How do I factor 16x^4 - x^2y^2 + y^4?

[moe darklight]

[SOLVED] reviewing pre... factoring

Homework Statement

I bought a Schaum's with Precalculus questions; figured I'd review my pre. I'm not as rusty as I thought I'd be... but I'm screwing up this question for some reason:

16x^{4}-x^{2}y^{2}+y^{4}

The Attempt at a Solution

4x^{2}-xy+y^{2}

4x^{2}+4xy+y^{2}-xy-4xy

(2x+y)^{2}-3xy

?? I'm doing something wrong here.

 

[Gib Z]

Complete the square: (4x^2 + y^2) = 16x^4 + 8x^2y^2 + y^4.

So now express what you have as a difference between the perfect square, and another number =]

 

[moe darklight]

how did you get (4x^2 + y^2) from (2x+y)^{2}? wouldn't it be (4x^2 + y^2)^{2}? ... I'm guessing it's a typo, or else I'm really lost

ok, 16x^4 + 8x^2y^2 + y^4 leaves me with (4x^{2}+y^{2})^{2}-3xy ... but wouldn't that 3xy have to be a 3xy^{2} for me to be able to do a difference of a square? ... right now it's an a^{2}-b

EDIT: post #4

ugh, things like this frustrate me. I'll be doing just fine, and then a simple question like this comes along that I get all wrong... I wish I'd taken math in high school

 

[moe darklight]

wait... I square all of 4x^{2}+4xy+y^{2}-3xy and get

(4x^{2}+y^{2})^{2}-9xy^{2}

(4x^{2}+y^{2}-3xy)(4x^{2}+y^{2}+3xy)

right? ... you know, maybe the doctor's right and I do need Ritalin after all .

 

[Gib Z]

Well your first mistake is wrongly reducing 16x^{4}-x^{2}y^{2}+y^{4} to 4x^{2}-xy+y^{2}. What I am sure you meant was that 16x^{4}-x^{2}y^{2}+y^{4} =4u^{2}-uv+v^{2} where u= 4x^2 and v=y^2.

Instead, from post 2, we can see what you have is (4x^{2}+y^{2})^{2}-9xy^{2}, which you did factor properly =]

 

[moe darklight]

o boy there we go. thanks