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Math Help - Finding a polynomial with degree, zeroes, and multiplicity

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    Finding a polynomial with degree, zeroes, and multiplicity

    Question:
    Find a polynomial f(x) of degree 6 such that 0,3 are zeroes of multiplicity 3 and f(2) = -24.

    Can someone explain this step-by-step? what exactly is "multiplicity"?
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    Moo
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    Hello,
    Quote Originally Posted by mwok View Post
    Question:
    Find a polynomial f(x) of degree 6 such that 0,3 are zeroes of multiplicity 3 and f(2) = -24.

    Can someone explain this step-by-step? what exactly is "multiplicity"?
    If a number a has a multiplicity b in a polynomial f(x), it means that (x-a)^b divides this polynomial.


    So here, f(x)=k (x-3)^3 (x-0)^3
    that's how it works.

    Now, substitute x=2 to find k !
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    I see.
    How about if it was:
    Find a polynomial f(x) of degree 5 such that 0,3 are zeroes of multiplicity 3 and f(2) = -24. (Notice the degree is now 5)

    How would you solve this one?
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    Quote Originally Posted by mwok View Post
    I see.
    How about if it was:
    Find a polynomial f(x) of degree 5 such that 0,3 are zeroes of multiplicity 3 and f(2) = -24. (Notice the degree is now 5)

    How would you solve this one?
    It is not possible.

    Note that if the degree of a polynomial is n, there are at most 5 roots for it.

    If a root a has a multiplicity of b, it's like you have b roots that are all equal to a.

    In fact, if a has a multiplicity of b, you have f(a)=f'(a)=f''(a)=\ldots=f^{(b-1)}(a)=0

    f^{(b-1)}(x) denots the (b-1)th derivative of f.




    I replied to someone here : http://www.mathhelpforum.com/math-he...tiplicity.html dunno if it can give you another view on it
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