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### Factoring a Multivariate Polynomial: Strategies Beyond Grouping

```Date: 11/11/2008 at 09:33:32
From: Greg
Subject: Factoring a Multivariate Polynomial

Hi, Dr. Math.

Could you help me factor this polynomial?

x^3 + y^3 + z^3 - 3xyz

Also, could you tell me the general techniques for factoring
multivariate polynomials?  (I'm looking for a technique other than
"grouping," which doesn't work here.)

And how can I tell when it's "fully factored?" Would I have a product
of binomials?

I simply have no idea how to factor multivariate polynomials like
this.  I tried a few things and they didn't work.

The only way I know how to factor multivariate polynomials is to put
the terms in "groups" and get a common factor out of each
"group."  (This is similar to factoring quadratic equations after
splitting the middle term.)  I tried that and I could not get any
common factors.  Also, I looked at the expression as a cubic equation
and tried to find possible roots of x using the rational root
theorem; it got really confusing and I couldn't find the roots.

I don't know what else to try.

```

```
Date: 11/15/2008 at 19:33:16
From: Doctor Jerry
Subject: Re: Factoring a Multivariate Polynomial

Hello Greg,

Thanks for writing to Dr. Math.

I noticed your question but didn't reply because I felt that, besides
the factorization, I had nothing useful to say.

The factorization is

x^3 + y^3 + z^3 - 3*x*y*z
= (x + y + z) (x^2 - x y + y^2 - x z - y z + z^2)

My thoughts were not clear at all.  I noticed that the given
polynomial was symmetric.  One expression that came to mind was x + y
+ z, and I tried that.

I have no background in factoring polynomials in several unknowns.

you have been able to do, and I will try to offer further suggestions.

- Doctor Jerry, The Math Forum
http://mathforum.org/dr.math/

```

```
Date: 11/15/2008 at 20:09:39
From: Greg
Subject: Thank you (Factoring a Multivariate Polynomial)

I checked my book and the answer is

(1/2)(x + y + z)[(x - y)^2 + (y - z)^2 + (x - z)^2]

I don't understand what you wrote. What is a symmetric polynomial,
and why would that imply a factor of x + y + z?

```

```
Date: 11/15/2008 at 22:25:39
From: Doctor Jerry
Subject: Re: Thank you (Factoring a Multivariate Polynomial)

Hello Greg,

Thanks for writing to Dr. Math.

A symmetric polynomial is one for which interchanging x and y, or x
and z, or y and z does not change the polynomial.  One might guess
that the factors will also be symmetric.  That's all I meant.

you have been able to do, and I will try to offer further suggestions.

- Doctor Jerry, The Math Forum
http://mathforum.org/dr.math/

```

```
Date: 11/15/2008 at 22:39:41
From: Doctor Pete
Subject: Re: Thank you (Factoring a Multivariate Polynomial)

Hi Greg,

This is another Math Doctor responding to your question.

First, I should point out that the factorization Doctor Jerry
provided is equivalent to the one in your book, although my
preference is for Doctor Jerry's form.  The reason is that
representing the second-degree factor as the sum of squares is not
themselves -- it is not a "more factored" expression.  Also, the
writing of the second-degree factor in such a way then puts a
constant factor of 1/2 out in the front, which again is unnecessary.
However, they are both legitimate ways of rendering a factorization
of the original polynomial.

Second, a polynomial is symmetric if it is invariant under any
permutation of its variables.  For example, in your case, we have

F[x,y,z] = x^3 + y^3 + z^3 - 3xyz.

There are six distinct permutations of the variables x, y, z:

{x,y,z}
{x,z,y}
{y,x,z}
{y,z,x}
{z,x,y}
{z,y,x}.

If we applied the function to each of these permutations, then we
would still have the same polynomial; i.e.,

F[y,z,x] = F[x,y,z] = F[z,x,y] = ....

This is what we mean by "invariant under any permutation."  Other
examples of symmetric polynomials include

x + y,
x^2 y + y^2 z + z^2 x, and
xyz + yzw + zwx + wxy,

where the number of variables is 2, 3, and 4, respectively.  Doctor
Jerry mentioned symmetry because -- as a general heuristic approach
to factorization -- if a symmetric polynomial can be factored, its
factors are, themselves, likely to be symmetric.  For example,

x^2 + 2xy + y^2 = (x + y)^2

is symmetric, and its factors are symmetric.  Also,

x^3 + y^3 = (x + y)(x^2 - xy + y^2)

is a symmetric polynomial with symmetric factors.

When we look at the structure of F[x,y,z] = x^3 + y^3 + z^3 - 3xyz,
we might try to factor it by comparing it to other expressions that
have easy factorizations that we already know.  For example,

(x + y + z)^3 = x^3 + y^3 + z^3 + 3(x^2 y + y^2 z + z^2 x)
+ 3(x y^2 + y z^2 + z x^2)
+ 6xyz.

This implies

F[x,y,z] = (x + y + z)^3 - 3(x^2 y + y^2 z + z^2 x)
- 3(x y ^2 + y z^2 + z x^2)
- 9xyz.

We can then try to pull out a factor of x + y + z from the remaining
terms, because the first term has such a factor.  We observe that

(x + y + z)(xy + yz + zx) = x^2 y + y^2 z + z^2 x
+ x y^2 + y z^2 + z x^2
+ 3xyz,

so we now have

F[x,y,z] = (x + y + z)^3 - 3(x + y + z)(xy + yz + zx)
= (x + y + z)[(x + y + z)^2 - 3(xy + yz + zx)].

At this point, we have successfully pulled out one factor.  The
remaining term,

(x + y + z)^2 - 3(xy + yz + zx),

simplifies to

x^2 + y^2 + z^2 - (xy + yz + zx),

which gives the result that Doctor Jerry provided.

The lesson is that factorization is something of a guessing game.
There do exist algorithms that computers use to factor polynomials.
These are generally difficult for humans to apply to arbitrary cases,
however; so you have to try different approaches to see if they solve
the puzzle -- and there's no guarantee that any of them will work.

- Doctor Pete, The Math Forum
http://mathforum.org/dr.math/

