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**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:

*r*^{4}-2*r*^{2}-3*r*-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(-1)^{2}-3(-1)-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
*r*^{4}-2*r*^{2}-3*r*-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}-2*r*^{2}-3*r*-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
In summary, we have found that the roots of
*r*^{4}-2*r*^{2}-3*r*-2=0

are -1, 2, and .

**Multiple roots**

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
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
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.

**The ***n*th root of a complex number

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
That is, if *x* and *y* are the real an imaginary parts of *w*, we want to
find and is such a way that
and .
In other words, and 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 and .

2. *w*=-1. Now *x*=-1 and *y*=0. We have and .

3. *w*=1+*i*. We have and .

The point of using polar form is that it is very easy to take the *n*th root.
We find
Since is an angle, it is not unique; it is determined only up to a
multiple of . That is, in the polar form of *w*, we could have written
instead of .
Although, we have
if *n*>1. To find all *n*th roots of *w*, you must consider the representations
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 *n*th roots of *w* are then

Example:
*r*^{3}=-1.

We put -1 into polar form
For the third roots, we find

We could have found the same result from the factorization
*r*^{3}+1=(*r*+1)(*r*^{2}-*r*+1).

The quadratic formula gives
for the roots of the second factor.

**WARNING**

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 . In general, the equation *r*^{n}=*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 *r*^{n}=*w* has an
*n*-fold root.

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*Michael Renardy*

*1998-03-04*