What Causes Discrepancies Between Rational Root Test and Modulo p Methods?

In summary: The part "if it is reducible in Q[x]" refers to checking if a given polynomial is reducible in the field of rational numbers. If it is reducible, then you can express it in terms of irreducible polynomials.
  • #1
Joe20
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Hi all,

I have done the question in two methods. The first method is done by rational root test and the second method is by modulo p (theorem is as attached). It seems that my answers for both methods do not tally.

1. Where have I done wrong in the attached for the methods? Which is the correct presentation of answer for this question (i.e. rational test method or modulo p ?
2. How do I tell when to use rational root test method or modulo p method? When modulo p method not applicable?


Your advise is greatly appreciated. Thanks.
 

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  • #2
Your Theorem 6.2.13 says that if $\deg \overline{f}(x) = \deg f(x)$ and $\overline{f}(x)$ is irreducible in $\Bbb{Z}_p[x]$ then $f(x)$ is irreducible in $\Bbb{Q}[x]$. The theorem does not tell you anything at all about what might happen when $\overline{f}(x)$ is reducible in $\Bbb{Z}_p[x]$. So the fact that your $\overline{f}(x)$ is reducible in $\Bbb{Z}_3[x]$ does not imply that ${f}(x)$ is reducible in $\Bbb{Q}[x]$. That would amount to saying that the converse of Theorem 6.2.13 is true, which is certainly not the case.

The technique of looking at $\overline{f}(x)$ in $\Bbb{Z}_p[x]$ is a one-way test. You can use it to prove irreducibility of $f(x)$, but you cannot use it to prove reducibility.

Your argument using the rational root test is correct, and shows that $f(x)$ is irreducible in $\Bbb{Q}[x]$.
 
  • #3
Thank you for the advice.

I have one more question. Can you kindly help me to check if the argument or reasoning stated in the last part of the rational root test method is correct? ( from this " f(x) has no linear factor, hence 2x^2+2x -3 ... till hence 2x^4+8x^3+5x^2-7x-3 = (2x^2+2x -3)(x^2+3x+1).

Thanks.
 
  • #4
Alexis87 said:
Thank you for the advice.

I have one more question. Can you kindly help me to check if the argument or reasoning stated in the last part of the rational root test method is correct? ( from this " f(x) has no linear factor, hence 2x^2+2x -3 ... till hence 2x^4+8x^3+5x^2-7x-3 = (2x^2+2x -3)(x^2+3x+1).

Sorry, I forgot that your rational root investigation actually showed that $2x^4+8x^3+5x^2-7x-3$ is reducible in $\Bbb{Q}[x]$. You correctly showed that $2x^4+8x^3+5x^2-7x-3 = (2x^2+2x -3)(x^2+3x+1).$
 
  • #5
I am somehow confused by the question with this statement " if it is reducible in Q[x], express it as a product of irreducible polynomials in Q[x]. "

1. Does the part "if it is reducible in Q[x]" refers to any factorization [i.e. like the case of 2x^4+8x^3+5x^2−7x−3=(2x^2+2x−3)(x^2+3x+1) ] without the need to solve for the x value to check if it rational or irrational? Then the next question is how does this apply to reducible in Q[x]?

2. The next part " express it as a product of irreducible polynomials in Q[x]. " refers to (2x^2+2x−3) and (x^2+3x+1) being degree 2 and no roots to each quadratic polynomial?

3. So it also implies that if the polynomial is irreducible, then I can't even get to any form of factorization, hence needless to say any product of polynomials. Am I correct to say that?

Thanks.
 
  • #6
Opalg said:
Sorry, I forgot that your rational root investigation actually showed that $2x^4+8x^3+5x^2-7x-3$ is reducible in $\Bbb{Q}[x]$. You correctly showed that $2x^4+8x^3+5x^2-7x-3 = (2x^2+2x -3)(x^2+3x+1).$
I am somehow confused by the question with this statement " if it is reducible in Q[x], express it as a product of irreducible polynomials in Q[x]. "

1. Does the part "if it is reducible in Q[x]" refers to any factorization [i.e. like the case of 2x^4+8x^3+5x^2−7x−3=(2x^2+2x−3)(x^2+3x+1) ] without the need to solve for the x value to check if it rational or irrational? Then the next question is how does this apply to reducible in Q[x]?

2. The next part " express it as a product of irreducible polynomials in Q[x]. " refers to (2x^2+2x−3) and (x^2+3x+1) being degree 2 and no roots to each quadratic polynomial?

3. So it also implies that if the polynomial is irreducible, then I can't even get to any form of factorization, hence needless to say any product of polynomials. Am I correct to say that?

Thanks.
 

FAQ: What Causes Discrepancies Between Rational Root Test and Modulo p Methods?

1. What is the Rational Root Test?

The Rational Root Test is a method used to determine possible rational roots of a polynomial equation with integer coefficients. It states that if a polynomial has a rational root, then that root must be a factor of the constant term over the leading coefficient.

2. How is the Rational Root Test used to solve polynomial equations?

The Rational Root Test is used to find the possible rational roots of a polynomial equation. Once these possible roots are identified, they can be tested using synthetic division to see if they are actually roots of the equation. Any roots that are found can then be used to factor the equation and solve for the remaining roots.

3. What is Modulo p?

Modulo p is a mathematical operation that involves dividing a number by a positive integer p and finding the remainder. This operation is often used in number theory and cryptography, as it allows for efficient calculations involving large numbers.

4. How is Modulo p used in the Rational Root Test?

Modulo p is used in the Rational Root Test to determine whether a potential root is actually a root of the polynomial equation. By taking the remainder of the polynomial when divided by p, we can determine if the potential root is a factor of the polynomial. If the remainder is 0, then the potential root is a factor and therefore a root of the equation.

5. What is the significance of using Modulo p in the Rational Root Test?

The use of Modulo p in the Rational Root Test is significant because it narrows down the potential rational roots of a polynomial equation. By limiting the possible roots to those that are factors of the constant term, we can avoid testing a large number of potential roots, making the process more efficient. Additionally, Modulo p can also be used to find the prime factorization of the leading coefficient, which can further narrow down the possible roots.

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