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Polynomial division homework

  1. Nov 18, 2015 #1
    1. The problem statement, all variables and given/known data
    How many pairs of solutions make x^4 + px^2 + q = 0 divisable by x^2 + px + q = 0

    2. Relevant equations
    x1 + x2 = -p
    x1*x2= q


    3. The attempt at a solution
    I tried making z = x^2 and replacing but got nowhere. I figure 0,1,-1 are 3 numbers that fit but im not sure whats being asked also. How many pairs of x1 and x2 make the polynomial 1 divisible by polynomial 2. How to start solving this?
     
  2. jcsd
  3. Nov 18, 2015 #2

    HallsofIvy

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    The second polynomial can be written as [itex](x- a)(x- b)= x^2- (a+ b)x+ ab[/itex] with a+ b= -p and ab= q. It divides the first if and only if the first polynomial has those same factors: is of the form [itex](x- a)(x- b)(x- c)(x- d)= x^4- (a+ b+ c+ d)x^3+ (ab+ ac+ ad+ bc+ bd+ cd)x^2- (abc+ acd+ bcd)x+ abcd[/itex] with a+ b+ c+ d= 0, ab+ ac+ ad+ bc+ bd+ cd= p, abc+ acd+ bcd= 0, and abcd= q. See what you can make of those 6 equations.
     
  4. Nov 18, 2015 #3

    Samy_A

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    Just a guess:

    Maybe the question is:
    For how many pairs (p,q) is the polynomial ##x^4+px^2+q## divisible by the polynomial ##x^2+px+q##?
    Where divisible means that the quotient is a polynomial too.

    Admittedly, that doesn't explain the "=0" in the question, so I'm far for certain.

    EDIT: didn't see the post above before I posted, so just ignore this.
     
    Last edited: Nov 18, 2015
  5. Nov 18, 2015 #4

    PeroK

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    This is a good question and will stretch your understanding of polynomials. First, I assume that you've added the "=0". It's important to know the difference between a polynomial and a polynomial equation. In this case you are dealing with polynomials. You are not trying to find the values of ##x## for which the polynomial is 0.

    I would rephrase the question as:

    Find ##p## and ##q## such that ##x^2 + px + q## divides ##x^4 + px^2 + q##

    Can you see the first step?
     
  6. Nov 19, 2015 #5
    It is true what you said. I added = 0. It shouldnt be there. I dont seem to have any idea on how to start, maybe you could help me with some subtle hint without giving me the step?
     
  7. Nov 19, 2015 #6

    PeroK

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    In general, the definition of ##A## divides ##B## is that there exists some ##C## such that ##B = AC##
     
  8. Nov 19, 2015 #7

    ehild

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    Do what you wrote in the title: Polynomial Long Division. Divide x4+px2+q with x2+px+q, and find the values of p and q so as the remainder is zero.
     
  9. Nov 21, 2015 #8
    I havent done long division before so i checked it out online and tried to do it.Could you check this?
     

    Attached Files:

  10. Nov 21, 2015 #9

    ehild

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    You have sign errors at the last stage.
     
  11. Nov 21, 2015 #10
    I dont see it
     
  12. Nov 21, 2015 #11

    SammyS

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    (Your reply to ehild:
    )

    upload_2015-11-21_12-18-8.png
    Those are the last three lines of the image you posted.

    You need to subtract the part enclosed in the red parentheses.
     
  13. Nov 22, 2015 #12
    (p^2 + p - q)(- px - q) + q(px + 1)= 0
    This must be true for some p and q so that there is no remainder. But how do i find those values? There are three variables here.
     
  14. Nov 22, 2015 #13

    ehild

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    They must be true for any value of x. Collect the x terms and the constant terms separately, in form A(p,q)x +B(p,q)≡0. Both A and B are equal to zero. So you have two equations with the parameters p,q.
     
  15. Nov 22, 2015 #14
    Well i get x(-p^3 - p^2 + 2pq) + (-qp^2 - qp + q^2 + q) = 0
    So like you said A and B are 0. I set them to 0 and solved for p but got some cubic equation so i supposed three solutions there. But i cant solve for q.
     
  16. Nov 22, 2015 #15

    vela

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    I suggest you avoid polynomial division altogether and use HallsofIvy's suggestion back in post #2. He made a slight error in the coefficient of the linear term, however. It should be (abc + abd + acd + bcd).

    Consider the cases q = 0 and ##q \ne 0## separately.
     
  17. Nov 22, 2015 #16
    That seems more complicated. Any chance we could try resolving this polynomial division? How to check the solutions after the last part of the division? I separated the cases in the x(-p^3 - p^2 + 2pq) + (-qp^2 - qp + q^2 + q) = 0. I tried to eliminate one variable by substituting p for q but gotten -2p^3 + 2p^2 - p = 0, and when i try the other way around i get something like p^2 + p = q - 1
     
  18. Nov 22, 2015 #17

    ehild

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    You can factorize both equations, getting p(-p^2-p+2q)=0 and q(-p^2-p+q+1)=0. What solutions are possible?
     
    Last edited: Nov 22, 2015
  19. Nov 22, 2015 #18
    p and q = 0
    q = 0 and p either of two solutions
    p = 0 and q = -1
    p = q = 1
    That makes it 5. Is that it?
     
  20. Nov 22, 2015 #19

    ehild

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    Make a table for the possible values of p and q.
    It looks that you counted p=0 q=0 twice, and one solution is still missing. What are the solutions if neither p nor q are zero?
     
    Last edited: Nov 22, 2015
  21. Nov 22, 2015 #20
    How did i count p and q to be 0 twice? I counted both to be 0, two cases were in the first q is 0 and p p1 or p2 ( making the first expression 0) and the only case were p and q are non zero but still make zero are p = q = 1. I see no other way
     
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