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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?

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Take a look at the graph of the function. – anon Jul 28 '11 at 10:23
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. – Arturo Magidin Jul 28 '11 at 16:50
$-2x$ is also a possibility... – Aryabhata Jul 28 '11 at 23:33

$x^2+2x+1$ has two square roots, $x+1$ and $-(x+1)$. Similarly, $x^2-2x+1$ has two square roots, $x-1$ and $-(x-1)=1-x$. If you combine the first choice for each, you get $(x+1)+(x-1)=2x$; if you combine the first choice for the first term with the second choice for the second term, you get $(x+1)+(1-x)=2$. The remaining two combinations yield two more results: $-(x+1)+(x-1)=-2$, and $-(x+1)+(1-x)=-2x$. However, none of this is correct, because by convention $\sqrt{y}$ always denotes the non-negative square root of $y$. Thus, $\sqrt{x^2+2x+1}$ is actually $|x+1|$, and $\sqrt{x^2-2x+1}=|x-1|$, so that the correct simplification is $$|x+1|+|x-1|.$$ If you want to get rid of the absolute values, you’ll have to break the real line into pieces and use a multi-part definition of the function. And if you do this, you’ll see how $2$, $2x$, etc. actually come into the picture. (Literally: a graph should prove quite informative.)

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You can write it as

$$\sqrt{x^2 + 2x + 1} + \sqrt{x^2 - 2x + 1}=\sqrt{(x+1)^2} + \sqrt{(x-1)^2}$$

But note that this simplifies to

$$|x+1| + |x-1|$$

as squaring and taking the root eliminates the sign. So neither of your answer is correct (for a real $x$). Maybe you can figure out the intervals where something interesting happens on your own?

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We view the problem geometrically, or equivalently in terms of motion.

Suppose that we are travelling on the $x$-axis, and are now at the point $(x,0)$. Then $\sqrt{(x-1)^2}$ is our distance from the point $(1,0)$. Similarly, $\sqrt{(x+1)^2}=\sqrt{(x-(-1))^2}$ is our distance from the point $(-1,0)$.

It follows that $$\sqrt{(x+1)^2}+\sqrt{(x-1)^2}$$ is the sum of our distances from $(-1,0)$ and $(1,0)$.

Suppose that $x$ is anywhere between $-1$ and $1$. Then it is clear that the sum of our distances from $(-1,0)$ and $(1,0)$ is $2$. Any small motion to the right decreases our distance from $1$, but increases our distance from $-1$ by the same amount.

Suppose that $x>1$. Then our distance from $(1,0)$ is $x-1$. Our distance from $(-1,0)$ is $2$ more than that, so it is $x+1$. The sum is $2x$. The sum of the distances is increasing at twice the rate that the distance from $(1,0)$ is increasing.

Suppose finally that $x<-1$. By symmetry, the sum of the distances is the same as the sum for $|x|$. By the preceding paragraph, this sum is $2|x|$, which could also be written as $-2x$.

Generalization: We could make a "word problem" which is answered by the above calculation. Adam lives at $(-1,0)$ and Beth lives at $(1,0)$. They want to meet at the point $(x,0)$. What is the sum of the distances they must travel?

The problem can be generalized. Suppose that we have $n$ people, who live respectively at $(a_1,0)$, $(a_2,0)$, $\dots$, $(a_n,0)$, where $a_1\le a_2\le\cdots\le a_n$. What is the sum of their distances from the point $(x,0)$? The analysis is not much more complicated than for $2$ people.

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We know $\sqrt{x^2} = |x|$, so $$\sqrt{x^2 + 2x + 1} + \sqrt{x^2 - 2x + 1} = |x + 1| + |x-1|.$$

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