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Math Help - factor (x^4)+(2y^4)???????????

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    Member sluggerbroth's Avatar
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    factor (x^4)+(2y^4)???????????

    book shows answer as x^4+4x^2y^2+4y^4-4x^2y^2=(x^2+2y^2)^2-(2xy)^2=(x^2+2y^2-2xy)(x^2+2y^2+2xy) What is happening here and what steps am i missing???
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    Re: factor (x^4)+(2y^4)???????????

    Quote Originally Posted by sluggerbroth View Post
    book shows answer as x^4+4x^2y^2+4y^4-4x^2y^2=(x^2+2y^2)^2-(2xy)^2=(x^2+2y^2-2xy)(x^2+2y^2+2xy) What is happening here and what steps am i missing???
    x^4+4x^2y^2+4y^4-4x^2y^2

    {\color{red} (x^4+4x^2y^2+4y^4)} -4x^2y^2

    note these terms form a perfect square ...


    {\color{red} (x^2+2y^2)^2} - (2xy)^2

    now you have the difference of two squares ... a^2-b^2 = (a-b)(a+b)

    [(x^2+2y^2)-2xy] \cdot [(x^2+2y^2)+2xy]
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    Re: factor (x^4)+(2y^4)???????????

    What they have done is add and subtract 4x^2y which, of course, doesn't change the actual value. But that does change the form so that it we can think of it as x^4+4x^2y^2+4y^4-4x^2y^2= (x^2+ 2y^2)^2- 4x^2y^2 which is a "difference" of two squares. You probably know that a^2- b^2= (a- b)(a+ b) so that last is ([x^2+ 2y^2]+ 2xy)([x^2+ 2y^2]- 2xy)= (x^2+ 2xy+ 2y^2)(x^2- 2xy+ 2y^2).
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    Member sluggerbroth's Avatar
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    Re: factor (x^4)+(2y^4)???????????

    they just add ( and subsequently subtract ) 4x^2y^2 because it forms a perfect square?
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    Re: factor (x^4)+(2y^4)???????????

    another example ... let's say you wish to factor y = x^4 + 1

    x^4 + 2x^2 + 1 - 2x^2

    (x^2 + 1)^2 - (\sqrt{2} x)^2

    [(x^2+1) - \sqrt{2} x][(x^2 + 1) + \sqrt{2} x]

    (x^2 - \sqrt{2} x + 1)(x^2 + \sqrt{2} x + 1)
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