Rules for Factorisation? 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?
 A: $x^2+2x+1$ 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.)
A: 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?
A: 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.
A: We know $\sqrt{x^2} = |x|$, so 
$$\sqrt{x^2 + 2x + 1} + \sqrt{x^2 - 2x + 1} = |x + 1| + |x-1|.$$
