Question

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

Answer 1

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)

Answer 2

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.

Answer 3

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

Answer 4

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

Answer 5

yes got it :)

Answer 6

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

Answer 7

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

Answer 8

What is common there?

Answer 9

the x?

Answer 10

50% correct :)

Note that \(x^4 = (x^2)(x^2)\)

Answer 11

ah ok i see now

Answer 12

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

Answer 13

Patience, @callisto will lead you there...

Answer 14

What is the common factor then?

Answer 15

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

Answer 16

the common factor is x^4 right?

Answer 17

No o_o

\[x^4 - x^2y^2\]\[=(x^2)(x^2) - x^2y^2\]Try again?

Answer 18

y?

Answer 19

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

Answer 20

2?

Answer 21

No :(

Answer 22

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

Answer 23

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?

Answer 24

honestly no. is it x?

Answer 25

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

Answer 26

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

Answer 27

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?

Answer 28

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

Answer 29

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

Answer 30

2 numbers/letters on each side?

Answer 31

What letter is common?

Answer 32

x

Answer 33

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?

Answer 34

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

Answer 35

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?

Answer 36

x

Answer 37

One x only?

Answer 38

no 2 x so x^2?

Answer 39

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?

Answer 40

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

Answer 41

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

Answer 42

x^4-x^2y^2?

Answer 43

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

Answer 44

x

Answer 45

One x only?

Answer 46

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

Answer 47

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.

Answer 48

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.

Answer 49

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

Answer 50

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

Answer 51

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.

Answer 52

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

Answer 53

What is the answer?

Answer 54

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

Answer 55

Now, expand it.

Answer 56

but thats the answer no?

Answer 57

Just expand it, you'll know :)

Answer 58

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

Answer 59

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?

Answer 60

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

Answer 61

Be honest and fair to everyone...

Answer 62

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

Answer 63

Same :)

Answer 64

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

Answer 65

Welcome, good luck!

Answer 66

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

Answer 67

But don't cheat in the exam...

Answer 68

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 \[(x-2)(x+2)(x+y)(x-y)\]

Answer 69

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.

Answer 70

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.

Answer 71

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 :-)