Upper & Lower Bounds Real Zeros-Polynomial Division, Proofs?

[Saturnine Zero]

Homework Statement

Upper Bound[/B]
If all of the numbers in the final line of the synthetic division tableau are non-positive, prove for $f(b)<0$, no real number $b > c$ can be a zero of $f$

Lower Bound
To prove the lower bound part of the theorem, note that a lower bound for the negative real zeros of $f(x)$ is an upper bound for the positive real zeros of $f(−x)$. Applying the upper bound portion to $f(−x)$ gives the result.

Do you see where the alternating signs come in?

Homework Equations

$p(x)=d(x)q(x)+r(x)$
$f(x)=(x − c)q(x)+r$

The Attempt at a Solution

I've been struggling for three weeks with this topic trying to properly understand it and to formulate a sensible and reasonably short question. I'm hoping someone can tolerate this long post and let me know if I'm heading in the right direction.

Upper Bound
My book shows a proof for the upper bound for $f(b)>0$ which makes sense to me, but just states there is a simlilar proof for $f(b)<0$. So I've attempted to follow the proof in the book and adapt it as follows:

Suppose $c>0$ is divided into f and the resulting last line in the division tableau contains all non-positive numbers.

This means $f(x)=(x-c)q(x)-r$, where the coefficients of the quotient polynomial and the remainder are non-positive.

If $b>c$, then $f(b)=(b-c)q(b)-r$, where $(b-c)$ is positive and $q(b)$ is negative and $r≤0$.

Hence $f(b)<0$ which shows b cannot be a zero of $f$. Thus no real number $b>c$ that can be a zero of $f$ as required.
​

Does this show $c$ is an upper bound? That is to have $f(b)=0$, $b$ would need to be less than it is so that $b=c$ and $b-c$ sum to zero, so $c$ is an upper bound?

Lower Bound
For the lower bound part. I've struggled more. I really didn't see where the alternating signs came in. But I figured this morning I would default to using a numerical example, and I think I'm closer now.

I think the alternating sings of the coefficients come into it because the odd/even powers of the polynomial change the signs in $f(-x)$ and this results in a quotient with coefficients of all the same sign.

So my current understanding is, if the quotient polynomial of $f(x)$ alternates signs, when $f(-x)$ is divided by the negative of the divisor, all the coefficients of quotient polynomial of $f(-x)$ will always have all the same sign, and the upper bound proof can be used.

Choosing a more negative divisor for $f(x)$ would make each coefficient in the quotient either more positive or more negative alternatively, so not approaching a zero. And then dividing by the negative of that divisor for $f(-x)$ would make the coefficients in the quotient either more positive or more negative, so not approaching a zero. Therefore that divisor for $f(x)$ is a lower bound.

I'd like to know if my upper bound proof is correct and if I am on the right track for understanding the lower bound part?

Sorry for the long post, I'm not sure how to cut it down any further.

 

[SammyS]

Homework Statement

Upper Bound[/B]
If all of the numbers in the final line of the synthetic division tableau are non-positive, prove for $f(b)<0$, no real number $b > c$ can be a zero of $f$

Lower Bound
To prove the lower bound part of the theorem, note that a lower bound for the negative real zeros of $f(x)$ is an upper bound for the positive real zeros of $f(−x)$. Applying the upper bound portion to $f(−x)$ gives the result.

Do you see where the alternating signs come in?

Homework Equations

$p(x)=d(x)q(x)+r(x)$
$f(x)=(x − c)q(x)+r$

The Attempt at a Solution

I've been struggling for three weeks with this topic trying to properly understand it and to formulate a sensible and reasonably short question. I'm hoping someone can tolerate this long post and let me know if I'm heading in the right direction.

Upper Bound
My book shows a proof for the upper bound for $f(b)>0$ which makes sense to me, but just states there is a simlilar proof for $f(b)<0$. So I've attempted to follow the proof in the book and adapt it as follows:

Suppose $c>0$ is divided into f and the resulting last line in the division tableau contains all non-positive numbers.

This means $f(x)=(x-c)q(x)-r$, where the coefficients of the quotient polynomial and the remainder are non-positive.

If $b>c$, then $f(b)=(b-c)q(b)-r$, where $(b-c)$ is positive and $q(b)$ is negative and $r≤0$.

Hence $f(b)<0$ which shows b cannot be a zero of $f$. Thus no real number $b>c$ that can be a zero of $f$ as required.
​

Does this show $c$ is an upper bound? That is to have $f(b)=0$, $b$ would need to be less than it is so that $b=c$ and $b-c$ sum to zero, so $c$ is an upper bound?
...

Please state the entire problem.

What is ƒ ? What are b and c ? What is being divided by what ?

 

[Saturnine Zero]

Please state the entire problem.

What is ƒ ? What are b and c ? What is being divided by what ?

OK, thanks for looking, I will try to be a bit clearer:

• Suppose $f$ is a polynomial of degree $n ≥ 1$

• If $c > 0$ is synthetically divided into $f$ and all of the numbers in the final line of the division tableau have the same signs, then $c$ is an upper bound for the real zeros of $f$ . That is, there are no real zeros greater than $c$.

So $f$ is the polynomial and $c$ is a possible real zero and is the divisor in the synthetic division, so $f(x)$ is being divided by $(x-c)$, and $b$ would any real number greater than $c$.

I hope that's better.

 

[Ray Vickson]

OK, thanks for looking, I will try to be a bit clearer:

• Suppose $f$ is a polynomial of degree $n ≥ 1$

• If $c > 0$ is synthetically divided into $f$ and all of the numbers in the final line of the division tableau have the same signs, then $c$ is an upper bound for the real zeros of $f$ . That is, there are no real zeros greater than $c$.

So $f$ is the polynomial and $c$ is a possible real zero and is the divisor in the synthetic division, so $f(x)$ is being divided by $(x-c)$, and $b$ would any real number greater than $c$.

I hope that's better.

Are you saying that
$$f(x) = (x-c)(a_{n-1} x^{n-1} + a_{n-2} x^{n-2} + \cdots + a_1 x + a_0) + r,$$
where $a_0, a_1, \ldots, a_{n-2}, a_{n-1}$ and $r$ all have the same sign?

 

[Saturnine Zero]

Are you saying that
$$f(x) = (x-c)(a_{n-1} x^{n-1} + a_{n-2} x^{n-2} + \cdots + a_1 x + a_0),$$
where $a_0, a_1, \ldots, a_{n-2}, a_{n-1}$ all have the same sign?

Yes, exactly that.

I was just using $q(x)$ as the quotient polynomial $(a_{n-1} x^{n-1} + a_{n-2} x^{n-2} + \cdots + a_1 x + a_0)$, and $r$ as the remainder as the book does.

I actually found a preview online of the exact two pages from the book which would make it clearer: Theorem 3.11.