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Homework Help: Help! Getting the roots of a polynomial

  1. Aug 4, 2010 #1
    1. The problem statement, all variables and given/known data

    I am trying to get the roots of:

    [tex](x-a)^3 - (x-a)b^2 - 2b^3 -2b^2(x-a)[/tex]

    which I know they are [tex] x = a-b [/tex] and [tex] x = a+2 b [/tex]

    the problem is, how can I reach that solution ?

    3. The attempt at a solution

    At first I thought of separating the independent term (the one without x's) and manipulating the expression in order to guess x, but I haven't been successful...

    [tex]x^3 - xb^2 - 3xb^2 - a^3 + 3ab^2 - 2b^3 = 0[/tex]

    [tex]x^3 - xb^2 - 3xb^2 = a^3 - 3ab^2 + 2b^3[/tex]

    How could I solve this?

  2. jcsd
  3. Aug 5, 2010 #2
    The first thing to notice is that x and a always appear together. I would solve for the quantity (x-a) first instead of x directly. With that in mind, let's substitute z = (x-a).

    When you are trying to guess the answer for a polynomial, factoring it is usually much better than expanding. In this particular case, there are two nice quantities you could factor out of the equation, either z=(x-a) or b^2. This yields:

    [tex]z (z^2 - 3 b^2) = 2 b^3 [/tex]

    [tex]b^2 (3 z + 2b) = z^3[/tex]

    You should be able to guess at least one solution for z from these equations. Then you can factor out a (z-solution) term and use the quadratic formula to find the remaining solutions. Once you have all solutions for z, undo the substitution to get x.
  4. Aug 5, 2010 #3


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    To add onto that, what you can try is the rational root theorem instead. Because for this question to be do-able without a calculator of some sort, the roots must be simple in some sense!

    Say we have a cubic x3+ax2+bx+12=0

    We would use the rational root theorem test by trying all values that are factors of 12, i.e. 1,2,3,4,6,12. In this sense, we can try and assume b is an integer, which means that the factors of -2b3
    Oh, btw, the cubic is z3-3b2z-2b3=0, where z=x-a

    Anyway, the possible simple factors of -2b3 are going to be b,-b,2b,-2b. There could be more using this logic, such as b3 but since by testing these, we are going to have to substitute this into z and the z3 will just make this term b9 and that's obviously not going to cancel with the others.

    So trying the above possible roots:

    z=b, b3-3b3-2b3[itex]\neq[/itex]0
    z=-b, -b3+3b3-2b3=0
    z=2b, 8b3-6b3-2b3=0
    z=-2b, -8b3+6b3-2b3[itex]\neq[/itex]0

    So this means we have roots z=-b,2b.

    This means to find the last root we factor these roots and make it equivalent to the initial cubic:


    Now equate.
  5. Aug 5, 2010 #4


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    Seems you have missed that you have two terms in (x-a)b2 that you should combine.

    Then you may also see that you could further simplify, or make it look easier and more obvious, by dividing by b3 and expressing in terms of the new unknown, call it X, X = (x - a)/b.

    After which I hope the whole factorisation should look fairly obvious, if not come back with your result after those operations.
    Last edited: Aug 5, 2010
  6. Aug 5, 2010 #5
    Thanks for your reply.

    However, factoring z = x-a I get the following:

    [tex]z (z^2 - z - 2 b^2) = 2 b^3 [/tex]

    Which is quite different to what you wrote...
  7. Aug 5, 2010 #6
    The z inside the parentheses should be a b^2. There's no z^2 term in the original polynomial.
  8. Aug 5, 2010 #7


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    Sorry - I am hasty and error prone. See my edited version. I think this is the simplest approach. Unless I still have another error. :biggrin::uhh:
  9. Aug 5, 2010 #8
    I like epenguin's approach. I totally missed that the powers of z and b always add up to three. Thanks for pointing that out!
  10. Aug 5, 2010 #9
    This is the form from which you WILL solve it :)

    z = (x-a)

    z^3 - z(3b^2) - 2b^3 = 0

    like Mentallic said, IF integer roots exist, they must be the integer divisors of - 2b^3.

    Try to find in this way at least ONE root, then you can lower the degree of the polynomial (e.g. by poly long division) and then solve a second degree polynomial :P
  11. Aug 5, 2010 #10
    I think epenguin's suggestion results in the simplest equation to solve:

    [tex]z^3 - 3 z - 2 = 0[/tex]

    with z = (x-a)/b
  12. Aug 5, 2010 #11
    Thank you very much everyone for your help.

    I finally reached the solution with z = (x-a), but I also tried epenguin's substitution (z=(x-a)/b) and everything gets nicely simplified.

  13. Aug 5, 2010 #12


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    Ahh yes epenguin that's a sharp eye you have there!
  14. Aug 5, 2010 #13


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    One can notice at once the factor 2 in the last two terms. The expression offers the grouping

    (x-a)((x-a)^2 - b^2 )- 2b^2(b +x-a).

    From here, it is straightforward to proceed.

  15. Aug 5, 2010 #14


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    That is quite clever too! I can honestly admit I didn't spot that before grouping, so I wouldn't even start with grouping it in the first place, but if I did... that is definitely the way to go.
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