Left-Factoring a grammar into LL(1)
Let's have a look at your grammar:

$\qquad \begin{align}
X &amp;\to aE \mid IXE \mid (X)E \\
E &amp;\to IE \mid BXE \mid \varepsilon \\
I &amp;\to \text{++} \mid \text{--} \\
B &amp;\to \text{+} \mid \text{-} \mid \varepsilon 
\end{align}$

Note that $X$ does not need left-factoring: all rules have disjoint FIRST sets¹.
If you want to make this obvious, you can drop $I$ and inline it:

$\qquad \begin{align}
X &amp;\to aE \mid \text{++}XE \mid \text{--}XE \mid (X)E \\
E &amp;\to \text{++}E \mid \text{--}E \mid BXE \mid \varepsilon \\
B &amp;\to \text{+} \mid \text{-} \mid \varepsilon 
\end{align}$

Similarly, we can inline $B$:

$\qquad \begin{align}
X &amp;\to aE \mid \text{++}XE \mid \text{--}XE \mid (X)E \\
E &amp;\to \text{++}E \mid \text{--}E \mid \text{+}XE \mid \text{-}XE \mid XE \mid \varepsilon 
\end{align}$

Now we see that we actually have to do left-factoring on $E$: we have obvious conflicts, and we get additional conflicts via $XE$. So, let's inline $X$ once at $XE$:

$\qquad \begin{align}
X &amp;\to aE \mid \text{++}XE \mid \text{--}XE \mid (X)E \\
E &amp;\to \text{++}E \mid \text{--}E \mid \text{+}XE \mid \text{-}XE \mid aEE \mid \text{++}XEE \mid \text{--}XEE \mid (X)EE \mid \varepsilon 
\end{align}$

And now we can left-factor as easily as in your example:

$\qquad \begin{align}
X &amp;\to aE \mid \text{++}XE \mid \text{--}XE \mid (X)E \\
E &amp;\to \text{+}P \mid \text{-}M \mid aEE \mid (X)EE \mid \varepsilon \\
P &amp;\to \text{+}E \mid XE \mid \text{+}XEE \\
M &amp;\to \text{-}E \mid XE \mid \text{-}XEE
\end{align}$

By now we can see that we are not getting anywhere: by factoring away $\text{+}$ or $\text{-}$ from the alternatives, we dig up another $X$ which again has both $\text{+}$ and $\text{-}$ in its FIRST set.

So let's have a look at your language. Via

$\qquad \displaystyle X \Rightarrow aE \Rightarrow^* aI^n E \Rightarrow aI^nBXE$

and 

$\qquad \displaystyle X \Rightarrow aE \Rightarrow^* aI^n E \Rightarrow aI^nIE$

you have arbitrarily long prefixes of the form $+^+$ which end differently, semantic-wise: an LL(1) parser can not decide whether any given (next) $\text{+}$ belongs
to a pair -- which would mean choosing alternative $IE$ -- or comes alone -- which would mean choosing $BXE$.

In consequence, it doesn't look like your language can be expressed by any LL(1) grammar, so trying to convert yours into one is futile.

It's even worse: as $BXE \Rightarrow BIXEE \Rightarrow^* BI^n X E^n E$, you can not decide to chose $BXE$ with any finite look-ahead. This is not a formal proof, but it strongly suggests that your language is not even LL. 

If you think about what you are doing -- mixing Polish notation with unary operators -- it is not very surprising that parsing should be hard. Basically, you have to count from the left and from the right to identify even a single $B$-$\text{+}$ in a long chain of $\text{+}$. If I think of multiple $B$-$\text{+}$ in a chain, I'm not even sure the language (with two semantically different but syntactically equal $\text{+}$) can be parsed deterministically (without backtracking) at all.




That would be the sets of terminals that can come first in derivations of a non-terminal/rule alternative.