roots of polynomial of 4th order

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?
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michaelbrewittConnect With a Mentor Commented:
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.
What do you mean by "this polynomial is 3D"?
xLeonAuthor Commented:
I meant polynomial of two variables, but the variable y plays no role. It looks like a wave.
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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

Sergio_HdezConnect With a Mentor Commented:
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.
aburrConnect With a Mentor Commented:
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
xLeonAuthor Commented:
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.
xLeonAuthor Commented:
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.

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.
xLeonAuthor Commented:
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.
Don't know where to go from here unfortunately.

Good luck.
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".
xLeonAuthor Commented:
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
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