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Math Help - find all real and imaginary roots

  1. #1
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    find all real and imaginary roots

    I need to find all real and imaginary roots of this equation:

    2x^7 - 2x^6 + 7x^5 - 7x^4 - 4x^3 + 4x^2 = 0

    Using synthetic division, I found that 1 is a root.

    (x - 1)(2x^6 + 7x^4 - 4x^2) = 0

    I'm not really sure how to proceed from here. Could someone point me in the right direction?
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  2. #2
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    Hello, absvalue!

    Find all real and imaginary roots: . 2x^7 - 2x^6 + 7x^5 - 7x^4 - 4x^3 + 4x^2 \:=\: 0
    First, factor out x^2 . . .

    . . x^2(2x^5 - 2x^4 + 7x^3 - 7x^2 - 4x + 4) \:=\:0


    Factor by grouping:

    . . x^2\,\bigg[2x^4(x-1) + 7x^2(x-1) - 4(x-1)\bigg] \:=\:0

    . . x^2(x-1)(2x^4 - 7x^2 - 4) \:=\:0

    . . x^2(x-1)(x^2-4)(2x^2+1) \:=\:0


    Then we have:

    . . x^2 \:=\:0 \quad\Rightarrow\quad\boxed{ x \:=\:0}


    . . x - 1 \:=\:0 \quad\Rightarrow\quad\boxed{ x \:=\:1}


    . . x^2 - 4 \:=\:0 \quad\Rightarrow\quad x^2 \:=\:4 \quad\Rightarrow\quad\boxed{ x \:=\:\pm2}

    . . . . 2x^2+1 \:=\:0 \quad\Rightarrow\quad x^2 \:=\:-\frac{1}{2} \quad\Rightarrow\quad\boxed{x \:=\:\pm\frac{i}{\sqrt{2}}}

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  3. #3
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    Hello absvalue
    Quote Originally Posted by absvalue View Post
    I need to find all real and imaginary roots of this equation:

    2x^7 - 2x^6 + 7x^5 - 7x^4 - 4x^3 + 4x^2 = 0

    Using synthetic division, I found that 1 is a root.

    (x - 1)(2x^6 + 7x^4 - 4x^2) = 0

    I'm not really sure how to proceed from here. Could someone point me in the right direction?
    Take out a common factor x^2 from your second factor; you then have a quadratic in x^2, which can be factorised in the usual way:

    (x-1)(2x^6+7x^4-4x^2) = x^2(x-1)(2x^4+7x^2-4)

    =x^2(x-1)(2x^2-1)(x^2+4)

    Can you complete it from here?

    Grandad
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  4. #4
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    Thanks! That makes sense now.
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