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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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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0Thank you for the _synopsis_. – 2012-06-28