roots of polynomial of 4th order

Posted on 2004-03-31
Last Modified: 2007-12-19
do you know
1) how to determine roots of polynomial a*x^4+b*x^3+c*x^2+d*x+e=0,
(some method for it), inclusive complex roots
2) this polynomial is 3D, does it make any difference in number of roots?
3) how to display this roots in 3D graph, inclusive complex roots

I'd like to compute and draw it in Maple. Could you write me convenient commands to this problem?
Question by:xLeon
  • 5
  • 3
  • 2
  • +3
LVL 84

Expert Comment

ID: 10725953
What do you mean by "this polynomial is 3D"?

Author Comment

ID: 10725993
I meant polynomial of two variables, but the variable y plays no role. It looks like a wave.
LVL 10

Expert Comment

ID: 10728316
Do you mean to say that the x-y-z relation is:

 z = a*x^4 + b*x^3 + c*x^2 + d*x + e + 0y

Free Tool: SSL Checker

Scans your site and returns information about your SSL implementation and certificate. Helpful for debugging and validating your SSL configuration.

One of a set of tools we are providing to everyone as a way of saying thank you for being a part of the community.

LVL 84

Assisted Solution

ozo earned 20 total points
ID: 10733289

Assisted Solution

Sergio_Hdez earned 40 total points
ID: 10739285
The root are four imaginary numbers, so they are four points inthe way (x,y), so no 3D is possible here.

You have only 3 possibilities in roots:

Four real solution (x,0): You have four points in the X axis.
A pair of real solution (x,0) and a pair of imaginary solutions (x,y) and (x,-y)
Two pairs of imaginary roots, each pair in the form (x,y) (x,-y)

Anyhow, they are always 4 points in a plane X-Y, no much 3D to display!

It doen's matter how many Y, Z or other variables are left unused in the formula: They don't existis for us as they are null in the formula.
LVL 27

Assisted Solution

aburr earned 40 total points
ID: 10753377
All quartic equations can be solved, but with difficulty. Put the equation in normal form (as you have written it,  but with the coefficient of the first (^4) term = 1, ie divide by a). The equation then becomes

x^4 + ax^3 + bx^2 + cx +d = 0

with the letters redefined. If that formula is set = 0 all the solutions (real or complex) agree with the solutions to

x^2 + (a + D)/2 +(y + [ay - c]/D) = 0


D = + or - (8y + a^2 - 4b)^1/2

and y is any real solution of the cubic equation

8y^3 - 4by^2 + (2ac - 8d)y + (4bd - [a^2]d - c^2) = 0

Author Comment

ID: 10776695
Sorry, I have been in a hospital for all the week.
Thank you everybody for partial answers. I will appreciate them later. So far nobody has written nothing in maple.
I must specify my question: The function is of one variable. Only the graph is 3d.
I can draw the graph. That is command
plot3d(function, x=a..b, y=c..d),
Relevant is how to compute the roots of the function and display all of the roots (inclusive complex) in the graph.
I will apprreciate partial answers too. But I must wait, if somebody know more.

Accepted Solution

michaelbrewitt earned 50 total points
ID: 10790659
Don't quite get this. The function is one dimensional. Only roots can be plotted on a 2-d graph i.e. with the x-axis real and the y-axis imaginary.

its easy to solve a polynomial in maple.

use something like

>solution =: [solve(x^3=1,x)];

solution[1], solution[2], solution[3] can then be plotted on a 2-d graph uising the real and imaginary parts of each solution.

The help in Maple is very good with lots of examples.

Author Comment

ID: 10793795
thanks to you  I have used:
 solution := [evalf(solve(poly=0,x))]
and we have gained the list of roots of our polynomial
(e.g.[5, 6, 1 + 5I, 3 + 7I]) but there are other problems.

I can't figure out
1) how to display the polynomial-function and all its roots in one 3d graph.
2) independently on the number of roots
3) we must probably extract the real and imaginary part from  the complex root  in order to display it.


