calculus - Confirmation of Proof: $f(x)= x^5 - x - 16$ has at most three real roots.

I've been struggling to find a way to solve this question for a while now.

I can prove that there is at least one real root using IVT but have no idea how to prove that there can exist more.

Currently my idea is to prove that there cannot be $5$ real roots somehow and then I can say that as complex numbers come with conjugates, there cannot only be one complex root and so there can be at most $3$ real roots.

I had another idea to assume that there are at most $2$ real roots. Then use the fact that complex conjugates come in pairs to say there must be at most 3 roots as we have already shown there are at least $2$. But then there would be nothing to account for if there are $5$ real roots.

Am I just overthinking this and there's an easier way to go about it?

Answer

We can actually do this directly. Note that between each pair of adjacent but distinct real roots, the polynomial must have a local maximum or a local minimum, since it has to turn around to get back to zero. Thus between each pair of adjacent and distinct real roots, the derivative of $f$ has a real root. So let's look at $f'(x)=5x^4-1$. We can see immediately that this has the four fourth roots of $\frac{1}{5}$ as its roots: $\frac{1}{\sqrt[4]{5}}$, $\frac{i}{\sqrt[4]{5}}$, $\frac{-1}{\sqrt[4]{5}}$, $\frac{i}{\sqrt[4]{5}}$. Since only two of these are real, $f$ can only have at most three distinct real roots.

All that remains is to check that $f$ has no repeated roots. We can do this by checking that $f$ and $f'$ are relatively prime. Dividing $f$ by $f'$, we get $f(x)=f'(x)(\frac{1}{5}x) -\frac{4}{5}x - 16$.

Then since the remainder is linear, it divides $f'$ if and only if they share a root, but the root of $-\frac{4}{5}x-16$ is the same as the root of $x+20$, and we know $-20$ is not a root of $f'(x)$. Thus $f$ and $f'$ are relatively prime, so $f$ has no repeated roots.

Hence $f$ has at most three real roots.