1. ## 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???

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

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

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

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

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

3. ## 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 $\displaystyle ([x^2+ 2y^2]+ 2xy)([x^2+ 2y^2]- 2xy)= (x^2+ 2xy+ 2y^2)(x^2- 2xy+ 2y^2)$.

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

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

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

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

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

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

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

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