Expert Comment

ID: 10794644
1) Why a 3d graph?? The polynomial has one variable => graph of f(x) on y-axis against x on x-axis. The real roots are then where f(x) cuts x-axis. Can't show imaginary roots on there too because f(x) is a real polynomial

2)not sure what you mean by this

3) if, say a:=3+8I then to extract real/imaginary parts use:

be careful to use "Re" and "Im" exactly i.e. "re" "im" wont work.

Author Comment

ID: 10796007
1) I don't know why, but it should be possible. That is very strange assignment, I can't figure out
2) I'm doing a Maplet: input is polynomial, first output are roots, and second output is 3d graph in Plotter with roots

I wouldn't ask if I would understand it.

Expert Comment

ID: 10796533
Don't know where to go from here unfortunately.

Good luck.

Expert Comment

ID: 10948995
I think I got the idea of what "3D" means here...

First, you have a 4th grade polinomy, with X being the variable, but X is NOT a real variable, it is an imaginary variable (as long as they ask your for the roots, they are supposing the roots exists, so the variable X have to be imaginary), so you have a function value in any (x,I*y) imaginary number, asociated with (x,y) point of the X-Y plane.

Defining z value as the value of the polinomy at any point (x,y), assuming it means polinomial value on (x+Iy), could give you almost what you are searching, a 3D grid, BUT z will be also an imaginary value, in the form zr+I*zi, or (zr, zi) if you prefer, so you have a 4th dimensional graph compossed by all the point in the form (x,y)(zr,zi) having P(x,y)=(zr,zi):

z = (zr+I*zi) = P(x,y) = P(x+y*I) = a*(x+y*I)^4+b*(x+y*I)^3+c*(x+y*I)^2+d*(x+y*I)+e

You need to take out one of those dimensions but without changing the roots, so I propose to define z value as the length of the vector (zr, zi), or the module of the imaginary number zr+I*zi (is the same), so z=sqrt(zr^2+zy^2)=|P(x,y)|

z = |P(x,y)| = |P(x+y*I)| = |a*(x+y*I)^4+b*(x+y*I)^3+c*(x+y*I)^2+d*(x+y*I)+e|

Now you have a 3D grid that represent the polinomy, and 4 points (x,y) that are the roots of the polinomy: They are just the points in the 3D grid where the grid lays on the X-Y plane, having z=0, so plotting them is just a matter of highlighting 4 points in the 3D grid.

There are also other ways to represent P(x,y) without using z=|P(x,y)|, but as they involve 4 dimensions, visualizing can be "obscure".

Author Comment

ID: 10951936
sorry, maybe I will dissapoint you, but 3d was meant as a 2d curve in a 3d space. As I have learned later from my schoolmates. This assignment was very vague and fast nobody didn't grasp it. I was moreover ill, without contact of my schoolmates.

thanks for your time

Featured Post

Free Tool: Site Down Detector

Helpful to verify reports of your own downtime, or to double check a downed website you are trying to access.

One of a set of tools we are providing to everyone as a way of saying thank you for being a part of the community.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Suggested Solutions

Title # Comments Views Activity
Loan to Value to interest rate relationship formula 1 188
Graph function 4 93
x2 – x2 = x2 – x2 conundrum 12 72
How to write an English digest paper 6 76
Foreword (May 2015) This web page has appeared at Google.  It's definitely worth considering! How to Know You are Making a Difference at EE In August, 2013, one …
We are taking giant steps in technological advances in the field of wireless telephony. At just 10 years since the advent of smartphones, it is crucial to examine the benefits and disadvantages that have been report to us.
Although Jacob Bernoulli (1654-1705) has been credited as the creator of "Binomial Distribution Table", Gottfried Leibniz (1646-1716) did his dissertation on the subject in 1666; Leibniz you may recall is the co-inventor of "Calculus" and beat Isaac…
Finds all prime numbers in a range requested and places them in a public primes() array. I've demostrated a template size of 30 (2 * 3 * 5) but larger templates can be built such 210  (2 * 3 * 5 * 7) or 2310  (2 * 3 * 5 * 7 * 11). The larger templa…

820 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question