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11 August, 19:57

What are the zeros of p (m) = (m^2-4) (m^2+1) jmap?

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  1. 11 August, 20:07
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    (m^2 - 4) (m^2 + 1) = 0

    (m - 2) (m + 2) (m^2 + 1) = 0

    (m - 2) = 0 gives m = 2

    (m + 2) = 0 gives m = - 2

    m^2 + 1 = 0

    m^2 = - 1

    m = i, - i

    The zeros are - 2, 2, i, - i
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