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Math Help - Polynomial Weird Roots question.

  1. #1
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    Polynomial Weird Roots question.

    If P(x) = x^5 + ax^2 + b has multiple roots. Prove that 3125b^5 = 108 a^3

    Umm... I just cant prove that..... is the question wrong or am i doing something wrong =S?


    thanks in advance.
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  2. #2
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    Hello,
    Quote Originally Posted by shinn View Post
    If P(x) = x^5 + ax^2 + b has multiple roots. Prove that 3125b^5 = 108 a^3

    Umm... I just cant prove that..... is the question wrong or am i doing something wrong =S?


    thanks in advance.
    I'll try. I did it on a paper and just noticed a tiny mistake, so I'll go freely while typing lol ! But you gotta check it because I'm sure I won't get the result you're given

    If P(x) has multiple roots (one or more or all roots are multiple ?), then there exists m such that P(m)=P'(m)=0

    0=m^5+am^2+b (1) and 0=5m^4+2am (2)
    From (2), we get m=0 or 5m^3+2a=0

    If m=0, then b=0 by (1).
    Let's say m \neq 0 otherwise that's not funny

    So 5m^3+2a=0 \implies m=-\left(\frac 25 a\right)^{\tfrac 13}

    Substituting in (1) :

    -\left(\frac 25\right)^{\tfrac 53} a^{\tfrac 53}+a*\left(\frac 25\right)^{\tfrac 23} a^{\tfrac 23}+b=0

    -\left(\frac 25\right)^{\tfrac 53} a^{\tfrac 53}+\left(\frac 25\right)^{\tfrac 23} a^{\tfrac 53}+b=0

    b=a^{\tfrac 53} \left(\left(\tfrac 25\right)^{\tfrac 53}-\left(\tfrac 25\right)^{\tfrac 23}\right)

    Take the 3rd power of both sides :

    b^3=a^5 \left(\left(\tfrac 25\right)^{\tfrac 53}-\left(\tfrac 25\right)^{\tfrac 23}\right)^3

    ----------------------------------------------------------------
    Remember that (x-y)^3=x^3-y^3+3(xy^2-x^2y)

    We have (I write it in the same order so that you can see) :
    x^3=\frac{2^5}{5^5} and y^3=\frac{2^2}{5^2}

    xy^2=\left(\frac 25\right)^{\tfrac 53} \cdot \left(\frac 25\right)^{\tfrac 43}=\left(\frac 25\right)^3=\frac{2^3}{5^3}

    and similarly x^2y=\dots=\frac{2^4}{5^4}
    ----------------------------------------------------------------


    b^3=a^5 \left(\frac{2^5}{5^5}-\frac{2^2}{5^2}+3 \cdot \frac{2^3}{5^3}-3 \cdot \frac{2^4}{5^4}\right)

    b^3=a^5 \cdot \left(-\frac{108}{3125}\right)


    \boxed{-3125b^3=108a^5}

    Which is too similar to be a big mistake
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  3. #3
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    Cheers, I've got the exact answer as u............... the question must be wrong.
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