# The Evolution of Polynomials

It’s always fascinating to see what’s going on in textbooks of the olden days, and National Treasure @mathsjem recently found a beauty of its type. Look at those whences! Check out the subjunctives! It thrills the heart, doesn’t it? ((If one were so inclined, one might even shiver in ecstasy.))

What caught my attention, though, was *evolution* - in this context, taking square or cube (or, presumably, higher) roots of an expression.

I know! There's a whole chapter on each of them. Looks complicated... pic.twitter.com/DWPH4e96Ip

— Jo Morgan (@mathsjem) January 3, 2018

I haven’t studied this method in detail - it looks similar in flavour to finding square roots by long division, which International Legend @colinthemathmo has discussed, at least for numbers .

However, it got me thinking: how would I find the cube root of $c(x) = 27 + 108x + 90x^2 + 80x^3 - 60x^4 + 48x^5 - 8x^6$?

### Quick and dirty

If we’re allowed to assume that $c(x)$ is a perfect cube, then it’s not tricky at all: it’s a degree-6 polynomial, so its cube root must be quadratic; the constant term is clearly 3 and the $x^2$ coefficient is just as clearly -2.

That means $c(x) = \br{ 3 + bx - 2x^2}^3$, for some value of $b$.

Assuming $x$ is small enough that we can ignore $x^2$ terms and higher, expanding the bracket with the binomial expansion gives $c(x) \approx 27 + 27bx$, and $b=4$.

Expanding the whole thing properly gives the same answer - which I leave as an exercise.

### Binomial expansion

As an alternative to assuming the form of the solution, it’s also possible to use the general binomial expansion (at least for small $x$) to figure out the cube root of $c(x)$.

Suppose we write $c(x) = 27 + x\br{108 + 90x + 80x^2 + …}$ and work out $\br{c(x)}^{\frac{1}{3}}$?

The general form for $\br{a+bx}^{\frac{1}{3}}$ is $a^{\frac{1}{3}} + \frac{1}{3}a^{-\frac{2}{3}}bx - \frac{1}{9}a^{-\frac{5}{3}}b^2x^2 + …$

That looks a mess; however, $a$ behaves quite nicely here and simplifies the whole thing down to $3 + \frac{1}{27} bx - \frac{1}{3^{7}}b^2 x^2 + …$

There’s a bit of book-keeping to do here, because $b$ is a bit of a monster, but stick with it: $\frac{1}{27} bx = 4x + \frac{10}{3}x^2 + …$, while $\frac{1}{3^7} b^2x^2 = \frac{108^2}{3^7} x^2 + …$.

What’s $\frac{108^2}{3^7}$? I don’t know off the top of my head. I do know that $108 = 2^2 \times 3^3$, so $108^2 = 2^4 \times 3^6$ …

#### A ninja!

*Whoosh*

“It’s 11,664, because $(100+k)^2 = 10,000 + 200k + k^2$.”

*Whoosh*

#### As I was saying

… which leaves us with $\frac{16}{3}$.

So, we have an answer – assuming we only need to go up to the $x^2$ term – of $3 + 4x + \br{\frac{10}{3}- \frac{16}{3}}x^2$, which works out to $3 + 4x - 2x^2$ again.

### Checking

The only problem left is checking whether we really *have* gone far enough. One option is to cube the answer, but you don’t really want to be doing that every time you extract a term.

Presumably, something in the original method given in Jo’s Victorian textbook checks for a remainder, but I don’t really feel like wading through it to figure it out. Perhaps you, dear reader, can enlighten us all?