Reducing a product-of-sums expression

Question

f = ($x_1$ + $x_3$ + $x_4$) * ($x_1$ + $\overline x_2$ + $x_3$) * ($x_1$ + $\overline x_2$ + $\overline x_3$ + $x_4$)

I've been working on this problem for a while but I cannot for the life of me figure out how to simplify the function without distributing everything. The following is what the answer shows.

1. f = ($x_1$ + $x_3$ + $x_4$) * ($x_1$ + $\overline x_2$ + $x_3$) * ($x_1$ + $\overline x_2$ + $\overline x_3$ + $x_4$)

2. f = ($x_1$ + $x_3$ + $x_4$) * ($x_1$ + $\overline x_2$ + $x_3$) * ($x_1$ + $\overline x_2$ + $x_3$ + $x_4$) * ($x_1$ + $\overline x_2$ + $\overline x_3$ + $x_4$)

3. ...

I just don't understand how they made the jump from step 1 to step 2. Everything after step 2 makes sense to me. If anyone could explain where the third term of line 2 came from I would greatly appreciate it.

Answer 1

The change from step 1 to 2 is the term $(x_1+\bar x_2+x_3+x_4)$, which is precisely an $\bar x_2$ more than the first term $(x_1+x_3+x_4)$. Then it's sufficient to show that $$a*(a+b)*c=a*c$$ which seems obvious to me.