IS f(x)= x^4 + x^3 -x^2 - 2x -2 irreducable over Q?

  • Thread starter Niall101
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In summary, the conversation showed that there is no solution to bd=-2 in the field F, and that the polynomial is reducible.
  • #1
Niall101
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IS f(x)= x^4 + x^3 -x^2 - 2x -2 irreducable over Q??

IS f(x)= x^4 + x^3 -x^2 - 2x -2 irreducable over Q??

Attempt:

Ive shown that this has no linear factors as +-1 +-2 are not roots.

Then I let F(x)=(x^2+ax+b)(x^2+cx+d)

And I end up with
a +c=1
d+ac+b=-1
ad+bc=-2
bd=-2

But i can seam to show why no a b c d exist (or Exist)

Is there an easier approach?

I have an exam in 12 hours so I hope soeone can help!

Thanks in advance!

Niall
 
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  • #2


There aren't many solutions to bd = -2...
 
  • #3


hey thanks for your reply I managed to get it. It is reducable. :)
 
  • #4


Sorry for another question. If i have F(x)=x^3-5 Is it sufficiant to show it have no linear factors by plugging in +-5 +-1 and showing not equal to zero? I was thinking because the only factors could be linear or of degree 3. Thanks again.
 
  • #5


The only limit on the ways that polynomials can factor is that the degrees of the factors add up to the degree of the polynomial. In particular, there are cubics that factor into a linear and a quadratic.
 
  • #6


Over Q tho are +-5 +-1 The only possible roots of x^3-5?

cool thanks so much!

Let A = cuberoot(5),
and let
F = a + bA + cA in R
a; b; c in Q
Find the multiplicative inverse of the element 2 +A +A^2 in the field F.

For this I Know There is an element in the field which multiplied by this give the Multiplicitive identity. Is there a way of doing this by taking powers of 2 +A +A^2?

I really appreciate the help!

Edit: I like your site with the comics made me smile
 
Last edited:
  • #7


The method you suggest sounds like you're planning on using linear algebra to compute the minimal polynomial of your field element. That will work, but I doubt it's the most efficient way to do it by hand.
 
  • #8


Yeah i think that's what i was trying to do. Have to do this in an exam later. But We haven't covered minimal polynomials yet so he probably wants a different method. Could you point me in the right direction?
 
  • #9


What about the direct approach? Write down an indeterminate field element, and solve the equation that says your field element times the general field element equals 1.
 
  • #10


great thanks I got it now. Thanks for your help!
 

1. What does it mean for a polynomial to be irreducible over Q?

For a polynomial to be irreducible over Q, it means that it cannot be factored into two polynomials with rational coefficients. In other words, there are no factors of the polynomial that are also polynomials with rational coefficients.

2. How can I determine if a polynomial is irreducible over Q?

One way to determine if a polynomial is irreducible over Q is to use the Rational Root Theorem. This theorem states that if a polynomial with integer coefficients has a rational root, then that root must be a factor of the constant term. So, if we can find all the rational roots of a polynomial and none of them are factors, then the polynomial is irreducible over Q.

3. Can a polynomial with degree greater than 2 be irreducible over Q?

Yes, a polynomial with degree greater than 2 can be irreducible over Q. In fact, there are many polynomials with degree greater than 2 that are irreducible over Q. However, all polynomials with degree 1 (linear) and degree 2 (quadratic) are reducible over Q.

4. Are there other methods to determine if a polynomial is irreducible over Q?

Yes, there are other methods to determine if a polynomial is irreducible over Q. Some common methods include the Eisenstein Criterion, the Descartes' Rule of Signs, and the Perron's Criterion. These methods can be used if the polynomial does not have any rational roots.

5. Is f(x) = x^4 + x^3 - x^2 - 2x - 2 irreducible over Q?

Yes, f(x) = x^4 + x^3 - x^2 - 2x - 2 is irreducible over Q. This can be determined by using the Rational Root Theorem and finding that the only possible rational roots are ±1 and ±2, none of which are actual roots of the polynomial.

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