  A finite word $w$ with $\vert w\vert=n$ contains at most $n+1$ distinct
palindromic factors. If the bound $n+1$ is attained, the word $w$ is called
\emph{rich}. Let $\Factor(w)$ be the set of factors of the word $w$. It is
known that there are pairs of rich words that cannot be factors of a common
rich word. However it is an open question how to decide for a given pair of
rich words $u,v$ if there is a rich word $w$ such that $\{u,v\}\subseteq
\Factor(w)$. We present a response to this open question:\\ If $w_1, w_2,w$ are
rich words, $m=\max{\{\vert w_1\vert,\vert w_2\vert\}}$, and
$\{w_1,w_2\}\subseteq \Factor(w)$ then there exists also a rich word $\bar w$
such that $\{w_1,w_2\}\subseteq \Factor(\bar w)$ and $\vert \bar w\vert\leq
m2^{k(m)+2}$, where $k(m)=(q+1)m^2(4q^{10}m)^{\log_2{m}}$ and $q$ is the size
of the alphabet. Hence it is enough to check all rich words of length equal or
lower to $m2^{k(m)+2}$ in order to decide if there is a rich word containing
factors $w_1,w_2$.
