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Math Help - Roots of a polynomial equation

  1. #1
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    Roots of a polynomial equation

    I'm pretty lost here.. and simply unclear on what exactly I should be doing.
    If someone could walk me through the steps to getting this answer, I'd extremely appreciate it.


    Find the roots of the polynomial equation: x^3 - 4x^2 + 6x - 4 = 0.

    I've used synthetic division to find -2 as a root and get the depressed polynomial x^2 - 2x + 2.. and then used the quadratic formula to find what I think are the correct zeroes of that equation. (which are: 1+i, 1-i.)

    Here's where I'm unsure of what to do. Thanks for your help!
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  2. #2
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    Quote Originally Posted by Savior_Self View Post
    I've used synthetic division to find -2 as a root and get the depressed polynomial x^2 - 2x + 2.. and then used the quadratic formula to find what I think are the correct zeroes of that equation. (which are: 1+i, 1-i.)
    Well done!
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  3. #3
    A riddle wrapped in an enigma
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    Quote Originally Posted by Savior_Self View Post
    I'm pretty lost here.. and simply unclear on what exactly I should be doing.
    If someone could walk me through the steps to getting this answer, I'd extremely appreciate it.


    Find the roots of the polynomial equation: x^3 - 4x^2 + 6x - 4 = 0.

    I've used synthetic division to find -2 as a root and get the depressed polynomial x^2 - 2x + 2.. and then used the quadratic formula to find what I think are the correct zeroes of that equation. (which are: 1+i, 1-i.)

    Here's where I'm unsure of what to do. Thanks for your help!
    Hi Savior_Self,

    -2 is not one of your roots. (x - 2) is a factor of your original cubic function, but the root is +2.

    Your depressed quadratic is correct, though.

    All you need to do now is state your roots: {2, 1+i, 1-i}.
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  4. #4
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    Ahhh, actually, -2 was a typo.. I have +2 written down on the paper.

    But good to know I was close. Thanks guys.
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