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As all of us know, the easiest method currently available in finding the zeroes of a polynomial is based in factorisation of polynomials.

I tried to find a faster way of calculating the zeroes of a quadratic polynomial, but ended up getting a trivial rewrite of the quadratic formula :

If $f(x) = ax^2+bx+c$, then the zeroes of the polynomial $f(x) = \frac{-b}{2a} \pm \sqrt{f(\frac{-b}{2a})\times \frac{-1}{a}} $

Looking at a linear polynomial $ax + b$, $x = \frac{-b}{a}$ is its zero.

Observing the above forms, we can see that the denominator of the zeroes get multiplied by the polynomial's degree (in both the polynomials) and an extra term comes where the multiplicative inverse (reciprocal) of the degree is its power (as in the case of the quadratic polynomial).

Now, my doubt is : can the zeroes of any polynomial be found using such forms (as given above)? Maybe some $\frac{-b}{na} (n = \text{degree of the polynomial})$ can be used to deduce the zeroes faster ? Or are all the formulae to find the zeroes of a polynomial of a specific degree (be it $2,3,4,5.... $etc.) based on such forms ?

I have only heard of the ways to calculate the zeroes of the polynomials of degree $1$ to $3$ and nothing more, since I am a tenth grader. That's why I am asking this.

Spectre
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  • I won't be here always, so I apologise as early as possible for any late comment I make. – Spectre Sep 17 '20 at 03:34
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    "the easiest method currently available in finding the zeroes of a polynomial is based in factorisation of polynomials" I can understand how you would be lead to think this, but it is not true. The best* general methods are numerical, meaning that they are algorithms that obtain approximate roots. You can read much more here: https://en.wikipedia.org/wiki/Root-finding_algorithms#Roots_of_polynomials Since you know some calculus, you may find this interesting to work through: https://en.wikipedia.org/wiki/Newton%27s_method (*"best" really depends on what you want the roots for.) – Elle Najt Sep 17 '20 at 03:37
  • Thanks, @LorenzoNajt ... If you are confident you may post that comment as an answer. – Spectre Sep 17 '20 at 03:39
  • Well, @LorenzoNajt , going through iterations may take time, right ? What if we want the answers as fast as we can ? – Spectre Sep 17 '20 at 03:41
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    Newton's method is actually incredibly fast outside of pathological cases, although of course you program a computer to do it. There is a whole field of study dedicated to these methods, called Numerical Analysis, which I unfortunately don't know a lot about. – Elle Najt Sep 17 '20 at 03:48
  • I actually meant to do paper-pen calculations. BTW, thanks for your inclination to reply. – Spectre Sep 17 '20 at 10:54
  • @Spectre Generally, I'm thinking of using the Argument Principle (https://en.wikipedia.org/wiki/Argument_principle). You can find help here (https://math.stackexchange.com/questions/3368268/help-with-finding-the-number-of-roots-with-an-interval-using-complex-contour-int). Notify me if you want me to explain more in detail. – DUO Labs Sep 17 '20 at 12:38
  • @LorenzoNajt Isn't Newton's method bad if you reach extrema? – DUO Labs Sep 17 '20 at 12:40
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    @Spectre You could also check this out: https://en.wikipedia.org/wiki/Root-finding_algorithms#Roots_of_polynomials – DUO Labs Sep 17 '20 at 13:00
  • Thanks , friends ! – Spectre Sep 17 '20 at 14:46
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    @DUO Yeah, that's a good point. There are a lot of failure modes of Newton's method (i.e. 'pathological cases' :p ). I didn't mean to suggest that it's perfect, but it's a good one to learn. – Elle Najt Sep 17 '20 at 16:20
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    I thank you all for clearing my doubt!!! – Spectre Sep 17 '20 at 16:57
  • You might be interested in Unknown Quantity: A Real and Imaginary History of Algebra by John Derbyshire – Acccumulation Sep 18 '20 at 00:42
  • Thanks, @Acccumulation .... I will try to go trough it... – Spectre Sep 18 '20 at 02:22
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    Wow, 1k views in a day, that's popularity :) – A-Level Student Sep 18 '20 at 08:04
  • Not really.... That is a problem of high research importance. Why don't you try it ? I am 15 and trying to do a bit of research on it (mathematically, I mean. This is an unofficial research that I am doing). – Spectre Sep 18 '20 at 08:18

3 Answers3

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Now, my doubt is : can the zeroes of any polynomial be found using such forms (as given above)?

You are asking a very deep question - one of the main problems of mathematics in the 18th and 19th century was to figure out if the roots of any given polynomial can be calculated by means of a formula (or formulas) that depends only on the coefficients of a polynomial and uses only the operations of multiplication, division, addition, subtraction and taking any $n$-th root. We say that such formulas are in terms of radicals.

It was known that for polynomials of degree $1$, $2$, $3$ and $4$ formulas in terms of radicals exists, and you may look them up on e.g. Wikipedia. The formula for degree $4$ polynomials is especially elaborate.

However, for degree $5$ and higher such a formula in terms of radicals was not known, and in fact it was proven by Ruffini, and later Abel, that such a formula does not exist.

