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Math Help - Please explain this question

  1. #1
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    Please explain this question

    The number a is called a double zero (or a zero of multiplicity 2) of the polynomial P if

    P(x) = (x -a)^2 q(x) and q(a) not = 0.

    Prove that if a is a double zero of P, then a is a zero of both P and P', and P"(a) not = 0.

    Recall that a Zero of a polynomial function P is one number a is subset to R such that P(a) = 0. You may assume without proof that q is a polynomial function.


    I do not understand the question at all, can anyone explain it and give some hints to prove it?
    Last edited by 450081592; November 8th 2009 at 07:52 PM.
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  2. #2
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    Quote Originally Posted by 450081592 View Post
    The number a is called a double zero (or a zero of multiplicity 2) of the polynomial P if

    P(x) = (x -a)^2 q(x) and q(a) not = 0.

    Prove that if a is a double zero of P, then a is a zero of both P and P', and P"(a) not = 0.


    I do not understand the question at all, can anyone explain it and give some hints to prove it?

    Let a be a double zero of a polynomial P

    Then, P(x)=(x-a)^2q(x) and q(a)\not = 0

    P(a)=(a-a)^2q(a)=0\cdot q(a)=0

    P'(x)=2(x-a)q(x)+(x-a)^2q'(x) by the product rule

    P'(a)=2(a-a)q(a)+(a-a)^2q'(a)=

    Then take the second derivative and show that when you plug in a you don't get zero
    Last edited by artvandalay11; November 10th 2009 at 07:34 AM.
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  3. #3
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    Can you tell me what are we trying to prove here, I really dont understand the question
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  4. #4
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    Quote Originally Posted by 450081592 View Post
    Can you tell me what are we trying to prove here, I really dont understand the question

    You are told that a is a double zero and told exactly what that means. Using that information, you are supposed to show that for ANY polynomial and ANY point satisfying this double zero property, the function evaluated at the point is 0, as is the derivative, but the second derivative is not.
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  5. #5
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    Quote Originally Posted by artvandalay11 View Post
    You are told that a is a double zero and told exactly what that means. Using that information, you are supposed to show that for ANY polynomial and ANY point satisfying this double zero property, the function evaluated at the point is 0, as is the derivative, but the second derivative is not.

    so we wanna prove that the function and its derivative are both 0 at a centain point, or any points, did you show me the solution or the thought.
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  6. #6
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    i showed you most of the solution and asked you to do the final part... do you understand the solution?
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    Quote Originally Posted by artvandalay11 View Post
    i showed you most of the solution and asked you to do the final part... do you understand the solution?

    I understand what you did, but I dont know why you are doing that , is all I need to do is to find the derivative? Are we treating a as a constant number here? Is that suppose to be q(a) and q'(a)?
    Last edited by 450081592; November 9th 2009 at 03:56 PM.
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  8. #8
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    Quote Originally Posted by 450081592 View Post
    I understand what you did, but I dont know why you are doing thatall I need to do is to find the derivative?

    All I did was find the derivative and plug in "a" to show it's zero


    I also plugged in "a" to show that P is zero at "a"
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  9. #9
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    I have no idea how to find the second derivative of the function, hints please1


    I got something like this  10q^2x^4-24aq^2x^3+18a^2q^2x^2-4a^3q^2x

    which is  10q^2a^4-24q^2a^4+18q^2a^4-4q^2a^4 which will equals to zero
    Last edited by 450081592; November 9th 2009 at 09:02 PM.
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  10. #10
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    Where did those "p" and "q" terms come from? In your original statement, P and q were functions, not constants.

    If P(x)= (x-a)^2q(x), P'= 2(x-a)q(x)+ (x-a)^2q'(x). Notice that each term contains a factor of "(x-a)" and so P'(a)= 0. Now differentiate again, using the product rule on each term.
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  11. #11
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    Hey 450081592, are you a first year student at Carleton University?
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  12. #12
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    Quote Originally Posted by Sterwine View Post
    Hey 450081592, are you a first year student at Carleton University?

    Haha, exactly, are you too?
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