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What Geezer wrote is not entirely correct. Abel showed that there exist polynomials of any degree greater than or equal to 5 whose roots can not be expressed by radicals and the four basic arithmetic operations. This does not mean that a given polynomial of degree greater than or equal to 5 cannot be solved by radicals and the 4 basic arithmetic operations. For instance, x^5 + 3x^3 + 2x^2 + 3x can be factored as x(x^2 + x + 1)(x^2 - x + 3) and so all of its roots are expressible by radicals and the 4 basic arithmetic operations (use the quadratic formula).

To know exactly when a polynomial's roots can be expressed in this nice way, one must go past the work of Abel to the work of Galois. This will take you into group and field theory, so if you have a couple years...

2006-06-28 08:18:22 · answer #1 · answered by Anonymous · 1 0

It is impossible to have a formula that produces the roots of a 7th deg, polynomial. In fact, it was proved by the Norwegian mathematician Abel that no such formula can exist for polynomials of 5th degree or higher. If the coefficients are rational (whole numbers or fractions), you can see if there are any rational roots by testing candidates from a finite list of possibilities (h. s. algebra). Otherwise, approximation methods, such as Newton's method(elem, calculus) give approx. roots.

2006-06-28 14:58:34 · answer #2 · answered by geezer 2 · 0 0

By hand would be tough lots of factoring and if it wasnt a clean fraction Ouch its gunna be rough

my advice use a graphing calculator and solve for the roots will take 3 seconds

TI 83 or HP 49 both will do it

2006-06-28 14:44:13 · answer #3 · answered by Aaron G 2 · 0 0

7 or 5 or 6..etc doesn't matter..methods are same..
try finding few roots using the factor theorem..if the expression is simple..then simplify it by dividing with (x- root found out).
If this is not the case you can use successive susbtitution , newton raphson or a solver to find roots.
There are many such numerical methods.
I can't think of any analytical method, other than using factor theorem.

2006-06-28 14:42:28 · answer #4 · answered by Vivek 4 · 0 0

depending on the polynomial, you may not be able to find the roots like you are with the quadratic, cubic, or quartic methods. If you are able to find three roots, then you would be able to use the quartic method of finding the roots, it's not pretty, but it works.

Here is a link on solving the quartic:

http://mathworld.wolfram.com/QuarticEquation.html

But remember, you may not be able to get it down to this point.

2006-06-28 14:48:21 · answer #5 · answered by Eulercrosser 4 · 0 0

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