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Thread: factor theorem

  1. #1
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    factor theorem

    Find all the real roots of the equation :

    $\displaystyle
    2x^6+x^4-8x^2+4=0
    $
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by thereddevils View Post
    Find all the real roots of the equation :

    $\displaystyle
    2x^6+x^4-8x^2+4=0
    $

    the possible root is $\displaystyle \pm 2 , \pm 1,\pm 4 , \pm \frac{1}{2} $

    but no one of the possible root is a root of the polynomial that you wrote

    there is no real root can be found without calculator see what I have when I put them in the calculator I think there is wrong in your equation

    factor theorem-root.jpg

    Edited : - correct something yeongil mentioned
    Last edited by Amer; Jun 11th 2009 at 01:21 AM.
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  3. #3
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    Quote Originally Posted by Amer View Post
    the possible root is $\displaystyle \pm 2 , \pm 1,\pm 4 $
    Minor correction: the possible rational roots are $\displaystyle \pm\frac{1}{2}, \pm 1, \pm 2, \pm 4$ . And as you discovered, none of the possible rational roots are roots of the polynomial.


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  4. #4
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    Btw the answer is $\displaystyle \pm\sqrt{2}$
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  5. #5
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    Quote Originally Posted by thereddevils View Post
    Btw the answer is $\displaystyle \pm\sqrt{2}$
    No, they are not. If you plug either $\displaystyle \sqrt{2}$ or $\displaystyle -\sqrt{2}$ into the original equation:
    $\displaystyle 2x^6+x^4-8x^2+4=0$
    you get 8 = 0, so neither solution works.

    However, if your original equation was supposed to be
    $\displaystyle 2x^6 + x^4 - 8x^2 {\color{red}-} 4 = 0$
    then yes, those two solutions work. In fact, it makes solving this equation much easier, as you can solve it by factoring by grouping:
    $\displaystyle \begin{aligned}
    2x^6 + x^4 - 8x^2 - 4 &= 0 \\
    x^4(2x^2 + 1) - 4(2x^2 + 1) &= 0 \\
    (x^4 - 4)(2x^2 + 1) &= 0 \\
    (x^2 - 2)(x^2 + 2)(2x^2 + 1) &= 0
    \end{aligned}$

    Set each factor equal to zero:

    $\displaystyle \begin{aligned}
    x^2 - 2 &= 0 \\
    x^2 &= 2 \\
    x &= \pm \sqrt{2}
    \end{aligned}$
    (the two real solutions)

    $\displaystyle \begin{aligned}
    x^2 + 2 &= 0 \\
    x^2 &= -2 \\
    x &= \pm i\sqrt{2}
    \end{aligned}$
    (two non-real solutions)

    $\displaystyle \begin{aligned}
    2x^2 + 1 &= 0 \\
    2x^2 &= -1 \\
    x^2 &= -\frac{1}{2} \\
    x &= \pm \frac{\sqrt{2}}{2}i
    \end{aligned}$
    (two more non-real solutions)

    So it looks like you copied the problem wrong?


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  6. #6
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    Quote Originally Posted by yeongil View Post
    No, they are not. If you plug either $\displaystyle \sqrt{2}$ or $\displaystyle -\sqrt{2}$ into the original equation:
    $\displaystyle 2x^6+x^4-8x^2+4=0$
    you get 8 = 0, so neither solution works.

    However, if your original equation was supposed to be
    $\displaystyle 2x^6 + x^4 - 8x^2 {\color{red}-} 4 = 0$
    then yes, those two solutions work. In fact, it makes solving this equation much easier, as you can solve it by factoring by grouping:
    $\displaystyle \begin{aligned}
    2x^6 + x^4 - 8x^2 - 4 &= 0 \\
    x^4(2x^2 + 1) - 4(2x^2 + 1) &= 0 \\
    (x^4 - 4)(2x^2 + 1) &= 0 \\
    (x^2 - 2)(x^2 + 2)(2x^2 + 1) &= 0
    \end{aligned}$

    Set each factor equal to zero:

    $\displaystyle \begin{aligned}
    x^2 - 2 &= 0 \\
    x^2 &= 2 \\
    x &= \pm \sqrt{2}
    \end{aligned}$

    $\displaystyle \begin{aligned}
    x^2 + 2 &= 0 \\
    x^2 &= -2 \\
    x &= \pm i\sqrt{2}
    \end{aligned}$

    $\displaystyle \begin{aligned}
    2x^2 + 1 &= 0 \\
    2x^2 &= -1 \\
    x^2 &= -\frac{1}{2} \\
    x &= \pm \frac{\sqrt{2}}{2}i
    \end{aligned}$

    So it looks like you copied the problem wrong?


    01
    oh thanks a lot .. nope , i copied it down correctly .. i think its the book's mistake

    Btw i hv a question here .. For questions like this where the leading coefficient is not 1 , is it a must for me to reduce the coefficient to 1 before i find the factors : example in this case

    $\displaystyle 2x^6+x^4-8x^2+4=0$

    $\displaystyle x^6+\frac{1}{2}x^4-4x^2+2=0$ , is this step necessary

    Before i start substituting the factors of the constant term to find the factors of the equation .

    THanks .
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  7. #7
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    Quote Originally Posted by thereddevils View Post
    oh thanks a lot .. nope , i copied it down correctly .. i think its the book's mistake

    Btw i hv a question here .. For questions like this where the leading coefficient is not 1 , is it a must for me to reduce the coefficient to 1 before i find the factors : example in this case

    $\displaystyle 2x^6+x^4-8x^2+4=0$

    $\displaystyle x^6+\frac{1}{2}x^4-4x^2+2=0$ , is this step necessary

    Before i start substituting the factors of the constant term to find the factors of the equation .

    THanks .
    Absolutely not.. It will just make your work difficult
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