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Thread: Factorisation

  1. #1
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    Factorisation

    Can someone tell me if i have factorized correctly:

    Equation is this:$\displaystyle x^4-6x^2+1$
    First i completed the square:
    $\displaystyle x^4-6x^2+1$
    $\displaystyle x^4-6x^2+1-9x^2$
    $\displaystyle (x^2-3)^2-(3x)^2$
    $\displaystyle (x^2-3-3x)(x^2-3+3x)$
    Now i use the difference of two squares where $\displaystyle a^2-b^2=(a-b)(a+b)$
    $\displaystyle x^2-3x-3$
    $\displaystyle (x-3)^2-(\sqrt3)^2$
    $\displaystyle (x-3-\sqrt3)(x-3+\sqrt3)$
    $\displaystyle x^2+3x-3$
    $\displaystyle (x+3)^2-(\sqrt3)^2$
    $\displaystyle (x+3-\sqrt3)(x+3+\sqrt3)$
    Therefore answer is:
    $\displaystyle (x-3-\sqrt3)(x-3+\sqrt3)(x+3-\sqrt3)(x+3+\sqrt3)$


    P.S
    Last edited by Paymemoney; Dec 3rd 2009 at 11:11 PM.
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  2. #2
    Super Member Bacterius's Avatar
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    Use the \sqrt{number} command to make a square root
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  3. #3
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    Quote Originally Posted by Bacterius View Post
    Use the \sqrt{number} command to make a square root
    oh ok
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  4. #4
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    i got this

    $\displaystyle x^4-6x^2+1=(x^2-3)^2-{\sqrt{8}}^2 = (x^2-3-\sqrt{8})(x^2-3+\sqrt{8})$
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  5. #5
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    Quote Originally Posted by Paymemoney View Post
    Can someone tell me if i have factorized correctly:

    Equation is this:$\displaystyle x^4-6x^2+1$
    First i completed the square:
    $\displaystyle x^4-6x^2+1$
    $\displaystyle x^4-6x^2+1-9x^2$
    $\displaystyle (x^2-3)^2-(3x)^2$
    $\displaystyle (x^2-3-3x)(x^2-3+3x)$
    Now i use the difference of two squares where $\displaystyle a^2-b^2=(a-b)(a+b)$
    $\displaystyle x^2-3x-3$
    $\displaystyle (x-3)^2-(\sqrt3)^2$
    $\displaystyle (x-3-\sqrt3)(x-3+\sqrt3)$
    $\displaystyle x^2+3x-3$
    $\displaystyle (x+3)^2-(\sqrt3)^2$
    $\displaystyle (x+3-\sqrt3)(x+3+\sqrt3)$
    Therefore answer is:
    $\displaystyle (x-3-\sqrt3)(x-3+\sqrt3)(x+3-\sqrt3)(x+3+\sqrt3)$


    P.S
    I'd use a dummy variable $\displaystyle X = x^2$ to turn it into a quadratic.

    So $\displaystyle x^4 - 6x^2 + 1 = X^2 - 6X + 1$

    $\displaystyle = X^2 - 6X + (-3)^2 - (-3)^2 + 1$

    $\displaystyle = (X - 3)^2 - 8$

    $\displaystyle = (X - 3)^2 - (\sqrt{8})^2$

    $\displaystyle = (X - 3)^2 - (2\sqrt{2})^2$

    $\displaystyle = (X - 3 + 2\sqrt{2})(X - 3 - 2\sqrt{2})$


    Now remembering that $\displaystyle X = x^2$

    $\displaystyle = (x^2 - 3 + 2\sqrt{2})(x^2 - 3 - 2\sqrt{2})$.


    You could now use the difference of two squares again.


    $\displaystyle = [x^2 - (3 - 2\sqrt{2})][x^2 - (3 + 2\sqrt{2})]$

    $\displaystyle = \left[x^2 - \left(\sqrt{3 - 2\sqrt{2}}\right)^2\right]\left[x^2 - \left(\sqrt{3 + 2\sqrt{2}}\right)^2\right]$

    $\displaystyle = \left(x + \sqrt{3 - 2\sqrt{2}}\right)\left(x - \sqrt{3 - 2\sqrt{2}}\right)\left(x + \sqrt{3 + 2\sqrt{2}}\right)\left(x - \sqrt{3 + 2\sqrt{2}}\right)$.


    Ugly, yes, but fully factorised, yes.
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  6. #6
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    ^
    i understand the first part, however i don't understand why $\displaystyle 3-2\sqrt2$ is square rooted?
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  7. #7
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    Quote Originally Posted by Paymemoney View Post
    ^
    i understand the first part, however i don't understand why is $\displaystyle = \left[x^2 - \left(\sqrt{3 - 2\sqrt{2}}\right)^2\right]\left[x^2 - \left(\sqrt{3 + 2\sqrt{2}}\right)^2\right]
    $ is square rooted?
    Because $\displaystyle 3 - 2\sqrt{2} = \left(\sqrt{3 - 2\sqrt{2}}\right)^2$ and $\displaystyle 3 + 2\sqrt{2} = \left(\sqrt{3 + 2\sqrt{2}}\right)^2$.

    We had to do this to create another difference of two squares.
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  8. #8
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    Quote Originally Posted by Prove It View Post
    Because $\displaystyle 3 - 2\sqrt{2} = \left(\sqrt{3 - 2\sqrt{2}}\right)^2$ and $\displaystyle 3 + 2\sqrt{2} = \left(\sqrt{3 + 2\sqrt{2}}\right)^2$.

    We had to do this to create another difference of two squares.
    ok thanks
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