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Math Help - Roots of Polynomial

  1. #1
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    Roots of Polynomial

    Let x=a be a root fo the quartic polynomial P(x) = x^4 + Ax^3 + Bx^2 + Ax + 1 where (2+B)^2=/ 4A^2

    Show that x=1/a is a root
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  2. #2
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    Quote Originally Posted by Lukybear View Post
    Let x=a be a root fo the quartic polynomial P(x) = x^4 + Ax^3 + Bx^2 + Ax + 1 where (2+B)^2=/ 4A^2

    Show that x=1/a is a root
    It would be helpful if you clarified what should be in place of the typo I marked in red above.
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  3. #3
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    Does not equal to is =/=

    sorry.
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    So we have P(a) = a^4 + Aa^3 + Ba^2 + Aa + 1 = 0.

    In order for \frac{1}{a} to be defined, we cannot have a=0. Setting a = 0 we see that the above equation reduces to 1 = 0, so we can safely say that a \neq 0.

    Notice that P\left(\frac{1}{a}\right) = \left(\frac{1}{a}\right)^4 + A\left(\frac{1}{a}\right)^3 + B\left(\frac{1}{a}\right)^2 + A\left(\frac{1}{a}\right) + 1 = \frac{1}{a^4}+\frac{A}{a^3}+\frac{B}{a^2}+\frac{A}  {a}+1.

    Multiply both sides by a^4 to remove fractions.

    a^4\cdot P\left(\frac{1}{a}\right)=a^4 + Aa^3 + Ba^2 + Aa + 1

    Look familiar?

    I'm still shaky on the restriction (2+B)^2 \neq 4A^2, which seems unnecessary. I'll look into it but maybe someone with keener insight can elucidate. Or perhaps it's just unnecessary.
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  5. #5
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    Ive pretty much got that. But the only thing i dont understand is that if we multiply a^4 to the polynomial, it changes dosent it. Since a^4 is not a roots of unity. Or are we to argue that since a^4 P(1/a) = 0 then p(1/a) = 0 via division?

    The restriction of (2+B)^2 =/=4A^2 is i think unncessary. Think that is another part of the question.
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  6. #6
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    Quote Originally Posted by Lukybear View Post
    Or are we to argue that since a^4 P(1/a) = 0 then p(1/a) = 0 via division?
    Yes. To put it another way, whenever we have xy = 0 then either x = 0 or y = 0, or both.
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