By viewing the polynomials as a difference of two squares, factorise the following polynomial: $$x^4+x^2+1.$$ I searched but couldn't find a way to solve this
Edit: By using Hans Lundmark hint, I get:
$$(x^2+1)^2-x^2$$
Is it fully factorized?
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3$\begingroup$ Now use $a^2-b^2=(a+b)(a-b)$. $\endgroup$– MacavityFeb 15, 2015 at 12:45
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2 Answers
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Here's a hint: $(x^4+2x^2+1)- x^2$.
answered Feb 15, 2015 at 12:19
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2$\begingroup$ @Mr.Mohamed Factorized means written as a product, this isn't. You got a difference of squares, why not go all the way? $\endgroup$– orionFeb 15, 2015 at 12:40
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$\begingroup$ @ Mohamed: note that $x^4+x^2+1=x^4+x^2+x^2-x^2=x^2+2x^2+1-x^2$ so you have the result in @Hans answer. $\endgroup$ Feb 15, 2015 at 13:10
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We don't factorization:
$$\dfrac2{x^4+ax^2+1}=\dfrac{x^2+1}{x^4+ax^2+1}-\dfrac{x^2-1}{x^4+ax^2+1}=\dfrac{1+\dfrac1{x^2}}{x^2+a+\dfrac1{x^2}}+\dfrac{1-\dfrac1{x^2}}{x^2+a+\dfrac1{x^2}}$$
Now as $\displaystyle\int\left(1\pm\dfrac1{x^2}\right)dx=x\mp\dfrac1x,$
for the first integral, choose $x-\dfrac1x=u$ and $x+\dfrac1x=v$ for the second
Use $x^2+\dfrac1{x^2}=\left(x+\dfrac1x\right)^2-2=\left(x-\dfrac1x\right)^2+2$
answered Aug 28, 2019 at 11:12
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$\begingroup$ @farruhota, No, that will be closed as duplicate,right? $\endgroup$ Aug 28, 2019 at 12:11
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$\begingroup$ Well, how does it help the OP here? Your answer is not a polynomial factorization and integration is not asked by OP... I see you and the user aryan bansal on the link posted similar answers almost at the same time. So, nevermind! $\endgroup$ Aug 28, 2019 at 12:27
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$\begingroup$ @farruhota, Sorry didn't see that :) $\endgroup$ Aug 28, 2019 at 12:32