Please help me factor $x^4 - y^4 -4x^2 + 4$ by grouping terms. Thank you.
Factoring by grouping: $x^4 - y^4 -4x^2 + 4$
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0Why do people needlessly down vote questions? – 2012-06-28
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1Try 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... – 2012-06-28
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0That's better :-) you tried any factoring before posting the question? – 2012-06-28
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0@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. – 2012-06-28
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0So 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! – 2012-06-28
3 Answers
$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
$$x^4-y^4-4x^2+4=(x^4-4x^2+4)-y^4=(x^2-2)^2-(y^2)^2=(x^2-y^2-2)(x^2+y^2-2)$$
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0Thanks for the **answer**. – 2012-06-28
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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0which I see is just a synopsis of your own answer :-) – 2012-06-28
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0Thank you for the _synopsis_. – 2012-06-28