Polynomial division for olympiads

This is part of a series of articles based on my high school research paper from 2021 covering nonstandard approaches to solving olympiad problems. The problems from the Latvian Open Mathematics Olympiad mentioned in the article can be found in the LU NMS archive.

As is well known, expressions like $x^2-7x+10$ and $2x^2+8x+6$ can be factored by first finding their roots. The roots of these two expressions are $\{2,5\}$ and $\{-3,-1\}$, respectively, so they can be rewritten as $(x-2)(x-5)$ and $2(x+3)(x+1).$

More complex polynomials like $3x^3-4x^2-5x+2$ can also be factored. Let’s explore how this can be done and look at some problems from the Latvian Open Mathematics Olympiad where this method is useful.

Before continuing, it’s worth reviewing the terminology.

Rational Root Theorem

In most olympiad problems, the polynomials have three nice properties: there is only one variable, the coefficients are all integers, and there is at least one rational root.

All integers are rational. Any fraction where an integer is divided by another integer is rational. But other numbers, such as $\sqrt2$ or $\pi$, are not rational.

If these assumptions about the polynomial hold, we can use the Rational Root Theorem, which states:

If $\frac rs$ is a root of a polynomial, then the constant term is divisible by $r$, and the leading coefficient is divisible by $s$.

For a simple example, let’s revisit $3x^3-4x^2-5x+2.$ The constant term is $2$, and the leading coefficient is $3$. It seems that the possible values for $r$ are $1$ or $2$, and for $s$ they are $1$ or $3$, which tells us that $\frac rs$ could be $\tfrac13,\tfrac23,1$, or $2$.

But $r$ and $s$ can also be negative, so there are actually eight rational root candidates in total: \[-2,-1,-\frac23,-\frac13,\frac13,\frac23,1,2.\] We can evaluate $3x^3-4x^2-5x+2$ for each of these values of $x$ and discover that $-1,\tfrac13,2$ are all roots. Therefore, \[3x^3-4x^2-5x+2=3(x+1)(x-\tfrac13)(x-2).\]

Polynomial division

But not all roots are rational, and sometimes the list of candidates is so long that finding more than one root takes a lot of time.

Let’s look at $x^3-10x^2+27x-12.$ Applying the Rational Root Theorem, we can determine that one of the roots is $4$ and that there are no other rational roots.

So let’s try dividing the polynomial $x^3-10x^2+27x-12$ by the binomial $x-4$. One of the quickest ways to do this is to split the middle terms of the polynomial so that you get binomials easily divisible by $x-4$.

\begin{aligned} &\,\frac{x^3{\color[rgb]{0.9,0.1,0.3}-10x^2}\quad+27x\quad-12}{x-4} \\ =&\,\frac{ {\color[rgb]{0,0.9,0}x^3-4x^2}\quad-6x^2{\color[rgb]{0.9,0.1,0.3}+27x}\quad-12}{x-4} \\ =&\,\frac{ {\color[rgb]{0,0.9,0}x^3-4x^2}\quad{\color[rgb]{0,0.9,0}-6x^2+24x}\quad{\color[rgb]{0,0.9,0}+3x-12}}{x-4} \\ =&\,\frac{ {\color[rgb]{0,0.9,0}x^2(x-4)}\quad{\color[rgb]{0,0.9,0}-6x(x-4)}\quad{\color[rgb]{0,0.9,0}+3(x-4)}}{x-4} \\ =&\,x^2-6x+3. \end{aligned}

There are also other approaches, for example, the one presented in this video.

We get $x^3-10x^2+27x-12=(x-4)(x^2-6x+3)$. The factor $(x^2-6x+3)$ can be further split up as $(x-3-\sqrt6)(x-3+\sqrt6)$ with the help of the quadratic formula.

Olympiad examples

Latvian Open Mathematics Olympiad 2017/2018, Form 11, Problem 5.
The roots of the equation $x^3-44x^2+623x-2860=0$ are the side lengths of a triangle. Calculate the area of this triangle!

In this problem, we can use the Rational Root Theorem to find all three roots. It’s easier to find one root and then divide the polynomial, since we could find the remaining two sides using Vieta’s formulas.

In both cases, we find the side lengths to be $11,13,20$. Then the area of the triangle, which is $66$, can be quickly found using Heron’s formula: \[\text{area}=\sqrt{p(p-a)(p-b)(p-c)},\quad\text{where}~~p=\frac{a+b+c}{2}.\]

Latvian Open Mathematics Olympiad 2016/2017, Form 9, Problem 5.
Solve in natural numbers the equation $x^3+(x+1)^3=(x+3)^3+1$.

Here you can expand all the parentheses to get $x^3-6x^2-24x-27=0$. According to the Rational Root Theorem, the only natural numbers that could be solutions are $1,3,9,27$, and among them, only $9$ is a root.

Final thoughts

It must be admitted that this topic hasn’t appeared much in Latvian olympiads in recent years. There may be two reasons for that:

• The research paper was written in 2021, which was before the Skola2030 education revamp in Latvia, and now polynomial division is a topic in the Mathematics II curriculum.
• For older students, polynomial division was the Regional Olympiad topic in 2022, with at least one guaranteed problem on it, so currently there isn’t much motivation or many good ideas for including the topic again.

Nonetheless, knowing how to do polynomial division is certainly a valuable skill. As seen in the example problems, if it’s likely to be useful, it’s usually quite obvious, and solving such problems becomes significantly easier.