5. [6] Let $a, b$, and $c$ be real numbers such that
$$
\begin{aligned}
a+b+c & =100, \\
a b+b c+c a & =20, \text { and } \\
(a+b)(a+c) & =24 .
\end{aligned}
$$

Compute all possible values of $b c$.
Answer: $\quad 224,-176$
Solution: We first expand the left-hand-side of the third equation to get $(a+b)(a+c)=a^{2}+a c+$ $a b+b c=24$. From this, we subtract the second equation to obtain $a^{2}=4$, so $a= \pm 2$.
If $a=2$, plugging into the first equation gives us $b+c=98$ and plugging into the second equation gives us $2(b+c)+b c=20 \Rightarrow 2(98)+b c=20 \Rightarrow b c=-176$.
Then, if $a=-2$, plugging into the first equation gives us $b+c=102$, and plugging into the second equation gives us $-2(b+c)+b c=20 \Rightarrow-2(102)+b c=20 \Rightarrow b c=224$.
Therefore, the possible values of $b c$ are $224,-176$.