polynomial question
asdf
exists a polynamial F(x) such that
(x-a)F(x) = P(x)
An example:
x^n - 1 has root 1 how do you know that x-1 divides x^n-1 resulting in
another polynomial. Do you have to do long division ?
José Carlos Santos
> How do you prove that if a is a root of a polynamial P(x) then there
> exists a polynamial F(x) such that
>
> (x-a)F(x) = P(x)
>
> An example:
Given any two polynomials P_1(x) and P_2(x), there are polynomials Q(x)
and R(x) such that
1) P_1(x) = P_2(x)*Q(x) + R(x);
2) the degree of R(x) is smaller than the degree of P_2(x).
You can compute Q(x) and R(x) by long division.
Now, apply this to P(x) and x - a. There is a polynomial Q(x) and there
is a number _r_ such that P(x) = (x - a)Q(x) + r. But then
0 = P(a) = (a - a)Q(a) + r = r.
So, P(x) = (x - a)Q(x).
Best regards,
Jose Carlos Santos
Arturo Magidin
The result is usually called the Factor Theorem in general.
You need to show that given any constant a, and any polynomial P(x),
there is a polynomial q(x) such that (x-a)q(x)+P(a) = P(x). You can
argue by induction on the degree of P, but the argument essentially is
using long division:
If P is constant, then set q(x)=0; Then P(x)=k=P(a)=P(a)+(x-a)0, and
you are done.
Assume the result is true for polynomials of degree k, and let P be a
polynomial of degree k+1. Write
P(x) = a_{k+1}x^{k+1} + ... + a_1*x + a_0
with a_{k+1} different from zero. Then
R(x) = P(x) - a_{k+1}x^k*(x-a) is a polynomial of degree at most k, so by
the induction hypothesis there exists a polynomial Q(x) such that
R(x) = (x-a)Q(x) + R(a).
What is R(a)? SInce R(x) = P(x) - a_{k+1}x^k*(x-a), we have R(a) =
P(a)-a_{k+1}a^k(a-a) = P(a). Sop
R(x) = (x-a)Q(x) + P(a).
Now substitute back in the value of R(x). We have
P(x) - a_{k+1}x^k*(x-a) = (x-a)Q(x) + P(a)
Solving for P(x) we get
P(x) = (x-a)Q(x) + (x-a)a_{k+1}x^k + P(a)
Let q(x) = Q(x) + a_{k+1}x^k. Then
P(x) = (x-a)q(x) + P(a).
Now, assume that P(a) = 0. Then you have P(x)=(x-a)q(x), the result
you want.
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
mag...@math.berkeley.edu
Arturo Magidin
=?ISO-8859-1?Q?Jos=E9_Carlos_Santos?= <jcsa...@fc.up.pt> wrote:
>On 28-03-2006 14:13, asdf wrote:
>
>> How do you prove that if a is a root of a polynamial P(x) then there
>> exists a polynamial F(x) such that
>>
>> (x-a)F(x) = P(x)
>>
>> An example:
>
>Given any two polynomials P_1(x) and P_2(x), there are polynomials Q(x)
>and R(x) such that
>
>1) P_1(x) = P_2(x)*Q(x) + R(x);
>
>2) the degree of R(x) is smaller than the degree of P_2(x).
Careful there. I saw no assumption that the polynomials were over a
field.
Suppose you are working in Z[x]. If P_1(x) = 3x+2 and P_2(x)=2x+3,
what are Q(x) and R(x)?
You can prove the same result in general if you add the assumption
that the leading coefficient of P_2 is a unit (or 1); that is all that
is needed here.
The World Wide Wade
Lemma: If m is a natural number, then there exists a polynomial q
such that x^m - a^m = (x-a)q(x). Proof: It's true for m = 1.
Suppose true for m. Write x^(m+1) - a^(m+1) = x^(m+1) - x^m*a +
x^m*a - a^(m+1) = x^m*(x-a) + a(x^m - a^m) = x^m*(x-a) +
a[(x-a)q(x)] = (x-a)(x^m + aq(x)).
Now suppose p(x) = c_n*x^n + ... + c_1*x + c_0. If p(a) = 0, then
p(x) = p(x) - p(a) = c_n*(x^n-a^n) + ... + c_1*(x-a). By the
lemma, (x-a) can be factored out of each summand, which implies
p(x) = (x-a)q(x) for some polynomial q.
asdf
The World Wide Wade wrote:
>
> Lemma: If m is a natural number, then there exists a polynomial q
> such that x^m - a^m = (x-a)q(x). Proof: It's true for m = 1.
> Suppose true for m. Write x^(m+1) - a^(m+1) = x^(m+1) - x^m*a +
> x^m*a - a^(m+1) = x^m*(x-a) + a(x^m - a^m) = x^m*(x-a) +
> a[(x-a)q(x)] = (x-a)(x^m + aq(x)).
>
> Now suppose p(x) = c_n*x^n + ... + c_1*x + c_0. If p(a) = 0, then
> p(x) = p(x) - p(a) = c_n*(x^n-a^n) + ... + c_1*(x-a). By the
> lemma, (x-a) can be factored out of each summand, which implies
Thanks. I was using this to show that 2 polynomials that have the same
n roots and that are both of degree n are the same polynomial.
LuckyOne
another polynomial. Do you have to do long division ?
Long division will convince you. By the way more properly you should
say
x^n - 1 = 0
The solutions of this equation will lie on the unit circle in R^2 when
thought of as the complex plane. They will be evenly spaced by
360deg/n.
