Results 1 to 5 of 5

Thread: factor (x^4)+(2y^4)???????????

  1. #1
    Member sluggerbroth's Avatar
    Joined
    Apr 2012
    From
    Chicago
    Posts
    108

    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???
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,216
    Thanks
    3701

    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???
    $\displaystyle x^4+4x^2y^2+4y^4-4x^2y^2$

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

    note these terms form a perfect square ...


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

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

    $\displaystyle [(x^2+2y^2)-2xy] \cdot [(x^2+2y^2)+2xy]$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,714
    Thanks
    3003

    Re: factor (x^4)+(2y^4)???????????

    What they have done is add and subtract $\displaystyle 4x^2y$ which, of course, doesn't change the actual value. But that does change the form so that it we can think of it as $\displaystyle x^4+4x^2y^2+4y^4-4x^2y^2= (x^2+ 2y^2)^2- 4x^2y^2$ which is a "difference" of two squares. You probably know that $\displaystyle a^2- b^2= (a- b)(a+ b)$ so that last is $\displaystyle ([x^2+ 2y^2]+ 2xy)([x^2+ 2y^2]- 2xy)= (x^2+ 2xy+ 2y^2)(x^2- 2xy+ 2y^2)$.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member sluggerbroth's Avatar
    Joined
    Apr 2012
    From
    Chicago
    Posts
    108

    Re: factor (x^4)+(2y^4)???????????

    they just add ( and subsequently subtract ) 4x^2y^2 because it forms a perfect square?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,216
    Thanks
    3701

    Re: factor (x^4)+(2y^4)???????????

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

    $\displaystyle x^4 + 2x^2 + 1 - 2x^2$

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

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

    $\displaystyle (x^2 - \sqrt{2} x + 1)(x^2 + \sqrt{2} x + 1)$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: Jul 19th 2011, 11:45 AM
  2. factor analysis , extract second factor loading
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: Jun 1st 2011, 05:17 AM
  3. factor analysis , extract second factor loading
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: May 30th 2011, 05:29 AM
  4. factor
    Posted in the Algebra Forum
    Replies: 15
    Last Post: Dec 8th 2009, 12:38 PM
  5. Factor
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Apr 26th 2008, 11:03 AM

Search Tags


/mathhelpforum @mathhelpforum