Factoring by grouping: $x^4 - y^4 -4x^2 + 4$

Please help me factor $x^4 - y^4 -4x^2 + 4$ by grouping terms. Thank you.

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Why do people needlessly down vote questions? – Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ Jun 28 '12 at 8:02
Try to be more polite and describe what you tried to do in order to solve the question. People here want to help you so you'll be able to do that yourself next time and not to get orders from you and be an answers machine... – Jozef Jun 28 '12 at 8:22
That's better :-) you tried any factoring before posting the question? – Jozef Jun 28 '12 at 8:26
@Jozef Yes, I tried a few ways but the $y^4$ term kept getting in the way. I never thought of putting it aside and working with the the $x^4 - 4x^2 + 4$ first. – Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ Jun 28 '12 at 8:30
So you should write down this in the question so people would see that you tried, and what was the problem.Do that the next time, In addition to "thank you" and "please' and you won't get downvotes. +1 from me. good luck! – Jozef Jun 28 '12 at 8:41

Group all the $x$ terms together and all the $y$ terms together: $$(x^4-4x^2+4)-y^4=(x^2-2)^2-(y^2)^2=(x^2-y^2-2)(x^2+y^2-2)$$ using $a^2-b^2=(a-b)(a+b)$.

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which I see is just a synopsis of your own answer :-) – robjohn Jun 28 '12 at 21:45
Thank you for the synopsis. – Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ Jun 28 '12 at 22:26

$$x^4-y^4-4x^2+4=(x^4-4x^2+4)-y^4=(x^2-2)^2-(y^2)^2=(x^2-y^2-2)(x^2+y^2-2)$$

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Thanks for the answer. – Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ Jun 28 '12 at 7:57

$x^4 - y^4 -4x^2 + 4$

$= (x^4 -4x^2 + 4) - y^4$

$= [(x^2)^2 -2(2x^2) + 2^2] - (y^2)^2$

The expression $(x^2)^2 -2(2x^2) + 2^2$ has the form: $a^2 -2ab + b^2 = (a - b)^2$

where $a=x^2$ and $b=2$

Thus, $(x^2)^2 -2(2x^2) + 2^2 = (x^2 - 2)^2$

Hence

$[(x^2)^2 -2(2x^2) + 2^2] - (y^2)^2$

$= (x^2 - 2)^2 -(y^2)^2$

which is a difference of two squares i.e. has the form: $a^2-b^2 = (a+b)(a-b)$

where $a= x^2-2$ and $b = y^2$

Thus, $(x^2 - 2)^2 -(y^2)^2 = (x^2 -2 + y^2)(x^2 -2 - y^2)$

$\therefore$ $x^4 - y^4 -4x^2 + 4 = (x^2 -2 + y^2)(x^2 -2 - y^2)$ Answer

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