```

```
Date: 11/16/2008 at 09:39:03
From: Greg
Subject: Thank you (Factoring a Multivariate Polynomial)

Thank you everyone for the help!

```

```
Date: 11/16/2008 at 18:36:41
From: Doctor Vogler
Subject: Re: Thank you (Factoring a Multivariate Polynomial)

Hi Greg,

This is a third Math Doctor.  I was watching your question because
I'm interested in the same subject myself, although my knowledge of
the matter is less than I'd like.

Unlike polynomials in a single variable, multivariate polynomials do
not always factor into linear terms, even over the complex numbers.
In fact, *most* multivariate polynomials are irreducible.  So yours
was one of the ones that does not factor.

Your polynomial is also special in two other ways.  Doctors Jerry and
Pete mentioned that it is symmetric, which is true.  Not all
symmetric polynomials factor into symmetric factors.  For example,

(x + y)(x + z)(y + z)

is symmetric in three variables, but none of the factors is.
However, a symmetric polynomial that factors into terms of different
degrees will have symmetric factors.  The reason is that permuting
the variables doesn't have to leave all of the factors unchanged, but
it does have to leave the collection of factors unchanged, which
means that the most it can do is rearrange the factors.  (Look at how
swapping x and y, for example, will rearrange the factors in my
example polynomial.)

Your polynomial is also homogeneous, which means that every term in
the polynomial has the same degree.  For example,

x^3 + 2xy^2 + y^3

is homogeneous of degree 3.  A homogeneous polynomial always factors
into homogeneous factors (of smaller degree when the polynomial is
not irreducible).

But homogeneous polynomials are really polynomials in one fewer
variable than you see.  For example, a homogeneous polynomial f
(x,y,z) of degree d satisfies the equation

f(rx, ry, rz) = r^d * f(x, y, z),

which means that

f(x, y, z) = z^d * f(x/z, y/z, 1),

so that the equation

f(x, y, z) = 0

is the same as

f(x/z, y/z, 1) = 0.

So we can consider this to be a polynomial in the two variables u = x/
z and v = y/z.

Anyway, back to the point of factoring multivariate polynomials: we
are left with the question of how do you actually do it.  Well, there
are a variety of ad-hoc techniques like those Doctors Jerry and Pete
have described.  Those kinds of things usually work well when you're
doing this by hand and you really expect to get a factorization.
They are less useful for determining that your polynomial is
irreducible. There are other techniques that can determine that your
polynomial is irreducible.  And there are still others that can do
either one.  But some of them require a lot of work.  I am certain
that there are sophisticated algorithms which will work efficiently
even for rather large polynomials, although I am unfortunately not
familiar with such techniques.  Let me describe one which -- though
quite messy, involving lots of variables and equations -- is sort of
like the last method described (for one-variable polynomials) at

http://mathforum.org/library/drmath/view/68494.html

If a degree-3 polynomial factors, then it must factor into a product
of a degree-1 and a degree-2 polynomial.  (There are more cases for
higher degrees, but for degrees 2 and 3, there is only one
possibility.)  So you could argue that if your polynomial is
irreducible, then it must equal

(rx + sy + tz)(ax^2 + bxy + cy^2 + dxz + eyz + fz^2)

for some numbers (r, s, t, a, b, c, d, e, f).  In fact, since both of
these are homogeneous, you can assume that two of these are 1.

For example, you can assume that r = 1 and a = 1 if your polynomial
has 1 for the coefficient of x^3 (otherwise, you just divide your
polynomial by the coefficient of x^3).  Then you multiply out the
above product, gather like terms, and for each one you get an
equation that says that its coefficient must be equal to the

Continuing my example above, if you multiply through, you'll get sc
for the coefficient of y^3; and since your polynomial has y^3 with
coefficient 1, that means that sc = 1.  Similarly, you'll get te + sf
for the coefficient of yz^2; and since your polynomial has no yz^2
term, that means that te + sf = 0.  It turns out that since you have
more than one variable, this will give you more equations than
variables, which makes things easier to solve.

But that's still a lot of equations to work out.

back and show me what you have been able to do, and I will try to
offer further suggestions.

- Doctor Vogler, The Math Forum
http://mathforum.org/dr.math/

```

```
Date: 11/17/2008 at 16:02:07
From: Greg
Subject: Thank you (Factoring a Multivariate Polynomial)

Thank you!
```
Associated Topics:
High School Polynomials

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