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\section*{How to find roots of polynomials}

{\bf Integer roots}
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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.
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Example:
$$r^4-2r^2-3r-2=0.$$
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.
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{\bf Reducing the degree}
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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 
$$r^4-2r^2-3r-2=0$$
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
$$(r^4-2r^2-3r-2)/(r+1)=r^3-r^2-r-2,$$
and
$$(r^3-r^2-r-2)/(r-2)=r^2+r+1.$$
Therefore, the remaining roots must solve
$$r^2+r+1=0.$$
The quadratic formula gives 
$$r=-{1\over 2}\pm i{\sqrt{3}\over 2}.$$
In summary, we have found that the roots of
$$r^4-2r^2-3r-2=0$$
are $-1$, 2, $-1/2+i\sqrt{3}/2$ and $-1/2-i\sqrt{3}/2$.
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{\bf Multiple roots}
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It can happen that a polynomial contains the same linear factor more than once.
For instance,
$$r^3-r^2-r+1=(r-1)^2(r+1).$$
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 $n$th degree has $n$ 
roots.
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{\bf The $n$th root of a complex number}
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We want to find the roots of the equation
$$r^n=w,$$
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.
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Examples:
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1. $w=1$. In this case $x=1$ and $y=0$. We can choose $\rho=1$ and $\theta=0$.
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2. $w=-1$. Now $x=-1$ and $y=0$. We have $\rho=1$ and $\theta=\pi$.
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3. $w=1+i$. We have $\rho=\sqrt{2}$ and $\theta=\pi/4$.
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The point of using polar form is that it is very easy to take the $n$th 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 $n$th roots of $w$, you must consider the representations
$$w=\rho\exp(i(\theta+2k\pi))$$
where $k$ takes the integer values $0$, $1$, ..., $n-1$ (if you go further,
then $k=n$ will give you the same as $k=0$). The $n$th roots of $w$ are then
$$\sqrt[n]{\rho}\exp(i(\theta+2k\pi)/n).$$
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Example:
$$r^3=-1.$$
We put $-1$ into polar form
$$-1=\exp(i\pi)=\exp(3i\pi)=\exp(5i\pi).$$
For the third roots, we find
$$r_1=\exp(i\pi/3)=\cos(\pi/3)+i\sin(\pi/3)={1\over 2}+i{\sqrt{3}\over 
2},$$
$$r_2=\exp(3i\pi/3)=\exp(i\pi)=-1,$$
$$r_3=\exp(5i\pi/3)=\cos(5\pi/3)+i\sin(5\pi/3)={1\over 2}-i{\sqrt{3}\over 
2}.$$
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We could have found the same result from the factorization
$$r^3+1=(r+1)(r^2-r+1).$$
The quadratic formula gives
$${1\over 2}\pm i{\sqrt{3}\over 2}$$
for the roots of the second factor.
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{\bf WARNING}
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Beware of the difference between
$$(r+1)^3=0$$
and 
$$r^3+1=0.$$
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 $r^n=w$
always has $n$ {\bf DIFFERENT ROOTS}. Be prepared for a whipping with the
cat of nine tails if you should ever say that the equation $r^n=w$ has an
$n$-fold root.
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