# Reviewing pre factoring

1. Nov 22, 2007

### moe darklight

[SOLVED] reviewing pre... factoring

1. The problem statement, all variables and given/known data

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:

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

3. 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.

2. Nov 23, 2007

### 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 =]

3. Nov 23, 2007

### 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 :rofl:

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 highschool :grumpy:

Last edited: Nov 23, 2007
4. Nov 23, 2007

### 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 .

Last edited: Nov 23, 2007
5. Nov 23, 2007

### Gib Z

Well your first mistake is wrongly reducing $$16x^{4}-x^{2}y^{2}+y^{4}$$ to $$4x^{2}-xy+y^{2}$$. What Im 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 =]

6. Nov 23, 2007

### moe darklight

o boy there we go. thanks