# 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?

• I'm having trouble from your question of discerning whether you want to show that there are at least three real roots or at most three real roots. I'm going to defer to your title though and assume you mean at most three real roots. – jgon Jan 29 '18 at 3:30
• yes sorry, i mean at most three real roots – Alb P Jan 29 '18 at 3:32
• Ok, good, just checking. You just say things like "there must be at least 3 real roots" a few times. – jgon Jan 29 '18 at 3:33
• yeah, just edited the question to fix that up – Alb P Jan 29 '18 at 3:34
• Do you know Descartes' rule of signs? en.wikipedia.org/wiki/Descartes%27_rule_of_signs – kimchi lover Jan 29 '18 at 3:36

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.

• Thank you! This was a clear explanation and was very useful! – Alb P Jan 29 '18 at 3:52

You can use Descartes' Sign Rule. http://mathworld.wolfram.com/DescartesSignRule.html

For positive roots, start with the sign of the coefficient of the lowest (or highest) power. Count the number of sign changes n as you proceed from the lowest to the highest power (ignoring powers which do not appear). Then n is the maximum number of positive roots. Furthermore, the number of allowable roots is n, n-2, n-4, ....

...For negative roots, starting with a polynomial f(x), write a new polynomial f(-x) with the signs of all odd powers reversed, while leaving the signs of the even powers unchanged. Then proceed as before to count the number of sign changes n. Then n is the maximum number of negative roots.

$$f\left( x \right) = {x^5} - x - 16$$ ^^^ 1 sign change, so there is a maximum of 1 positive, real roots. $$f\left( { - x} \right) = {\left( { - x} \right)^5} - \left( { - x} \right) - 16 = - {x^5} + x - 16$$ ^^^ 2 sign changes, so there is a maximum of 2 negative, real roots. So there are a maximum of 2+1 = 3 real roots.

Here's a proof of Descartes' Rule of Signs.

If it has more than $3$ it has $5$, since complex roots come in conjugate pairs...

So multiply $(x-a)(x-b)(x-c)(x-d)(x-e)$ and check the implications...

Using change of sign rule, I.e, 1) number of positive roots of a polynomial equation p(x) cannot exceed the number of changes in the signs of given equation p(x). 2) number of negative roots of a polynomial equation cannot exceed the number of changes in the signs of equation obtained by p(-x)

Applying this, we have atmost 1 positive and 2 negative roots of given equation.

So the equation cannot have 5 real roots.

Just to give a different approach tailored to the specific polynomial in question, note that

$$f(x)=x^5-x-16=x(x^4-1)-16$$

If $x\lt-1$, then $x(x^4-1)\lt0$, so $f(x)\lt-16\lt0$. for all $x\in(-\infty,-1)$. If $|x|\le1$, then $|x^4-1|\le1$ as well, so $|x(x^4-1)|\le1$, hence $f(x)\le-15\lt0$ for all $x\in[-1,1]$. Finally, if $x\gt1$, then $x(x^4-1)$ is positive and strictly increasing, hence can (and does) take the value $16$ exactly once. Thus $f(x)=x^5-x-16$ has exactly one real root, and that real root lies in $(1,\infty)$.