Factor the following polynomial completely. -x^2 y^2 + x^4 + 4y^2 - 4x^2

- anonymous

Factor the following polynomial completely. -x^2 y^2 + x^4 + 4y^2 - 4x^2

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- chestercat

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- anonymous

these are the options
(x- 2)(x- 2)(x+y)(x-y)
(x+ 2)(x+ 2)(x+y)(x-y)
(x+ 2)(x- 2)(x+y)(x-y)

- whpalmer4

You can always brute force it by multiplying out your answer choices. Or, take advantage of the fact that you've got (x+y)(x-y) = x^2-y^2 as a common factor, and divide your polynomial by that, then factor the remaining part.

- anonymous

It doesnt make sense to me at all... could you answer this question and walk me through it... it would help me understand it alot better

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## More answers

- Callisto

\[-x^2 y^2 + x^4 + 4y^2 - 4x^2\]Rearrange the terms:
\[=(x^4-x^2 y^2 ) -( 4x^2- 4y^2)\]Got this step?

- anonymous

yes got it :)

- Callisto

Now, factorize the term \(x^4 - x^2y^2\). First, take out the common factor. What is the common factor?

- anonymous

2? or x... i find the whole thing hard and confusing. my teacher didnt explain it very well at all.

- Callisto

What is common there?

- anonymous

the x?

- Callisto

50% correct :)
Note that \(x^4 = (x^2)(x^2)\)

- anonymous

ah ok i see now

- anonymous

you saw the options i have for this question right? I posted them right below the question

- whpalmer4

Patience, @callisto will lead you there...

- Callisto

What is the common factor then?

- anonymous

Im just checking as i dont understand how to get to the answer so :)

- anonymous

the common factor is x^4 right?

- Callisto

No o_o
\[x^4 - x^2y^2\]\[=(x^2)(x^2) - x^2y^2\]Try again?

- anonymous

y?

- Callisto

No :(
What is *common* in \((x^2)(x^2)\) and \(x^2y^2\) ?

- anonymous

2?

- Callisto

No :(

- anonymous

but all are to the power of 2. im really confused. haha thanks for your patience though

- whpalmer4

If I might step in:
x^4-x^2y^2 = x*x*x*x - x*x*y*y
can you see a common part in each?

- anonymous

honestly no. is it x?

- whpalmer4

you could divide out x*x from both of those terms, no?

- anonymous

yes you could. which means? i honestly am so confused on this! thanks guys though!

- Callisto

x*x*x*x - x*x*y*y
You take one x out from both terms, so it becomes
x (x*x*x - x*y*y)
***********************************************************
In (x*x*x - x*y*y), what is common?

- anonymous

i dont know. not even a clue honestly... is it that your multiplying?

- Callisto

Hmm :(
********************************************
Let say, we have x*x - x*y
What is common there?

- anonymous

2 numbers/letters on each side?

- Callisto

What letter is common?

- anonymous

x

- Callisto

Yup!
So, we take one 'x' out from the two terms, and then group the rest, i.e.
x*x - x*y
= x (x-y)
^Got this part?

- anonymous

i said x wayyy earlier... :/ ah well... and yes i have that bit :)

- Callisto

Hey! It's not the same as the previous one we were discussing :P
*********************************************************************
Now, we have x*x*x - x*x*y
What is common there?

- anonymous

x

- Callisto

One x only?

- anonymous

no 2 x so x^2?

- Callisto

Yes <3
So, we do this again, take x*x, that is x^2, out, and group the rest. Then we get this:
(x*x*x - x*x*y)
= x^2 ( x-y)
^Got this part?

- anonymous

yes got it! thanks so much for helping i do really appreciate it!

- Callisto

That is factorize by grouping.
How do you factorize this: x*x*x*x - x*x*y*y ?

- anonymous

x^4-x^2y^2?

- Callisto

What is common in x*x*x*x - x*x*y*y?

- anonymous

x

- Callisto

One x only?

- anonymous

no x^2? could you tell me the answer so i can work backwards? i actually find it easier.

- Callisto

If you want to work backwards, then as @whpalmer4 has said, you just need to expand the factors in each choice, you'll get the polynomial given in the question eventually.

- anonymous

but that doesnt make sense to me... if i knew which answer was right out of the ones given i could see what answer it was and then figure out how they got to it. i cant work backwards until i know the answer which i cant figure out.

- anonymous

like i cant work backwards because i cant figure out the answer.

