Hey Stormey.
You should be able to factorize a cubic polynomial exactly. Take a look at this:
Cubic function - Wikipedia, the free encyclopedia
Hey Stormey.
You should be able to factorize a cubic polynomial exactly. Take a look at this:
Cubic function - Wikipedia, the free encyclopedia
Hi chiro, and thanks for the help.
since I'm dealing with parameters a,b and c, the calculations became much too complicated and messy then I think they should be.
Isn't there another way to conclude that the polynomial of a lower degree will not satisfy the equation?
anyway, couple of things I thought that could work:
according to Cayley–Hamilton theorem, every square matrix satisfies its own characteristic equation, so I tried to solve this equation:
and tried to get some direction from there.
It yielded quite a nice, easy equation:
not sure what now, though.
abc are not the roots. If x3-ax2-bx-c had multiple roots, it would be of the form (x-g)^{2}(x-h) or (x-g)^{3} which have less than 3 constants.
abc could be determined st f(x) had the form (x-h)^{3}, for example, from the identity:
x^{3}-ax^{2}-bx-c = (x-g)^{3} which would give a,b,c in terms of g.
EDIT: Fundamental theorm of algebra says every nth degree polynomial has n roots, some of which may be complex.