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:

[tex]16x^{4}-x^{2}y^{2}+y^{4}[/tex]

The Attempt at a Solution

[tex]4x^{2}-xy+y^{2}[/tex]

[tex]4x^{2}+4xy+y^{2}-xy-4xy[/tex]

[tex](2x+y)^{2}-3xy[/tex]

?? I'm doing something wrong here.

 

[Gib Z]

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

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

 

[moe darklight]

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

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

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 [tex]4x^{2}+4xy+y^{2}-3xy[/tex] and get

[tex](4x^{2}+y^{2})^{2}-9xy^{2}[/tex]

[tex](4x^{2}+y^{2}-3xy)(4x^{2}+y^{2}+3xy)[/tex]

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

 

[Gib Z]

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

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

 

[moe darklight]

o boy there we go. thanks