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Math Help - [SOLVED] Miscellaneous Factors

  1. #1
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    [SOLVED] Miscellaneous Factors

    Q. Factorise the following:

    z^3 - 7z - z^2 + 7
    had a look at the answer and is (z - 1)(z^2 - 7)
    i tried the group in pairs method but i didn't get the answer, and now i have no idea on how to work it out.
    if anyone can show me the full working out that would be a great help, thanks
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  2. #2
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    Quote Originally Posted by waven View Post
    Q. Factorise the following:

    z^3 - 7z - z^2 + 7
    had a look at the answer and is (z - 1)(z^2 - 7)
    i tried the group in pairs method but i didn't get the answer, and now i have no idea on how to work it out.
    if anyone can show me the full working out that would be a great help, thanks
    It can be written as

    z^3-z^2 -7z + 7

    1z^2*(z-1) -7*(z-1)

    Take (z-1) out as common

    2 (z^2-7)*(z-1)

    Tell the step number where you had/have trouble

    Adarsh
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  3. #3
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    Quote Originally Posted by waven View Post
    Q. Factorise the following:

    z^3 - 7z - z^2 + 7
    had a look at the answer and is (z - 1)(z^2 - 7)
    i tried the group in pairs method but i didn't get the answer, and now i have no idea on how to work it out.
    if anyone can show me the full working out that would be a great help, thanks
    Let p(z) = z^3 - 7z - z^2 + 7. By inspection, p(1) = 0. Therefore (z - 1) is a factor of z^3 - 7z - z^2 + 7. Divide (z - 1) into the cubic to get the quadratic factor.
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