Solve this 4th degree polynomial?

Question

How do you find the roots for a 4th degree polynomial? (Or even third degree for that matter.) Here's the equation I need to solve:

r^4 - r^2 - 2r + 2 = 0

Answer 1

There are methods for solving cubic and quartic equations, e.g. Cardano's Rule for a cubic. I don't think that you will ever have to deal with them.

When all the coefficients are integers, as is the case with yours, then you can reduce the possibilities for rational solutions. A rational solution will take the form

(divisor of constant coefficient)/(divisor of leading coefficient).

So for your problem, you should look at +1, -1, +2, -2 (all possibilities of a divisor of 2 over a divisor of 1) to see if you can pick out a root. (r=1 is a solution.) Then, divide out the factor via long division or synthetic division, and work with what you have left. So, you would divide (r^4 - r^2 - 2r - 2) by (r -1) and work with the resulting cubic. Keep in mind that r=1 may work again on the cubic! Once you get it down to a quadratic, you have the Quadratic Formula at your disposal to get any remaining solutions.

Answer 2

The solutions are
r = 1 (double root)
r = -1 + i
r = -1 - i

Here are the formulas for finding the roots of a quartile polynomial:
http://www.josechu.com/ecuaciones_polinomicas/cuartica_solucion.htm

Answer 3

Do you mean factorise or locate the zeroes? In the two case, replace x^2 with the aid of u to grant 9u^2 - 5u + 4 it is easy to factorise, then sub decrease back u=x^2 and factorise each and each bracket