Dear Uncle Colin,

I have solved $16x^5 - 20x^3 + 5x + 1 = 0$ by letting $x = \cos(\theta)$ and considering $\cos(5\theta)$ – but how do I explain that there are only three roots?

My Understanding Limited To Interesting Problems Like Integration; Can’t Interpret Trigonometry Yet

Hi, MULTIPLICITY ((That’s a good one!)) and thanks for your message!

For the benefit of those playing along at home, let’s go through the steps fully.

The method for this relies on De Moivre’s theorem:

• Start with the fact that $\br{\cos(5\theta) + \i \sin(5\theta)} = \br{\cos(\theta) + \i \sin(\theta)}^5$
• Expand the bracket on the right using the binomial theorem: $\dots = \cos^5(\theta) + 5\i \cos^4(\theta) \sin(\theta) - 10 \cos^3(\theta) \sin^2(\theta) - 10\i\cos^2(\theta)\sin^3(\theta) + 5\cos(\theta)\sin^4(\theta) + \i \sin^5(\theta)$
• We only care about the real part:
• $\cos(5\theta) = \cos^5(\theta) - 10 \cos^3(\theta)\sin^2(theta) + 5\cos(\theta)\sin^4(\theta)$
• Let $c = \cos(\theta)$ to save on typing, and note that $\sin^2(\theta) = 1- c^2$
• … so we have $\cos(5\theta) = c^5 - 10c^3(1-c^2) + 5c\br{1-c^2}^2$
• … or $\cos(5\theta) = c^5 - 10c^3 + 10c^5 + 5c - 10c^3 + 5c^5$
• … or $\cos(5\theta) = 16c^5 - 20c^3 + 5c$

Hey! That looks familiar!

If we change our original variable to $c$, we now need to solve $\cos(5\theta) = -1$ for $\theta$, and find all of the values of $\cos(\theta)$ that result.

### Solving for $\theta$

The equation $\cos(5x) = k$ typically has ten solutions between $0$ and $2\pi$, but our quintic can only have, at most, five ((Because it’s a quintic.)). We can reconcile this by remembering that we’re ignoring the imaginary part: we could have picked $\cos(\theta) - \i \sin(theta)$ as the basis for our De Moivre expansion and got just the same expression for $\cos(5\theta)$. Each solution for $\theta$ “works” for both $\cos(\theta) + \i \sin(\theta)$ and $\cos(\theta) - \i \sin(\theta)$, so we can discard the lower half of the unit circle without losing any solutions.

If we let $\alpha = 5\theta$, and solve $\cos(\alpha) = -1$ for $0 \le \alpha \le 5\pi$ ((The upper limit needs to be inclusive; we’re discarding only the negative-imaginary parts.)), we get $\alpha = \pi, 3\pi, 5\pi$.

So $\theta = \frac{1}{5}\pi, \frac{3}{5}\pi$ and $\pi$.

But that’s only three solutions, when we wanted five!

“Ah, the line $y = -1$ is tangent to the curve $y=\cos(5\theta)$ there!”.

But that implies six solutions, doesn’t it?

It’s a bit subtle. If we consider $\cos(5\theta)=-1 + \epsilon$, for some small value of $\epsilon$, we get two solutions near the “internal” values of $\theta$ – but only one at $\theta = \pi$, because there’s no “back up” part of the curve. So while it feels like there should be six, there are only five!

### Solving for $c$

After that, it’s straightforward: $\cos\br{\frac{1}{5}\pi} = \frac{\phi}{2}$, $\cos\br{\frac{3}{5}\pi} = -\frac{1}{2\phi}$, and $\cos(\pi) = -1$.

So the solutions for $x$ are $\frac{\phi}{2}$ (twice), $-\frac{1}{2\phi}$ (twice) and -1.

### Other methods?

The solution of $x = -1$ is a fairly obvious one to the original problem.

Dividing out the factor of $(x+1)$ gives $16x^4 - 16x^3 - 4x^2 + 4x + 1 = 0$. Let’s call that left-hand side $f(x)$.

Now, generally, factorising quartics is possible-but-messy. This one looks like it should be nice, though! If I let $y = 2x-\frac{1}{2}$, I know that $y^4$ starts $16x^4 - 16x^3$ and it might be that things drop out cleanly from there.

So, $y^4 = 16x^4 - 16x^3 + 6x^2 - x + \frac{1}{16}$, so $y^4 = f(x) + 10x^2 - 5x - \frac{15}{16}$.

And we are lucky! $y^2 = 4x^2 - 2x + \frac{1}{4}$, so $\frac{5}{2}y^2 = 10x^2 - 5x + \frac{5}{8}$.

That means $y^4 - \frac{5}{2}y^2 = f(x) - \frac{25}{16}$… or $f(x) = y^4 - \frac{5}{2}y^2 + \frac{25}{16}$, which is a quadratic we can solve!

• Multiply by 16: $16y^4 - 40y^2 + 25 = 0$
• Complete the square: $(4y^2-5)^2 = 0$

Gosh! It’s a perfect square! We get a pair of double roots: $y = \frac{\sqrt 5}{2}$ and $y =-\frac{\sqrt 5}{2}$

But we want $x$, which is $\frac{2y+1}{4}$, giving $x = \frac{1 \pm \sqrt{5}}{4}$, each twice – and those correspond to the roots we got the other way.

(For clarity: I wouldn’t have attempted this if I hadn’t known it came out clearly, and don’t recommend it as a general method.)

Hope that helps!

- Uncle Colin

• Many thanks to @mrsouthernamths for the prompt, and for setting me straight when I made an error.