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Basically presented with this, simplify

\begin{aligned} {\Bigl(\sqrt{x^2 + 2x + 1}\Big) + \Bigl(\sqrt{x^2 - 2x + 1}\Big)} \end{aligned}

Possible factorisations into both

\begin{aligned} {\Bigl({x + 1}\Big)^2}, {\Bigl({x - 1}\Big)^2} \end{aligned}

\begin{aligned} {\Bigl({1 + x}\Big)^2} , {\Bigl({1 - x}\Big)^2} \end{aligned}

Hence when simplified, answer has two possibilities. One independent of x, and the other not.

( Simplified Answers: 2x, 2 ) 

Why is one independent and the other not? If such is equal to 2, why then when, say x=2, the answer does not simplify to 2?

  • 3
    Take a look at [the graph](http://www.wolframalpha.com/input/?i=plot+y%3DSqrt%5Bx^2%2B2x%2B1%5D%2BSqrt%5Bx^2-2x%2B1%5D+from+x%3D-2+to+2) of the function.2011-07-28
  • 0
    I would like to note that in general there are no "rules for factorization"; there's no algorithm/recipe that you can follow that guarantees that you will definitely obtain a "nice" factorization (among other things, because what counts as 'nice' depends on what you want it for). Rather, what you have are *heuristics*: general approaches that may or may not work.2011-07-28
  • 0
    $-2x$ is also a possibility...2011-07-28

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