $$n-1$$th and the $$n$$th zero of $$p$$, where by "first", "second" etc. I just order the zeroes from left to right on the real number line).

Which, when you think about it for a moment, implies that:

If $$p'$$ has $$n-1$$ different zeroes, then $$p$$ cannot have more than $$n$$ different zeroes

which is what is used in your case, where $$n=3$$.