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I am trying to factor the polynomial $$(a-1)x^2 + a^2xy+(a+1)y^2.$$ The problem previous to it in the book uses the method of factoring a polynomial of the form $$ax^2 + bx +c$$ by inspection, and the problem following it uses a formula related to cubes (I thought it's best you know). That said, I began by multiplying the coefficients of $x^2$ and $y$, but that did not yield something good.So I started again by taking $ax$, $x$, and $y$ as common, and that yielded nothing good. I would show some of my other work, but that would seem way too messy without proper formatting.

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    $\begingroup$ Then format it properly. $\endgroup$ – dfeuer Jul 15 '13 at 8:32
  • $\begingroup$ @dfeuer,hard,as I haven't got the necessary tool to do so on android. $\endgroup$ – rahul Jul 15 '13 at 8:34
  • $\begingroup$ yes you do. Just view the desktop site instead of the mobile one to get your posts to preview properly--nothing is required but a browser supporting Javascript and either CSS or SVG or MathML. $\endgroup$ – dfeuer Jul 15 '13 at 8:35
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  1. Convince yourself that it's going to be $$(rx+sy)(tx+uy)$$ where $r$, $s$, $t$, and $u$ are going to have formulas involving $a$.

  2. Note that $$rt=a-1,\quad su=a+1$$

  3. How can you get two things that multiply to $a-1$? Don't look for anything really fancy; what are the simplest possibilities?

  4. Same question for $a+1$.

  5. Now you have some possibilities for $r$, $s$, $t$, and $u$; see which combination gives you the right coefficient for $xy$. One will soon note that r=a-1 and u=a+1 works fine.Putting the values gives us the result required.Hence,we get, (ax-x+y)(x+ay+y)

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  • $\begingroup$ Thanks for the hint.it helped. $\endgroup$ – rahul Jul 15 '13 at 9:19
  • $\begingroup$ Good. If you want to write up and post a complete answer, you can do that. $\endgroup$ – Gerry Myerson Jul 15 '13 at 9:22
  • $\begingroup$ What,edit yours or create one by my own? $\endgroup$ – rahul Jul 15 '13 at 9:24
  • $\begingroup$ I had in mind writing up your own, but suit yourself. $\endgroup$ – Gerry Myerson Jul 15 '13 at 9:25
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Look first at the coefficient of $xy$, $a^2$. The simplest way to get this from existing coefficients is $(a+1)(a-1)+1$. So one factor should contain $(a-1)x$ and the other should contain $(a+1)y$. Then the other terms need to have coefficient one to make the $xy$ coefficient correct in the product. That this all works out so nicely is deliberate, because generally such expressions will not factor nicely.

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  • $\begingroup$ Thanks for putting it nicely. $\endgroup$ – rahul Jul 16 '13 at 10:27

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