How to find roots of polynomials

Integer roots
If the coefficients of a polynomial are integers, it is natural to look for roots which are also integers. Any such root must divide the constant term. We can often ``guess" one or more roots by trying all possibilities.
Example:

r⁴-2r²-3r-2=0.

If there is an integer root, it must divide -2. This leaves only four possibilities: 1, -1, 2, and -2. By plugging in, we find that

$$1^4-2\cdot 1^2-3\cdot 1-2=-5,$$

(-1)⁴-2(-1)²-3(-1)-2=0,

$$2^4-2\cdot 2^2-3\cdot 2-2=0,$$

(-2)⁴-2(-2)²-3(-2)-2=12.

Therefore -1 and 2 are roots, but 1 and -2 are not.
Reducing the degree
If a root has been found, you can reduce the degree of the polynomial to get a simpler problem for the remaining roots. For example, you know from above that

r⁴-2r²-3r-2=0

has the roots -1 and 2. This tells you that the polynomial must contain the factors r+1 and r-2. You can use the long division algorithm to find

(r⁴-2r²-3r-2)/(r+1)=r³-r²-r-2,

and

(r³-r²-r-2)/(r-2)=r²+r+1.

Therefore, the remaining roots must solve

r²+r+1=0.

The quadratic formula gives

$$r=-{1\over 2}\pm i{\sqrt{3}\over 2}.$$

In summary, we have found that the roots of

r⁴-2r²-3r-2=0

are -1, 2, $-1/2+i\sqrt{3}/2$ and $-1/2-i\sqrt{3}/2$.
Multiple roots
It can happen that a polynomial contains the same linear factor more than once. For instance,

r³-r²-r+1=(r-1)²(r+1).

Since the factor r-1 appears twice, we call 1 a double root of the polynomial, while -1 is a simple root. Note that

$${d\over dr}(r^3-r^2-r+1)=3r^2-2r-1$$

also has a root at r=1. This happens in general: If a polynomial P(r) as a k-fold root at r=c, then

P(c)=P'(c)=...=P^(k-1)(c)=0,

but

$$P^{(k)}(c)\neq 0.$$

If roots are counted by multiplicity (i.e. a double root counts twice, a triple root three times etc.), then a polynomial of nth degree has n roots.
The nth root of a complex number
We want to find the roots of the equation

rⁿ=w,

where w is a given number. w may be complex, but the following procedure is important even if w is real. The solution of the equation requires writing w in polar form

$$w=x+iy=\rho\exp(i\theta)=\rho\cos\theta+i\rho\sin\theta.$$

That is, if x and y are the real an imaginary parts of w, we want to find $\rho$ and $\theta$ is such a way that $x=\rho\cos\theta$ and $y=\rho\sin\theta$. In other words, $\rho$ and $\theta$ are polar coordinates of the point (x,y) in the Cartesian plane.
Examples:
1. w=1. In this case x=1 and y=0. We can choose $\rho=1$ and $\theta=0$.
2. w=-1. Now x=-1 and y=0. We have $\rho=1$ and $\theta=\pi$.
3. w=1+i. We have $\rho=\sqrt{2}$ and $\theta=\pi/4$.
The point of using polar form is that it is very easy to take the nth root. We find

$$r=\sqrt[n]{w}=\sqrt[n]{\rho}\exp(i\theta/n).$$

Since $\theta$ is an angle, it is not unique; it is determined only up to a multiple of $2\pi$. That is, in the polar form of w, we could have written

$$w=\rho\exp(i(\theta+2\pi))$$

instead of $w=\rho\exp(i\theta)$. Although, $\exp(i\theta)=\exp(i(\theta+2\pi))$we have

$$\exp(i\theta/n)\neq \exp(i(\theta+2\pi)/n)$$

if n>1. To find all nth roots of w, you must consider the representations

$$w=\rho\exp(i(\theta+2k\pi))$$

where k takes the integer values , 1, ..., n-1 (if you go further, then k=n will give you the same as k=0). The nth roots of w are then

$$\sqrt[n]{\rho}\exp(i(\theta+2k\pi)/n).$$

Example:

r³=-1.

We put -1 into polar form

$$-1=\exp(i\pi)=\exp(3i\pi)=\exp(5i\pi).$$

For the third roots, we find

$$r_1=\exp(i\pi/3)=\cos(\pi/3)+i\sin(\pi/3)={1\over 2}+i{\sqrt{3}\over 2},$$

$$r_2=\exp(3i\pi/3)=\exp(i\pi)=-1,$$

$$r_3=\exp(5i\pi/3)=\cos(5\pi/3)+i\sin(5\pi/3)={1\over 2}-i{\sqrt{3}\over 2}.$$

We could have found the same result from the factorization

r³+1=(r+1)(r²-r+1).

The quadratic formula gives

$${1\over 2}\pm i{\sqrt{3}\over 2}$$

for the roots of the second factor.
WARNING
Beware of the difference between

(r+1)³=0

and

r³+1=0.

The first equation has a triple root at -1, the second has three different roots: -1 and $1/2\pm i\sqrt{3}/2$. In general, the equation rⁿ=w always has n DIFFERENT ROOTS. Be prepared for a whipping with the cat of nine tails if you should ever say that the equation rⁿ=w has an n-fold root.