Find p in terms of k and then completely factor -k^3 + k^2pk - pk +p^2 = 0?

Find p in terms of k and then completely factor -k^3 + k^2pk - pk +p^2 = 0

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  • 1 decade ago
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    -k^3 + k^2pk - pk +p^2 = 0

    p^2 + pk (k^2 - 1) - k^3 = 0

    Solve for p by using the quadratic formula, with the usual coefficients being, a = 1, b = k(k^2 - 1), c = -k^3

    p = { -k(k^2 - 1) ± √ [k^2(k^2 - 1)^2 + 4k^3] } /2

    p = { -k(k^2 - 1) ± k√ [(k^2 - 1)^2 + 4k] } /2

    and that is nasty.

    So now I am in the realm of making up my own problem:

    The second term "k^2pk" looks very strange, so I will suppose it was meant to be "k^2p". So starting with

    -k^3 + k^2p - pk + p^2 = 0

    p^2 + p (k^2 - k) - k^3 = 0

    Solve for p by using the quadratic formula, with the usual coefficients being, a = 1, b = (k^2 - k) = k(k -1), c = -k^3

    p = { (-k^2 + k) ± √ [k^2(k - 1)^2 + 4k^3] } /2

    p = { (-k^2 + k) ± k√ [(k - 1)^2 + 4k] } /2

    p = [ (-k^2 + k) ± k√ (k^2 + 2k +1) ] /2

    p = [ (-k^2 + k) ± k√ (k +1)^2 ] /2

    p = [ (-k^2 + k) ± k(k +1) ] /2

    p = k, -k^2

    so by the factor theorem, the factorization is:

    (p - k) (p + k^2)

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