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?

$x^2+2x+1$ 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.)
