[QUOTE=wblipp;219036]Bob,

I'm pretty sure you intend for polynomials to always have integer coefficients and that Tomer doesn't know this.
[/QUOTE]

Huh? I wrote originally:
"Suppose that its roots are x_0, x_1, .... x_n.
a_i are integers."

This doesn't leave much room for doubt that the coefficients are
integers.
[QUOTE]

I also think he may benefit from some guidance about how substitution of variables works.
[/QUOTE]

He [b]claims[/b] to be ready for calculus and other university level
courses. If he needs help with first year algebra (i.e. use of variables as
you suggest) then it is clear that he is nowhere near ready.

I have repeatedly told him that he introduces too many 'free variables'
in his problems; often without defining them. He doesn't seem to get
that message.

To be fair to both of us: this all would be better in front of a blackboard.

[QUOTE=R.D. Silverman;219020]No. For any polynomial f(x) of arbitrary degree k, and any polynomial q(x)
of degree r < k, we can write

f(x) = q(x) * g(x) + r(x) where g(x) is a polynomial of degree
r-k and r(x) has degree less than q(x).

[/QUOTE]

Ooops! Clearly r-k should be k-r.

[quote=R.D. Silverman;219041]Huh? I wrote originally:
"Suppose that its roots are x_0, x_1, .... x_n.
a_i are integers."

This doesn't leave much room for doubt that the coefficients are
integers.
.[/quote]
I know f(x)'s coefficients are integers.
But what about g(x)'s coefficients? are these integers too?
g(x) is defined as a polynomial with roots x_0^-1, x_1^-1, .... x_n^-1.

[quote=R.D. Silverman;219020]No. For any polynomial f(x) of arbitrary degree k, and any polynomial q(x)
of degree r < k, we can write

f(x) = q(x) * g(x) + r(x) where g(x) is a polynomial of degree
r-k and r(x) has degree less than q(x).

In words: we can divide any polynomial by any other polynomial of
lesser degree leaving a polynomial remainder of even lesser degree than
the divisor.

It is not specific to division by just linear polynomials and has nothing to
do with the roots of the polynomial.

It is analagous to the division algorithm for integers. Any integer N in Z+
can be written as N = d*q + r where r < d and d is the divisor.
The same applies for polynomials.

Do you know how to prove this?[/quote]

In wolfram alpha it is called a polynomial remainder.

[quote=R.D. Silverman;219020]No. For any polynomial f(x) of arbitrary degree k, and any polynomial q(x)
of degree r < k, we can write

f(x) = q(x) * g(x) + r(x) where g(x) is a polynomial of degree
r-k and r(x) has degree less than q(x).

In words: we can divide any polynomial by any other polynomial of
lesser degree leaving a polynomial remainder of even lesser degree than
the divisor.

It is not specific to division by just linear polynomials and has nothing to
do with the roots of the polynomial.

It is analagous to the division algorithm for integers. Any integer N in Z+
can be written as N = d*q + r where r < d and d is the divisor.
The same applies for polynomials.

Do you know how to prove this?[/quote]

I think you are able to find infinitely many synthetic devisions for a given polynomial.

[QUOTE=blob100;219069]I know f(x)'s coefficients are integers.
But what about g(x)'s coefficients? are these integers too?
g(x) is defined as a polynomial with roots x_0^-1, x_1^-1, .... x_n^-1.[/QUOTE]

Yes, they are integers. Try expressing them in terms of the coefficients of f.

[QUOTE=blob100;219071]I think you are able to find infinitely many synthetic devisions for a given polynomial.[/QUOTE]


Synthetic division involves TWO given polynomials, not 'a given polynomial'.
The divisor polynomial is also given.

[quote=R.D. Silverman;219085]Synthetic division involves TWO given polynomials, not 'a given polynomial'.
The divisor polynomial is also given.[/quote]
OK.

[QUOTE=R.D. Silverman;219020]No. For any polynomial f(x) of arbitrary degree k, and any polynomial q(x)
of degree r < k, we can write

f(x) = q(x) * g(x) + r(x) where g(x) is a polynomial of degree
r-k and r(x) has degree less than q(x).

In words: we can divide any polynomial by any other polynomial of
lesser degree leaving a polynomial remainder of even lesser degree than
the divisor.

It is not specific to division by just linear polynomials and has nothing to
do with the roots of the polynomial.

It is analagous to the division algorithm for integers. Any integer N in Z+
can be written as N = d*q + r where r < d and d is the divisor.
The same applies for polynomials.

Do you know how to prove this?[/QUOTE]


Bob,

I think the process you have described is usually called polynomial division. I think "synthetic division" is usually used for the special problem of checking for divisibility by a linear factor by evaluating the polynomial at the root of the linear polynomial. That usage of "synthetic division" [B]does[/B] have something do with the roots of the polynomial.

Also, I can't tell if the final "this" refers to the result for integers or the result for polynomials.

William

[QUOTE=wblipp;219091]Bob,

I think the process you have described is usually called polynomial division. I think "synthetic division" is usually used for the special problem of checking for divisibility by a linear factor by evaluating the polynomial at the root of the linear polynomial. That usage of "synthetic division" [B]does[/B] have something do with the roots of the polynomial.

Also, I can't tell if the final "this" refers to the result for integers or the result for polynomials.

William[/QUOTE]

'This' refers to the result for polynomials. We had already proved the
result for integers in an earlier post.

The Wikipaedia says of synthetic division:

"It is mostly taught for division by binomials of the form
but the method generalizes to division by any monic polynomial. "

I think the words "mostly" and "generalizes" are relevant.

[QUOTE=R.D. Silverman;219101]"It is mostly taught for division by binomials of the form but the method generalizes to division by any monic polynomial."[/QUOTE]

Hmm. The thing Wikipedia calls synthetic division is VASTLY different from the thing I was taught as synthetic division. I was taught the thing at the first Google link,

[url]http://www.purplemath.com/modules/synthdiv.htm[/url]

which says

"Synthetic division is a shorthand, or shortcut, method of polynomial division in the special case of dividing by a linear factor -- and it [I]only[/I] works in this case."

I guess there must be multiple definitions of synthetic division.