Dear Uncle Colin,

I need to factorise $14\cos^2(x)+ 19\cos(x) + 6$ by hand. How would you do it?

- Stupid Polynomials Leaning Into Trigonometry

Hi, SPLIT, and thanks for your message!

In the olden days, before my students taught me better, I would have reached directly for the quadratic formula here. That’s still a good approach:

$\cos(x) = \frac{-19 \pm \sqrt{361 - 24\times 14}}{28}$, which works out to $\frac{-19\pm5}{28}$.

The solutions are $-\frac{6}{7}$ and $-\frac{1}{2}$, so the factorisation is $(7\cos(x)+6)(2\cos(x)+1)$ almost directly. In honesty, that’s probably less trouble than doing it “properly”.

If you do want to do it properly, I salute you.

You can substitute $y = 14\cos(x)$ and multiply the whole thing by 14 to get $y^2 + 19y + 84$, which factorises as $(y+7)(y+12)$. Converting back gives $\frac{1}{14}(14\cos(x) + 7)(14\cos(x)+12)$, and dividing out a 7 from the first bracket and a 2 from the second to get the same factorisation as before.

You can also use a little logic if you know that it factorises: it needs to be of the form $(a\cos(x) + b)(c\cos(x)+d)$, with $a$, $b$, $c$ and $d$ all integers. In fact, they need to be positive integers so that the coefficients work out.

Matching coefficients, we’ve got $ac = 14$, $ad + bc = 19$ and $bd = 6$.

Suppose $a$ is even – then $b$ can’t be even, or else the 19 equation wouldn’t work, so $b$ is odd, and $d$ is even (because 6 isn’t a multiple of 4).

That means $ad$ is a positive multiple of 4. It can’t be 4 (because $bc$ would be 15 and it doesn’t have a factor of 5), or 8 (because $bc$ would be 11). It could be 12 with $bc=7$, and it can’t be 16 (because at least one of them must be a multiple of 7).

So $ad=12$ and $bc=7$. Looking at the possibilities, that means $a=2$, $c=7$, $b=1$ and $d = 6$ – again, just like we had before.

Hope that helps!

- Uncle Colin