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Proof of a corollary of fundamental theorem of algebra

  1. Mar 22, 2012 #1
    1. The problem statement, all variables and given/known data

    Assuming the validity of the fundamental theorem of algebra, prove the corollary that:

    Every polynomial of positive degree n has a factorization of the form:

    [itex]P(x)=a_{n}(x-r_{1})...(x-r_{n})[/itex] where [itex]r_{i}[/itex] aren't necessarily distinct.

    2. Relevant equations

    Fundamental Theorem of Algebra: Every polynomial of positive degree with complex coefficients has at least one complex zero.

    3. The attempt at a solution

    Does this even require proof? Don't know where to begin...
     
  2. jcsd
  3. Mar 22, 2012 #2

    Dick

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    Sure it requires proof. Start by saying P(x) is has a root r1. Show (x-r1) divides P(x), so you can write P(x)=(x-r1)P1(x) where P1(x) has degree n-1. Now apply the same thing to P1(x). Etc. If you want to be more formal you prove it by induction.
     
  4. Mar 23, 2012 #3
    Attempt:

    (1) We have a polynomial [itex]P(x)=a(x-r_{1})[/itex]

    (2) [itex]x-r_{1}[/itex] is a factor if and only if [itex]r_{1}[/itex] is a zero (Remainder theorem)

    (3) [itex]P(r_{1})=a(r_{1}-r_{1})[/itex] which is zero, therefore [itex]r_{1}[/itex] is a zero and [itex]x-r_{1}[/itex] is a factor.

    At this point I have a question: by what theorem can now take the step: (4) [itex]P(x)=a(x-r_{1})(x-r_{n-1})[/itex] or should this be obvious? Because I don't understand why is this the case.
     
  5. Mar 23, 2012 #4

    Dick

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    P(x) is a polynomial of degree n. You've only got a polynomial of degree 1. Reread my last post.
     
  6. Mar 24, 2012 #5
    But I can't assume that [itex]P(x)[/itex] is a polynomial o degree n and than just write: [itex]P(x)=a(x-r_{1})[/itex] where do I indicate the degree? Should I write: [itex]P(x)=a(x-r_{1})^{n}[/itex] ? Or something?
     
    Last edited: Mar 24, 2012
  7. Mar 24, 2012 #6

    Dick

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    I would write it as [itex]P(x)=(x-r_{1})P_1(x)[/itex]. When you divide [itex]x-r_1[/itex] into [itex]P(x)[/itex] you are going to get another polynomial, not just a constant.
     
  8. Mar 24, 2012 #7
    Attempt #2:

    (1) We denote a polynomial of degree n as [itex]P(x)=a(x-r_{1})P_{1}(x)[/itex]

    (2) [itex]x-r_{1}[/itex] is a factor if and only if [itex]r_{1}[/itex] is a zero (Remainder theorem)

    (3) [itex]P(r_{1})=a(r_{1}-r_{1})P_{1}(r_{1})[/itex] which is (NOT?)zero, therefore [itex]r_{1}[/itex] is (NOT?)a zero and [itex]x-r_{1}[/itex] is (NOT?) a factor.

    It doesn't divide then? Or have I misunderstood smth?
     
  9. Mar 24, 2012 #8

    HallsofIvy

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    I would be inclined to do this as a "proof by induction" on n, the degree of the polynomial. You have done the "n= 1" case. Now show that "if a polynomial of degree k can be factored as linear factors, so can a polynomial of degree k+1".
     
  10. Mar 25, 2012 #9
    (1) We have a statement [itex]P_{1}[/itex] "a polynomial [itex]a(x-r_{1})[/itex] has expresion [itex]x-r_{1}[/itex] as a factor", which is equivalent to saying that it can be factored as [itex]a(x-r_{1})[/itex]???. Lets call the first polynomial [itex]P_{1}(x)[/itex] in accordance to first statement and denote [itex]P_{n}(x)[/itex] a polynomial corresponding to the statement [itex]P_{n}[/itex] for the polynomial [itex]a(x-r_{1})(x-r_{2})...(x-r_{n})[/itex]

    (2) Now, [itex]x-r_{1}[/itex] is a factor if and only if [itex]r_{1}[/itex] is a zero (Remainder theorem)

    (3) Since [itex]P_{1}(r_{1})=a(r_{1}-r_{1})[/itex] is equal to zero, therefore [itex]r_{1}[/itex] is a zero and [itex]x-r_{1}[/itex] is a factor of the polynomial.

    (4) Assume that statement [itex]P_{k}[/itex] is true.

    (5) [itex]P_{k+1}[/itex] is the statement: a polynomial [itex]a(x-r_{1})(x-r_{2})...(x-r_{k})(x-r_{k+1})[/itex] has expresion [itex]x-r_{k+1}[/itex] as a factor. Repeating (2) and (3) with [itex]r_{k+1}[/itex] we get [itex]P_{k+1}(k+1)=0[/itex]

    (6)Therefore every polynomial of positive degree n has a factorization of the form:

    [itex]P(x)=a_{n}(x-r_{1})...(x-r_{n})[/itex] where [itex]r_{i}[/itex] aren't necessarily distinct.

    Q.E.D. ??? I don't think so.

    Shouldn't I start with general rule for a polynomial: [itex]f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+...+a_{1}x+a_{0}[/itex] ???
     
    Last edited: Mar 25, 2012
  11. Mar 25, 2012 #10
    Can someone approve that this constitutes a proof of what I intended to prove? Maybe additional qualifications are needed - some steps are missing?
     
  12. Mar 25, 2012 #11

    Dick

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    If you are doing induction [itex]P_k[/itex] should be the statement "any polynomial of degree k can be expressed in the form [itex]a (x-r_1) (x-r_2) ... (x-r_k)[/itex]".

    You can do the k=1 case without using your theorem at all. [itex]f(x)=a_1 x + a_0[/itex] is a polynomial of degree 1. You can write that as [itex]f(x)=a_1 (x + \frac{a_0}{a_1})[/itex] Since your root is [itex]r_1=(-a_0/a_1)[/itex] that expresses f(x) in the form a(x-r). To prove [itex]P_k[/itex] implies [itex]P_{k+1}[/itex], pick a polynomial p(x) of degree k+1. The fundamental theorem of algebra says p(x) has a root, call it [itex]r_{k+1}[/itex]. So [itex]p(x)=(x-r_{k+1})*q(x)[/itex] where q(x) has degree k. Now use your induction hypothesis on q(x).
     
  13. Mar 25, 2012 #12
    (1) Assume the statement [itex]P_{k}[/itex]: "the polynomial [itex]q(x)[/itex] od degree k can be written as [itex]q(x)=a_{k}(x-r_{1})...(x-r_{k})[/itex]" is true.

    (2) Now the statement [itex]P_{k+1}[/itex] is about the polynomial [itex]p(x)=(x-r_{k+1})q(x)[/itex] which can be rewritten as: [itex]p(x)=(x-r_{k+1})a_{k}(x-r_{1})...(x-r_{k})[/itex].

    (3) Since we've assumed [itex]P_{k}[/itex] is true, and [itex]P_{k+1}[/itex] was shown to be true previously (I didn't write the statement, but its trivial) - QED (???)

    Can anyone give feedback, especially for the step (2).
     
    Last edited: Mar 25, 2012
  14. Mar 25, 2012 #13

    HallsofIvy

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    You need to state explicitely, and support,
    "If Pk+1 is a polynomial of degree k+1 and Pk+1(x)= (x- r)Q(x), then Q(x) is a polynomial of degree k".
     
    Last edited: Mar 25, 2012
  15. Mar 25, 2012 #14
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