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

**sluggerbroth** · May 12th 2012, 10:54 AM

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???

**skeeter** · May 12th 2012, 12:14 PM

Originally Posted by sluggerbroth

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 ... $a^2-b^2 = (a-b)(a+b)$

$[(x^2+2y^2)-2xy] \cdot [(x^2+2y^2)+2xy]$

**HallsofIvy** · May 12th 2012, 12:18 PM

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 $([x^2+ 2y^2]+ 2xy)([x^2+ 2y^2]- 2xy)= (x^2+ 2xy+ 2y^2)(x^2- 2xy+ 2y^2)$ .

**sluggerbroth** · May 12th 2012, 12:29 PM

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

**skeeter** · May 12th 2012, 01:00 PM

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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