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Factoring Polynomials [Abstract Algebra]

  1. Feb 20, 2016 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data

    1. Let g(x) = x^4+46.
    a) Factor g(x) completely in ℚ[x].
    b) Factor g(x) completely in ℝ[x].
    c) Factor g(x) completely in ℂ[x].

    2. Completely factor the given polynomial in ℤ_5.
    [4]_5 x^3 + [2]_5 x^2 + x + [3]_5

    2. Relevant equations
    ℚ = {m/n / m and n belong to Z, m is not divided by m}
    ℝ = {real numbers which include all rational and irrational numbers}
    ℂ = Complex

    A polynomial is completely factored over F if it is written as a product of monic irreducible polynomials in F[x] and an element of F.


    3. The attempt at a solution

    Initial thoughts: This looks easy! However, it's a bit tougher than it looked.

    We have x^4+46.
    To start, let's just factor it as I normally would without worry about the fields.

    With a little help from wolfram alpha, and a personal check to make sure we see that for the factorization over ℂ (part c) we get:
    (x+4throot(-46)) (x-4throot(-46)) (x-i*4throot(-46)) (x+i*4throot(-46))

    This should be the correct answer to c as it is in completely factored form.

    Now, we try to get the factorization over the reals, ℝ. part b:
    Since we cant use i, we try to multiply the factors a bit to get rid of i.
    We get : (x^2-4throot(-46)) (x^2+4throot(-46))

    Question: Is this ok over field ℝ? I have a negative under the square root, but I'm not using imaginary unit i.
    Answer: Likely not as this is an expression of a+ib.
    So, this doesn't work.


    At this point, I'm kind of lost. I feel like, over the field ℚ, the polynomial g(x) is completely factored.
    But I see to have come to the conclusion that over the field ℝ it is also completely factored.

    My final answers:
    1.a It is already completely factored
    1.b It is completely factored
    1.c I factored it above.

    As for part 2, the difficulty here is dealing with the mod 5.
    If we look at it like 4x^3+2x^2+x+3
    I would think it would be in the form:
    (x+)(x+)(x+)

    But I'm not really sure on how to start this one. I understand what the mod 5 does, but I'm not used to trying to factor x^3+x^2+x+a.
     
  2. jcsd
  3. Feb 20, 2016 #2

    Ray Vickson

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    Try writing ##p(x) = 4 x^3 + 2 x^2 + x+3## as either ##p(x) = (4 x^2+ ax + b)(x+c)## or as ##p(x) = (x^2 + ax + b)(4x+c)##, then solving for the simplest combinations of ##(a,b,c)## that work in each case.
     
  4. Feb 20, 2016 #3

    Mark44

    Staff: Mentor

    It's not clear to me what your notation above means.
    These look OK to me. Notationwise, you can write ##\sqrt[4]{-46}## as \sqrt[4]{-46}. The fourth roots of -46 will be arranged equidistant around a circle: at ##\pi/4, 3\pi/4, 5\pi/4, 7\pi/4##.
    Right. ##x^4 + 46## can't be factored over the reals.
    , post: 5385224, member: 504241"]1. The problem statement, all variables and given/known data

    1. Let g(x) = x^4+46.
    a) Factor g(x) completely in ℚ[x].
    b) Factor g(x) completely in ℝ[x].
    c) Factor g(x) completely in ℂ[x].

    2. Completely factor the given polynomial in ℤ_5.
    [4]_5 x^3 + [2]_5 x^2 + x + [3]_5

    2. Relevant equations
    ℚ = {m/n / m and n belong to Z, m is not divided by m}
    ℝ = {real numbers which include all rational and irrational numbers}
    ℂ = Complex

    A polynomial is completely factored over F if it is written as a product of monic irreducible polynomials in F[x] and an element of F.


    3. The attempt at a solution

    Initial thoughts: This looks easy! However, it's a bit tougher than it looked.

    We have x^4+46.
    To start, let's just factor it as I normally would without worry about the fields.

    With a little help from wolfram alpha, and a personal check to make sure we see that for the factorization over ℂ (part c) we get:
    (x+4throot(-46)) (x-4throot(-46)) (x-i*4throot(-46)) (x+i*4throot(-46))

    This should be the correct answer to c as it is in completely factored form.

    Now, we try to get the factorization over the reals, ℝ. part b:
    Since we cant use i, we try to multiply the factors a bit to get rid of i.
    We get : (x^2-4throot(-46)) (x^2+4throot(-46))

    Question: Is this ok over field ℝ? I have a negative under the square root, but I'm not using imaginary unit i.
    Answer: Likely not as this is an expression of a+ib.
    So, this doesn't work.


    At this point, I'm kind of lost. I feel like, over the field ℚ, the polynomial g(x) is completely factored.
    But I see to have come to the conclusion that over the field ℝ it is also completely factored.

    My final answers:
    1.a It is already completely factored
    1.b It is completely factored
    1.c I factored it above.