Toni Lassila
They need not do any such thing. The polynomial could be from some
commutative ring which has nothing to do with complex numbers. For
example, in Z_2[x] it has exactly one root for any n >= 1.
LuckyOne
commutative ring which has nothing to do with complex numbers. For
example, in Z_2[x] it has exactly one root for any n >= 1.
By the nature of the question I suspect the OP was thinking about a
polynomial over the reals...
Bill Dubuque
>asdf <qjohn...@yahoo.com> wrote:
>>
>> How do you prove that if a is a root of a polynamial P(x)
>> then there exists a polynamial F(x) such that
>>
>> (x-a)F(x) = P(x)
>
>
> You need to show that given any constant a, and any polynomial P(x),
> there is a polynomial q(x) such that (x-a)q(x)+P(a) = P(x). You can
> argue by induction on the degree of P, but the argument essentially
> is using long division:
>
> If P is constant, then set q(x)=0; Then P(x)=k=P(a)=P(a)+(x-a)0,
> and you are done. Assume the result is true for polynomials of
> degree k, and let P be a polynomial of degree k+1. Write
>
> P(x) = a_{k+1}x^{k+1} + ... + a_1*x + a_0
>
> with a_{k+1} different from zero. Then
>
> R(x) = P(x) - a_{k+1}x^k*(x-a) is a polynomial of degree at most k, so
> by the induction hypothesis there exists a polynomial Q(x) such that
>
> R(x) = (x-a)Q(x) + R(a).
>
> What is R(a)? SInce R(x) = P(x) - a_{k+1}x^k*(x-a), we have R(a) =
> P(a)-a_{k+1}a^k(a-a) = P(a). Sop
>
> R(x) = (x-a)Q(x) + P(a).
>
> Now substitute back in the value of R(x). We have
>
> P(x) - a_{k+1}x^k*(x-a) = (x-a)Q(x) + P(a)
>
> Solving for P(x) we get
>
> P(x) = (x-a)Q(x) + (x-a)a_{k+1}x^k + P(a)
>
> Let q(x) = Q(x) + a_{k+1}x^k. Then
>
> P(x) = (x-a)q(x) + P(a).
>
> Now, assume that P(a) = 0. Then you have P(x)=(x-a)q(x)
SIMPLER Linearity reduces polynomial to monomial case
i.e. x-y|x^n-y^n -> x-y|Sum Pi(x^n-y^n) = P(x)-P(y)
The monomial case has a trivial proof by induction
n+1 n+1 n n
x - y n x - y
namely --------- = x + y -------
x - y x - y
hence P in S <------ P in S := R[x,y]
n+1 n
FACTOR THEOREM is a special case: P(y) = 0, y in R
--Bill Dubuque
Proginoskes
... up to a constant factor. x^2 - 1 and 2x^2 - 2 have the same factors
and the same degree, but they aren't the same polynomial.
--- Christopher Heckman
Toni Lassila
Perhaps, but the result is true in a more general setting and
needlessly resorting to real or complex numbers can be a crutch in
abstract algebra.
Bill Dubuque
>
> How do you prove that for a polynomial P(X)
>
> P(a)=0 -> X-a|P(X) i.e. (X-a) Q(X) = P(X) for some Q(X)
For a=0 it specializes to the obvious case: X|P(X) <-> P(0)=0
If a!=0 reduce to a=0 by shift: X-a|P(X) <-> X|P(X+a) <-> P(a)=0 QED
This is the FACTOR THEOREM. See my prior posts for discussion:
http://google.com/groups/search?q=dubuque+%22factor+theorem%22
--Bill Dubuque
José Carlos Santos
>>> How do you prove that if a is a root of a polynamial P(x) then there
>>> exists a polynamial F(x) such that
>>>
>>> (x-a)F(x) = P(x)
>>>
>>> An example:
>> Given any two polynomials P_1(x) and P_2(x), there are polynomials Q(x)
>> and R(x) such that
>>
>> 1) P_1(x) = P_2(x)*Q(x) + R(x);
>>
>> 2) the degree of R(x) is smaller than the degree of P_2(x).
>
> Careful there. I saw no assumption that the polynomials were over a
> field.
Indeed. Whenever someone asks me a question about polynomials, I tend
to assume that they are over some field.
asdf
Proginoskes wrote:
> asdf wrote:
> > Thanks. I was using this to show that 2 polynomials that have the same
> > n roots and that are both of degree n are the same polynomial.
>
> ... up to a constant factor. x^2 - 1 and 2x^2 - 2 have the same factors
> and the same degree, but they aren't the same polynomial.
>
Yes that is correct. I was wondering in the complex plane how we knew
that z^n + 1 was same polynomial as ( z - exp(Pi*i/n)( z -
exp(3*Pi*i/n))....(z - exp( (2n-1)*Pi*i/n ) )
asdf
asdf
Bill Dubuque wrote:
> For a=0 it specializes to the obvious case: X|P(X) <-> P(0)=0
>
> If a!=0 reduce to a=0 by shift: X-a|P(X) <-> X|P(X+a) <-> P(a)=0 QED
>
Looking at this more it seems what is going on here is that the
polynomial P(X+a) has constant term P(a) which we know is 0 and thus
P(X+a) divides x. We can see the P(a) coming out of the polynomial if
we envision subsituting x with x+a in P(x). If P(X+a) divides x, then
P(X) divides x-a.