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Polynomial Proofs

  1. May 29, 2006 #1
    If [tex] n \geq 1 [/tex] and [tex] f(a) = 0 [/tex] for some real a , then [tex] f(x) = (x-a)h(x) [/tex], where h is a polynomial of degree [tex] n-1 [/tex]. So:

    [tex] f(a) = \sum_{k=0}^{n} c_{k}a^{k} = c_{0} + c_{1}a + c_{2}a^{2} + ... + c_{n}a^{n} = 0 [/tex]. In a hint it says to consider [tex] p(x) = f(x+a) [/tex]. So I expanded that and got: [tex] c_{0}+c_{1}(x+a)+c_{2}(x+a)^{2} + ... + c_{n}(x+a)^{n} [/tex]. So how do I use this to prove the above statement?
     
  2. jcsd
  3. May 29, 2006 #2

    AKG

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    If there exists a polynomial h(x) such that f(x) = (x-a)h(x), we say that (x-a) is a factor of f(x). In general, if f(x) and g(x) are any two polynomials, then if there exists h(x) such that f(x) = g(x)h(x), we say that g(x) is a factor of f(x). So the problem asks you to prove that if f(a) = 0, then (x-a) is a factor of f(x). Do you see that it suffices to prove that x is a factor of p(x), given that p(x) is defined as f(x+a)?

    Well you know that f(a) = 0, which gives:

    [tex]c_{0} + c_{1}a + c_{2}a^{2} + ... + c_{n}a^{n} = 0 [/tex]

    And you've written out p(x) as:

    [tex] c_{0}+c_{1}(x+a)+c_{2}(x+a)^{2} + ... + c_{n}(x+a)^{n} [/tex]

    Can you put these two facts together to show that x divides p(x), i.e. that x is a factor of p(x)?
     
  4. May 29, 2006 #3
    Well [tex] c_{0}+c_{1}(x+a)+c_{2}(x+a)^{2} + ... + c_{n}(x+a)^{n} = c_{0} + c_{1}x+c_{1}a + c_{2}x^{2} + 2c_{2}xa + c_{2}a^{2} + ... + c_{n}(x+a)^{n}[/tex]. I know that we can factor out the [tex] f(a) [/tex] and set it equal to 0. But then what is the new expression. Also could you explain why it suffices to prove that x is a factor of [tex] p(x) [/tex] if [tex] p(x) = f(x+a) [/tex]?

    Thanks
     
  5. May 29, 2006 #4
    You want to show that (x-a) is a factor of f(x). Do you see how this is related to showing that x is a factor of p(x)=f(x+a)? Or perhaps showing that t is a factor of f(t+a)?
     
  6. May 30, 2006 #5
    When I first looked at this, I didn't understand it either. But after some deep thinking, it started making sense.

    If (x - a) is a factor of f(x), the x is a factor of p(x). Why? Because if we assume that f(x) = (x - a) h(x) for some h(x), then p(x) = f(x + a) = (x + a - a)h(x) = x h(x).

    Do you understand now?

    I'll give you a hint to prove that x is a factor of p(x): p(x) = f(x + a) = f(x + a) - f(a) since f(a) = 0.
     
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