    As for part 2, the difficulty here is dealing with the mod 5.
    If we look at it like 4x^3+2x^2+x+3
    I would think it would be in the form:
    (x+)(x+)(x+)

    But I'm not really sure on how to start this one. I understand what the mod 5 does, but I'm not used to trying to factor x^3+x^2+x+a.[/QUOTE]
     
  5. Feb 20, 2016 #4

    RJLiberator

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    Ah, yes, how could I forget such things!

    Marvellous.

    We have:
    p(x) = (4x^2+(-2)x+3)(x+1)

    Which seems to be the most factorable possibility. To see if it works in mod[5] we multiply it out.

    And it does.

    However, I can't help but feel that this almost seems too easy here. Perhaps I need to factor it to:
    x(x(4x+2)+1)+3
     
  6. Feb 20, 2016 #5

    RJLiberator

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    So my thoughts were confirmed. Part 1 cannot be factored over Q or R. Is there any way I should show this?
     
  7. Feb 20, 2016 #6

    Samy_A

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    ##\mathbb Z_5## isn't that big, you could try to find all the roots.
     
  8. Feb 20, 2016 #7

    Mark44

    Staff: Mentor

    Sure. Suppose that ##x^4 + 46 = 0##. Clearly x = 0 is not a solution. If x > 0, there are clearly no solutions, as both ##x^4## and 46 are positive. The same is true if x < 0. This means there are no solutions to the equation, and hence, no real factors of ##x^4 + 46##.

    If there are no real factors, then a fortiori, there can't be rational factors, either.
     
  9. Feb 20, 2016 #8

    RJLiberator

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    Ok, beautiful. I understand part 1 very well now. We can show that there is no factors by x^4+46 = 0.
    Then, clearly the complex factorization works.

    For part two, they ask to completely factor the given polynomial in Z_5.
    I believe I have done so by:
    p(x) = (4x^2-2x+3)(x+1)

    However, in doing so, I haven't used the mod 5 fact at all. All I've done is factor it without any sort of recognition to Z_5 [x].
     
  10. Feb 20, 2016 #9

    Samy_A

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    You could "recognize" that you work in ##\mathbb Z_5## by checking which other elements of ##\mathbb Z_5## is are roots of the polynomial.
     
  11. Feb 20, 2016 #10

    RJLiberator

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    You mean multiples of 5?

    I could just do:
    (9x^2-2x+3)(x+1)

    And that would work mod 5.

    Is that what you mean by recognizing this element of the question?
     
  12. Feb 20, 2016 #11

    Samy_A

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    No.
    You have a polynomial in ##\mathbb Z_5[x]##.
    You want to find the zeros in ##\mathbb Z_5## of that polynomial.
    There may be fancy methods to do that, but as ##\mathbb Z_5## contains 5 elements, you could just evaluate the polynomial in each of these 5 elements to find all the zeros.
     
  13. Feb 20, 2016 #12

    RJLiberator

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    Ah, I see what you are saying, so
    let x be 1, 2, 3, 4, 5 and see which one makes the equation true:
    4x^3+2x^2+x+3 = 0

    So we have x = 1, 2, and 4 that make this equation true.

    But, er, not sure where to go from here.

    We found the roots.
     
  14. Feb 20, 2016 #13

    Samy_A

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    Now do what one always does: if ##a## is a root of a polynomial ##p## of degree 3, then ##p(x)=(x-a)q(x)##, where ##q## is a polynomial of degree 2.
    And so on.
    Just remember that you are working in ##\mathbb Z_5##.
    Since you found 3 roots (I found the same 3 roots, so that looks good), you are almost done.
     
  15. Feb 20, 2016 #14

    Mark44

    Staff: Mentor

    Not quite -- let x be 0, 1, 2, 3, and 4.
    ##5 \equiv 0 (\mod 5)##.
     
  16. Feb 20, 2016 #15

    RJLiberator

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    Excellent correction, Mark.
    Indeed. 0 does not work.

    p(x) = (x-2)*(4x^2+10x-4)
    this should work as well when considering mod 5.

    However, I feel I can continually do this over and over.
    Is this what you meant by recognizing it?
     
  17. Feb 20, 2016 #16

    Samy_A

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    Now factor the two other roots too. (I didn't check your second degree polynomial, as we don't need it.)
     
  18. Feb 20, 2016 #17

    RJLiberator

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    Wopps/ Edit
     
  19. Feb 20, 2016 #18

    RJLiberator

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    @Samy_A
    I finally understand what you meant, it is obvious and important. Sorry I was so dazed this morning.

    Here is what I did.
    1. Found the roots of the initial polynomial in mod 5. These were 1, 2, and 4.
    2. I divided the initial polynomial by one of the roots, I choose (x-1). I was sure to use the mod 5 distinction here.
    3. I found that (x-1)(4x^2+x+2) was a factor.
    4. I repeated this process for the second term.

    I ended up with the answer 4(x+1)(x-1)(x-2) as the completely factored form.

    I feel great about this answer.

    Thank you.
     
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