A bit later Galois proved the same fact in a very elegant way, creating in the process a theory that bears his name, Galois Theory, that 200 years later is still a very active research area, unfortunately beyond the reach of 10th grade mathematics.

However, if you are interested in a proof of this fact, there is a textbook aimed at high-school students in which the proof is presented through a series of problems that introduce basic group and field theory and complex analysis. The book is called

Abel’s Theorem in Problems and Solutions

by V.B. Alekseev (it's maybe worthwhile to note that the book is based on the lectures of V. Arnold who delivered them at one of the Moscow State Schools specializing in mathematics, so it will have a heavier maths course load than a typical American high school, though the material is still very accessible).

Randy Marsh
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    I think it is worth adding that there are formulae for higher order polynomials, just not using radicals. Eg using elliptic functions or theta functions. As a teenager the way the insolubility of the general quintic was put to me confused me quite a bit later until I realised this and elliptic functions, at least, are not all that weird or unnatural. – Francis Davey Sep 17 '20 at 21:13
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    (+1) Very nicely summarized. Well done! – Mark Viola Sep 23 '20 at 15:26
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Now, my doubt is : can the zeroes of any polynomial be found using such forms (as given above)? Maybe some $\frac{-b}{na} (n = \text{degree of the polynomial})$ can be used to deduce the zeroes faster ?

Kind of, yes!

As you know, the quadratic formula can be used to find the roots of any quadratic polynomial. And as you've noticed, the quadratic formula (as you've written it) consists of the term $\frac{-b}{na}$, where $n$ is the degree of the polynomial (which is 2), plus some variable stuff. (By "variable stuff," I mean the part that looks like $\pm \sqrt{\text{something}}$, where choosing a plus sign gives you one root and choosing a minus sign gives you the other one.)

Does the pattern continue? Yes, it does! There is also such a thing as the cubic formula, which can be used to find the roots of any polynomial of degree 3. The cubic formula is somewhat complicated, but it does in fact consist of the term $\frac{-b}{3a}$ plus some "variable stuff," just like the quadratic formula.

After that, there's the quartic formula, which is even more complicated, but which can be used to find the roots of almost any polynomial of degree 4. Just as you would expect, the quartic formula consists of $\frac{-b}{4a}$ plus some "variable stuff".

Unfortunately, there's no formula which can be used for finding the roots of any polynomial of degree 5—at least, there isn't one that uses only the "ordinary" kinds of functions that are used.

Nevertheless, the pattern does continue. The average of all of the roots of a polynomial is always $\frac{-b}{na}$, where $n$ is the degree of the polynomial, $a$ is the coefficient of $x^n$, and $b$ is the coefficient of $x^{n-1}$. (I actually didn't know this before—you've taught me something!)

Here is the proof. Any polynomial can be written as $a(x - r_1)(x - r_2)\cdots(x - r_n)$, where $n$ is the degree of the polynomial. If we multiply this out, it's possible to see that $b$ (that is, the coefficient of $x^{n-1}$) is $-a$ times the sum of all of the roots. As an equation:

$$b = -a \cdot (\text{sum of roots}).$$

By dividing both sides by $-na$, we see that

$$\frac{-b}{na} = \frac{\text{sum of roots}}{n}.$$

The right-hand side is, of course, the average of all of the roots.

Tanner Swett
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  • But using the average of all roots without having a few other roots won't give a proper solution, will it, @TannerSwett ? – Spectre Sep 18 '20 at 03:55
  • The part where the $n$-th root sign comes is what decides the roots , right ? The average of roots is a constant value, from which a few deviations of the power equal to the multiplicative inverse of the power yields the zeroes – Spectre Sep 18 '20 at 03:58
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Others have pointed out that there is no general solution for polynomials degree 5 and above. Of course, there are certain higher degree polynomials that are easily solvable, like $x^5-32=0$.

And there are methods that yield an inexact solution, such as numerical methods.

Here is an approach to the quadratic that is a bit off the beaten path:

re-arrange $ax^2+bx+c=0$ to get $x = \frac{-b}{a} - \frac{c}{ax}$. Now all you need to know to calculate $x$ is... $x$ itself. The cool thing is that when you assume a value for $x$, you plug that into the formula, and you get an improved estimate. Then use that improved estimate with the formula to get an even better one. What you end up with is called a CONTINUED FRACTION.

You can do something similar by using $x = \sqrt{\frac{-bx}{a} -{c}{a}}$. This gives you a continued radical approximation.

Spectre
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richard1941
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  • I can understand that.. but we need quicker ways, that can be used in general, @richard1941 – Spectre Sep 24 '20 at 04:26
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    Quicker ways exist. For the quadratic x^2-a, there is the well known Newton-Rapson method, x'= (x+a/x)/2. Where x' is an improved estimate of the root. It converges quadratically. A less well known method is lubricated like 3-in-1 oil: – richard1941 Sep 25 '20 at 05:38
  • Nice answer, sir ! – Spectre Sep 25 '20 at 05:48
  • But I am talking about all the polynomials and not a specific one. – Spectre Sep 25 '20 at 05:52
  • Plus, Newton Raphson method requires iterations, which is what I am not expecting to see , @richard1941 . – Spectre Sep 25 '20 at 05:52