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Simple Polynomial Factorization

  1. Mar 3, 2012 #1
    There is a theorem in algebra, whose name I don't recall, that states that given a polynomial and its roots I can easily factor it so for instance :

    [itex]p(x)=x^2-36[/itex] ,
    assuming that p(x) is a real function,

    [itex]p(0)=0 \Leftrightarrow x=6,-6 [/itex]
    then p(x) can be written as :

    I was trying to derive the good and old difference of two cubes expression using this factorization technique, then :

    let [itex] p(x)=x^3-a^3 [/itex] and [itex]p(x)=0 \Leftrightarrow x=a[/itex] then [itex] p(x)=(x-a)(x-a)(x-a)=(x-a)(x^2 -2xa+a^2) ,[/itex] while the formula should be [itex] (x-a)(x^2+ax+a^2)[/itex]

    What's wrong?
    Last edited: Mar 3, 2012
  2. jcsd
  3. Mar 3, 2012 #2


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    The problem is that polynomials have roots which aren't real numbers. While a is the only real root to x3-a3=0, if you let [itex] \omega = e^{2\pi i/3}[/itex], then [itex] a\omega[/itex] and [itex] a \omega^2[/itex] are also roots. So
    [tex]p(x) = (x-a)(x-a\omega)(x-a\omega^2)[/tex]
  4. Mar 3, 2012 #3
    Thank you, makes perfect sense now :) .
  5. Mar 3, 2012 #4


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    Or you can even try this:

    So you know that x=a is a root of p(x)=x3-a3, so we'll factorize that out and leave the other factor as a quadratic (because a linear equation multiplied by a quadratic is a cubic)

    So what we have is


    And we want to find what the values b,c,d are. Well, we can expand!



    And we also know that


    So we can now equate the coefficients of each power in x.

    [tex]p(x)=x^3+0x^2+0x-a^3\equiv bx^3+(c-ab)x^2+(d-ac)x-ad[/tex]


    [tex]bx^3=x^3, b=1[/tex]
    (from here we'll neglect the x terms because they'll always cancel)



    [tex]-ad=-a^3, d=a^2[/tex]

    Substituting d=a2 back into d-ac=0 gives us
    [tex]a^2-ac = a(a-c) = 0[/tex]

    So either a=0 or a=c, but we already know [itex]a\neq 0[/itex] because that was our first root, and if a=0 then we're trying to factorize just x3 which is a trivial case, so a=c then. Thus we can plug all the values in to finally get


    And you can use the quadratic formula to find the complex roots that Office_Shredder mentioned.

    Oh and as a tip, you can quickly and easily use the method of equating the coefficients if you use some common sense such as already realizing that b=1 and d=a2 because there is only one x3 and constant term each.
  6. Mar 3, 2012 #5
    Thank you Metallic.
    This approach definitely looks elegant, I'll try to use it to extend to the general case [tex]x^n-a^n[/tex] .
  7. Mar 3, 2012 #6


    Staff: Mentor

    I don't think this was touched on completely in the reponses following this post. What is wrong is that x3 - a3 does not factor into (x - a)3, as you show above.

    Also, (x - a)3 ≠ (x - a)(x2 - 2ax + a2) as you also show above. (It is true that x3 - a3 = (x - a)(x2 - 2ax + a2). )
  8. Mar 3, 2012 #7
    [tex]a[/tex] is triple solution if [tex]f(a)=f'(a)=f''(a)=0[/tex].
    In this case we have [tex]f(a)=a^3-a^3=0[/tex], then, [tex]f'(x)=3x^2[/tex], so [tex]f'(a)=3a^2\neq 0[/tex] (if [tex]a\neq 0[/tex]), so you can't say that [tex]a[/tex] is triple solution as you did (saying [tex]x^3-a^3=(x-a)(x-a)(x-a)[/tex].
    We supposed that [tex]a[/tex] isn't function by [tex]x[/tex].

    Sorry for bad English.
  9. Mar 3, 2012 #8
    I was justifying that based on an result on algebra that, as far as I can remember, says that any n-polynomial (n[itex]\geq[/itex]1) has exactly(not sure about this "exactly") n roots. Therefore I thought that the only real root "a" found should have multiplicity 3. But I'm pretty sure now that the theorem meant n complex roots, which is much more reasonable. I could probably use some review of algebra theorems.
  10. Mar 3, 2012 #9

    Char. Limit

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    Yes indeed, the theorem means "complex" roots. Other than that, you had it correct, though. A polynomial of nth degree has n complex roots, exactly.
  11. Mar 3, 2012 #10


    Staff: Mentor

    No, if a is a root of an n-th polynomial, that doesn't mean that the multiplicity has to be n. For the polynomial f(x) = x3 - a3, a is the only real root (hence of multiplicity 1).
  12. Mar 3, 2012 #11
    So, what you're saying is that I can show that :
    a is an n-ple solution [itex]\Rightarrow[/itex] [tex]f(a)=f'(a)=f''(a)=...=f^n(x)=0[/tex]
  13. Mar 3, 2012 #12


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    And the other way around is true also
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