Why Does the Rational Zeros Theorem Not Predict All Zeros?

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Discussion Overview

The discussion centers around the Rational Zeros Theorem and its limitations in predicting all possible zeros of polynomials, particularly focusing on the polynomial 3x^3+4x^2-7x+2. Participants explore the implications of the theorem, the nature of rational versus irrational roots, and the challenges of solving higher-degree polynomials.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Some participants note that the Rational Zeros Theorem only lists possible rational roots and does not account for irrational roots, as seen in the example of a zero at approximately -2.4.
  • Others argue that the theorem is effective for identifying rational roots, but it does not imply that all roots must be rational.
  • There is a suggestion that the theorem may be clumsy for solving polynomials, prompting inquiries about alternative methods or theorems.
  • Some participants defend the theorem's simplicity and elegance, stating that it adequately describes all possible rational roots.
  • Concerns are raised about the complexity of solving higher-degree polynomials, with mentions of the limitations of general solutions for quintic polynomials and the need for additional techniques like factoring and long division.
  • There is a discussion about the distinction between solving expressions and equations, with some participants emphasizing the importance of having an equal sign in equations.

Areas of Agreement / Disagreement

Participants generally agree that the Rational Zeros Theorem identifies possible rational roots but disagree on its effectiveness and the implications of its limitations regarding irrational roots. The discussion remains unresolved regarding the best methods for solving higher-degree polynomials.

Contextual Notes

Some participants express uncertainty about the applicability of the Rational Zeros Theorem to all polynomials, particularly those of degree five or higher, and highlight the lack of a general solution for such cases. Additionally, there is a focus on the necessity of distinguishing between expressions and equations in mathematical discussions.

Curd
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When solving 3x^3+4x^2-7x+2 the rational zeros theorem says there can only be a possibility of zeros at plus minus 1, 2, 1/3, and 2/3

but for some reason a zero appears at about -2.4 (or more accurately -1-squareroot of 2).

How can this be? It states that if there are any zeros they will be listed by the theorem. It does not state the possibility of zeros lying outside that theorem.
 
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The theorem lists all the possible rational roots. If a root is not rational, it won't be listed.
 
Curd said:
When solving 3x^3+4x^2-7x+2 the rational zeros theorem says there can only be a possibility of zeros at plus minus 1, 2, 1/3, and 2/3
The rational zeros theorem says that the any rational roots must be of the form±p/q. It doesn't say a thing about irrational roots.
 
Then it seems a rather clumsy thing to use for solving polynomials. Is there a better theorem out there for this purpose?
 
Curd said:
Then it seems a rather clumsy thing to use for solving polynomials. Is there a better theorem out there for this purpose?

How is it clumsy? It's very simple and elegant, and it completely describes all possible rational roots of the polynomial. If you want every root of a of a polynomial of degree 2 through 4, then just use the quadratic/cubic/quartic formulas. If you want every solution to a polynomial of degree >4 then too bad, a general solution doesn't exist.
 
Number Nine said:
How is it clumsy? It's very simple and elegant, and it completely describes all possible rational roots of the polynomial. If you want every root of a of a polynomial of degree 2 through 4, then just use the quadratic/cubic/quartic formulas. If you want every solution to a polynomial of degree >4 then too bad, a general solution doesn't exist.

I'm pretty sure you can factor a x^5 or greater polynomial out with a combination of rational zeros theorem, bounds on zeros theorem, factoring and long division. The book has been making me do that in this section.

It would just be nice if it were cleaner.

Well, I've only had to do one x^5 polynomial. But one can continue using those tools to get the answers.

It would still be nice if this were a bit less messy though.
 
How is it messy? :confused:

Usually in exercises with difficult polynomials either 1, 0, or -1 used to be a root, so you could simplify.
Now, you need to consider a few more possibilities.
And if there are too many, you can shift and/or scale to reduce the number.
 
Curd said:
I'm pretty sure you can factor a x^5 or greater polynomial out with a combination of rational zeros theorem, bounds on zeros theorem, factoring and long division. The book has been making me do that in this section.

It would just be nice if it were cleaner.

Well, I've only had to do one x^5 polynomial. But one can continue using those tools to get the answers.

It would still be nice if this were a bit less messy though.

actually, it is a well-known theorem due to abel and ruffini, that for quintic polynomials and higher, NO general solution by the methods you suggest exists. this does not preclude SOME quintics from being solvable, but some are not.

as an exercise, try to find a solution by radicals of x5-x-1.
 
Curd said:
When solving 3x^3+4x^2-7x+2 the rational zeros theorem says there can only be a possibility of zeros at plus minus 1, 2, 1/3, and 2/3
What do you mean by solving? What you have is an expression, and you cannot solve an expression.

Curd said:
I'm pretty sure you can factor a x^5 or greater polynomial out with a combination of rational zeros theorem, bounds on zeros theorem, factoring and long division. The book has been making me do that in this section.
Just because you have been solving quintics in the book does not mean that you can solve any quintic that's out there. Regarding lower degree polynomials, there exists a cubic formula and a quartic formula. You can find them easily enough through a search.
 
  • #10
eumyang said:
What do you mean by solving? What you have is an expression, and you cannot solve an expression.


Just because you have been solving quintics in the book does not mean that you can solve any quintic that's out there. Regarding lower degree polynomials, there exists a cubic formula and a quartic formula. You can find them easily enough through a search.


You can solve it for zero, which is what is meant when typically when people speak of solving equations. You can also solve it for many other numbers.
 
  • #11
Curd said:
You can solve it for zero, which is what is meant when typically when people speak of solving equations. You can also solve it for many other numbers.

What eumyang is saying is that in your post you didn't write down an equation. You wrote an expression. Equations must have an equal sign.

You can however say
"find the zeros of the polynomial 3x3+4x2-7x+2"
 
  • #12
Curd said:
You can solve it for zero, which is what is meant when typically when people speak of solving equations. You can also solve it for many other numbers.
No, as already mentioned, you can't solve an expression, and "solving it for zero" is meaningless. You can evaluate an expression at a particular value of the variable. Is this what you meant?

pwsnafu said:
What eumyang is saying is that in your post you didn't write down an equation. You wrote an expression. Equations must have an equal sign.

You can however say
"find the zeros of the polynomial 3x3+4x2-7x+2"

And this is the same as saying "find the solutions of the equation 3x3+4x2-7x+2 = 0.
 
  • #13
Emphasis added.
Curd said:
When solving 3x^3+4x^2-7x+2 the rational zeros theorem says there can only be a possibility of rational[/color] zeros at plus minus 1, 2, 1/3, and 2/3

<snip>

How can this be? It states that if there are any rational[/color] zeros they will be listed by the theorem. It does not state the possibility of zeros lying outside that theorem.
Any potential rational zeroes are listed by the theorem. It doesn't say anything at all about irrational zeroes.
 

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