- anonymous

i just need the answer i cant understand it but at least if i have the answer i can help myself out a bit

- Callisto

You can get the answer by expanding the options and compare what you've got with the polynomial given by the question, i.e. you can work out the answer by yourself - probably by the way you would understood the answer if you were given one.

- anonymous

i got it dont worry... i have the answer i found it out :)

- Callisto

What is the answer?

- anonymous

(x+ 2)(x- 2)(x+y)(x-y)

- Callisto

Now, expand it.

- anonymous

but thats the answer no?

- Callisto

Just expand it, you'll know :)

- anonymous

yes it is :) im doing my final math exam and really need to pass it so... thanks for your help!

- Callisto

You have to understand how to get it to pass the next exam..
Wait. You're not supposed to get help for your final exam, right?

- anonymous

well it doesnt say not to... and its not timed so

- Callisto

Be honest and fair to everyone...

- anonymous

im homeschooled... im not at an actual school...

- Callisto

Same :)

- anonymous

ah ok well thanks again for your help! i really appreciate it!

- Callisto

Welcome, good luck!

- anonymous

thanks if i need more help i shall post my question :)

- Callisto

But don't cheat in the exam...

- whpalmer4

Here's a slightly different way to tackle this problem.
The polynomial is \[-x^2y^2+x^4 + 4y^2-4x^2\]All the answer choices have \[(x+y)(x-y)\] in them, which when multiplied out gives\[x^2-xy+xy-y^2 = x^2-y^2\]
We can do long division on the polynomial, dividing it by \(x^2-y^2\) to simplify it. This works just like doing long division on numbers.
x^2 - 4
_______________________________
x^2 - y^2 | x^4 -x^2y^2 + 4y^2 - 4x^2
x^4 - x^2y^2
------------------------------
4y^2 -4x^2
4y^2 - 4x^2
--------------------------------
0
So now we have \[x^2-4\] as the remaining polynomial after dividing out \[(x-y)(x+y)\]
Looking at our choices, we can see that one of them has \((x-2)(x+2)\) as part of its product terms, which is good, because \((x-2)(x+2) = x^2-2x+2x-4 = x^2-4\) and that's exactly what we have!
Our answer is thus \[(x-2)(x+2)(x+y)(x-y)\]

- Callisto

A more "traditional approach"\[-x^2 y^2 + x^4 + 4y^2 - 4x^2\]\[=( x^4-x^2 y^2)- ( 4x^2 - 4y^2 )\]Take out the common factor for the first two terms and the last two terms:
\[=x^2(x^2-y^2)- 4( x^2 - y^2 )\]Take out the common factor for the two terms again\[=(x^2-y^2)(x^2- 4)\]\[=(x^2-y^2)(x^2- 2^2)\]Factorize the two factors using the identity \(a^2-b^2 = (a+b)(a-b)\)\[=(x-y)(x+y)(x^2- 2^2)\]
Try to factorize \(x^2-2^2\) using \(a^2-b^2 = (a+b)(a-b)\), and you'll get the answer.

- whpalmer4

If you wanted to work this completely backwards, brute forcing it by multiplying out all the answer choices to see which ones match, the job is made easier if you notice that you have at least one, sometimes two, products that form difference of squares. \[(x-a)(x+a) = x^2-ax+ax-a^2 = x^2-a^2\]with a little bit of experience you should recognize such situations instantly, either factoring or multiplying, and save yourself the effort.
With that in mind, I would be multiplying \((x+y)(x-y) = x^2-y^2\) and \((x+2)(x-2) = x^2-4\) first, then using them as building blocks to do the rest of the multiplication. For example, with the answer choice that turned out to be the correct one,
\[(x-y)(x+y)(x-2)(x+2) = (x^2-y^2)(x^2-4) = x^4-4x^2-y^2x^2+4y^2\]All steps shown, time to multiply was just the time it took to type 30 or 40 characters.

- whpalmer4

Compare that with doing it piece by piece:\[(x-y)(x+y)(x-2)(x+2) = (x^2-xy+xy-y^2)(x-2)(x+2) \]\[= (x^2-y^2)(x-2)(x+2) = (x^3-2x^2-xy^2+2y^2)(x+2) \]\[= x^4+2x^3-2x^3-4x^2-x^2y^2-2xy^2+2xy^2+4y^2\]\[ = 4x^4-4x^2-x^2y^2+4y^2 \]which is much more typing if nothing else :-)

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