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Quartic polynomial factoring

  1. Jun 15, 2013 #1
    1. The problem statement, all variables and given/known data

    p(x) = x^4+10x^3+26x^2+10x+1
    p(x) = a(x)b(x) where a(x) and b(x) are quadratic polynomials with integer coefficients. It is given that b(1) > a(1). Find a(3) + b(2).

    2. Relevant equations

    p(x) = x^4+10x^3+26x^2+10x+1


    3. The attempt at a solution

    I tried to factor the given quartic but failed . Then I used wolfram-alpha and found that the factors are -
    a](1+4x+x^2) and b](1+6x+x^2)
    Now the problem is really easy to solve after this ( just evaluating a] at 3 and b] at 2 and adding the results equals 39 which is the correct answer ) , But I want to know how to factor that quartic into two quadratics . Please give hints.
     
  2. jcsd
  3. Jun 15, 2013 #2

    mfb

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    If you look at the product of (c+bx+ax^2) and (e+dx+x^2) and compare it to your original polynomial, you can fix some factors quickly. The others are the result of calculations.
     
  4. Jun 15, 2013 #3

    haruspex

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    That's true, mfb, but when you see e.g. that c*e = 1 you have to conclude that c = e = ±1. c = 2, e = 1/2 isn't going to work, in my experience. But I don't know of a sound basis for that. Do you?
     
  5. Jun 16, 2013 #4
    Awesome idea! Thanks.
    Simplifying (c+bx+ax^2)(e+dx+fx^2)
    I get (ce) + (cd+be)x + (bd+ae+cf)x^2 + (ad+bf)x^3 + (af)x^4

    Now comparing the coefficients with x^4+10x^3+26x^2+10x+1
    (ce) =1 (the constant term) Now if you assume the coefficients will be positive integers then both c and e =1 .
    Also cd+be=10 which means d+b=10 ( since c and e =1 )
    From the coefficient of x^4 , both a and f =1 .
    Then from coefficient of x^2 we get bd+ae+cf=26 , but since a,e,c,f=1 we get bd=26-2=24.
    Now we just have to find two numbers b and d whose product is 24 and sum is 10 .
    So the answer is b=6 , d=4 and factors are -
    a](1+4x+x^2) and b](1+6x+x^2)

    Again thanks a lot , but anyone knows how to figure out if the coefficients (of factors) will be integers or not ?
     
  6. Jun 16, 2013 #5
    Yeah , that's what I also want to know.
     
  7. Jun 16, 2013 #6

    verty

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    I think this problem is as difficult as finding roots of a quartic polynomial. In the case of a cubic polynomial, we want to factor ##x^3 + bx^2 + cx + d## into ##(x^2 + px + q)(x + r)##, but this is just the same as finding the root x = -r and doing the division, and we know finding the root is difficult.

    In the quartic case, if there was an easy method to factor the quartic into quadratics, we could find the roots by solving the quadratics. But it is not easy to find the roots, so it must be difficult to factor the quartic polynomial.
     
  8. Jun 16, 2013 #7

    haruspex

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    That's not quite what I meant. If tasked with factoring a quartic with integer coefficients into two quadratics, it may be reasonable to assume that the problem poser has been kind enough that the factor polynomials will have rational coefficients. But does it then follow that they will have integer coefficients?
     
  9. Jun 16, 2013 #8

    mfb

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    If there is a solution with rational numbers, then there is a solution with integers.
    This is a result of Gauß' lemma.

    Apart from that, integers are always the first thing one should try.

    You do not need f (and I did not include one in my post) - you can scale (c+bx+ax^2) by an arbitrary factor, so you don't need the same option for the second factor.
     
  10. Jun 16, 2013 #9
    Oh , thanks. I didn't understand that before and thought that you missed it.
     
  11. Jun 16, 2013 #10

    Mentallic

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    If the solution were to be

    [tex](x^2+ax+c)(x^2+bx+1/c)[/tex]

    where [itex]c\neq 0,1[/itex] then expanding this would yield

    [tex]x^4+(a+b)x^3+(c+1/c+ab)x^2+(bc+a/c)x+1[/tex]

    Now, we're particularly interested in the coefficients of [itex]x^2[/itex] and [itex]x[/itex]. Since we assumed that [itex]c\neq 1[/itex], then either

    [tex]c+1/c+ab[/tex]

    is not an integer, or

    [tex]bc+a/c[/tex]

    is not an integer (or both aren't), and since our quartic has all integer coefficients, this means our assumption of the value c can take is wrong, hence it must be 1.
     
  12. Jun 16, 2013 #11

    epenguin

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    Before I say more there anything special you notice about this polynomial?
     
  13. Jun 17, 2013 #12
    Um , maybe the coefficients of both x and x^3 are same?
    I can't think of anything else.
    What is special about it?
     
  14. Jun 17, 2013 #13

    epenguin

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    The completely general unspecial quartic has 5 coefficients, you could say, so it's not just a1 and a3 being equal that make this one special, also a0 = a4.

    Divide by x2, group, and see if that suggests something you could express in solvable form.

    However it should be useful to you to check whether there was anything relevant in your book or course handouts etc. that you could have used.
     
    Last edited: Jun 17, 2013
  15. Jun 17, 2013 #14
    Actually this is not a Homework question , I just found it somewhere and tried to solve it.

    Okay , dividing by x^2 I get (x^2)+(10x)+(26)+(10/x)+(1/x^2).
    Now , I've no idea how should I group this.
     
  16. Jun 17, 2013 #15

    epenguin

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    OK, it is more suggestive if we had had algebraic coeffients, in whivch case we would have had the form

    x2 + ax + b + a(1/x) + 1/x2

    and you group by like coefficients.
     
    Last edited: Jun 17, 2013
  17. Jun 17, 2013 #16
    Can you please explain the term bx^2? I guess it should be just b.
    Can I group like this - a(x+1/x)+ x^2 + b + 1/x^2 ?
    How will that benefit me?
     
  18. Jun 17, 2013 #17

    epenguin

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    Right - just b, original now corrected.

    So I should have said the algebraic form more completely, though I pointed out above that a0 = a4. This is preventing you recognising that 1 is a coefficient of two terms.

    So you have got

    (x2 + 1/x2) + 10(x + 1/x) + 26

    You can express the first bracket in terms of the second.
     
    Last edited: Jun 17, 2013
  19. Jun 17, 2013 #18

    Office_Shredder

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    Bonus points if anyone can solve the problem without calculating what a(x) and b(x) are (I make no guarantee that it's possible but I feel like there's always some crazy trick with polynomials)
     
  20. Jun 18, 2013 #19

    haruspex

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  21. Jun 18, 2013 #20
    Let me try.
    (x2 + 1/x2) + 10(x + 1/x) + 26
    = ((x+1/x)2-2) + 10(x + 1/x) + 26
    =(x+1/x)2+ 10(x + 1/x)+24

    Am I right? Also what to do now? ( Sorry for asking you so many questions , but I'm very curious)
     
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