Explaining the intuition for why finding roots of polynomials is hard

Question

I'm currently teaching a mini-seminar to high school students, most of whom have at most a background in Algebra/Algebra II (in the US high school system) about finding roots of polynomials.

In general, doing so is rather hard, and by the Abel-Ruffini theorem, there doesn't exist a radical (and field operation) based formula for solving general degree five polynomials and up.

I'd like to convey the intuition for why polynomials are so hard to crack, and here's my idea so far.

When I learned the Abel-Ruffini theorem, my intuition was that for "most" polynomials (with full Galois groups), the roots were so symmetric that they almost "protect" each other: i.e., knowing one will unlock the rest.

As an analogy, a symmetric knot on your earbuds is far harder to untangle than an asymmetric knot as it's hard to know where to start. Or maybe Sudoku is a good analogy, as the unknown cells almost protect each other, and symmetric Sudokus (by which I mean ones in which unknown cells are all tied together) are much harder to solve than ones with "gimme" cells.

As a final analogy, I was thinking about systems of linear equations, and how systems with a few variables as far easier to solve than the rest makes for a much easier system in general.

However, all of the above is handwaving nonsense. Is this intuition meaningful? Is there a way of presenting these ideas without confusing the kids?

Answer 1

There is a closed form expression for the roots of a quintic, it is just that the solutions do not always simplify to expressions involving field operations and radicals.

For example, using Wolfram Alpha, one root of the equation $x^5-x+a=0$ is given by $$x_1 =a\phantom{.}{}_4F_3\left(\frac15, \frac25, \frac35, \frac45;\frac12, \frac34,\frac54;\frac{3125a^4}{256}\right)$$ where $_4F_3$ is a generalized hypergeometric function. The other four roots have similar expressions. This approach involves the Bring radical and the reversion of a Taylor series, e.g., finding the Taylor series for $f^{-1}(a)$ where $f(x)=x-x^5$. See this reference.

The exact form solutions for a quintic in terms of hypergeometric functions simplify to expressions in radials only in special circumstances, which can be quantified by algebraic properties of the Galois group of the polynomial. In order for the solutions to simplify in terms of radicals, the Galois group must deviate from the generic case of the full symmetric group $S_5$.

It turns out that by substitutions, polynomials of degree $n=2,3,4,5$ can be reduced to the form $x^n+a_1x+a_0$. For $n=2$ this is called completing the square. This reduction is not possible in general for $n\ge6.$ However, there are formulas for the roots of a general polynomial in terms of theta-functions. You can see these formulas here.

Answer 2

I advocate for an explanation of why the strategies for quadratics don't immediately work for higher orders. When finding the roots of a quadratic, there are two predominant strategies: factoring, and somehow taking its square root.

Using Viete's formulas, we can get sum and product of roots. This is fine for quadratic since we have two roots and two equations, but for higher orders this more or less results in guessing and checking. It's also entirely possible for a real polynomial to have zero rational roots, which evades the rational root theorem. See example

The other strategy of taking the $n$th root requires cleverly manipulating the formula into a perfect power. All quadratics can be written as $0 = \text{multiplier} (\text{power}) + \text{constant}$, but it's not as simple for higher powers because instead of a constant it's another function of $x$, which is not very useful if you are trying to take its $n$th root.

Not sure if you absolutely need an "analogy" to explain the intuition, but I'll edit my answer once I think of an appropriate one.

Answer 3

We can rescale a polynomial so its leading coefficient is $1$ without changing the roots. So let's focus on polynomials with leading coefficient $1$. If its degree is $n$ then it has $n$ coefficients after the leading term. Writing down that polynomial amounts to saying what the $n$ non-leading coefficients are.

When you are given a polynomial of degree $n$ (with leading coefficient $1$) and want to find its roots, you are essentially told the $n$ non-leading coefficients and want to get the $n$ roots from that information. Let's see what system of equations this amounts to for some small values of $n$, starting with $n = 2$.

$n = 2$: when $$ x^2 + bx + c = (x-r)(x-s) = x^2 - (r+s)x + rs, $$ we have $$ b = -r-s, \ \ c = rs. $$ We are told $b$ and $c$ and want to figure out $r$ and $s$. So we have to find $r$ and $s$ such that $$ r + s = -b, \ \ rs = c. $$ This is not a system of linear equations in $r$ and $s$, but with a little algebraic trickery we can "break the symmetry" in $r$ and $s$ to extract $b$ and $c$: $$ (r-s)^2 = r^2 - 2rs + s^2 = (r+s)^2 - 4rs = b^2 - 4c \Longrightarrow r-s = \pm \sqrt{b^2-4c}. $$ Thus we have the system of linear equations $$ r + s = -b, \ \ r - s = \pm\sqrt{b^2-4c}. $$ This can be solved for $r$ and $s$ by standard methods of algebra.

$n = 3$: when $$ x^3 + bx^2 + cx + d = (x-r)(x-s)(x-t) = x^2 - (r+s+t)x + (rs+rt+st)x - rst, $$ we have $$ b = -r-s-t, \ \ c = rs+rt+st, \ \ d = -rst. $$ We are told $b$, $c$, and $d$ and want to figure out $r$, $s$ and $t$. So we have to find $r$, $s$, and $t$ such that $$ r+s+t = -b, \ \ rs +rt+st = c, \ \ rst = -d. $$ As in degree $2$, we are faced with a nonlinear system of equations in the unknowns $r$, $s$, and $t$. With a lot more algebraic work, this can be converted into a system of $3$ linear equations in $r$, $s$, and $t$ using cube roots of unity. But already we see the level of algebra looks more difficult than in degree $2$.

I am going to stop here, because I think it is already clear the math is getting hard and we're already in degree $3$.