Let $a_1,a_2,a_3,b_1,b_2,b_3\in \mathbb{R}$, $a_1,a_2,a_3>0$, $b_1<b_2<b_3$.
Prove that $\frac{a_1}{x-b_1}+\frac{a_2}{x-b_2}+\frac{a_3}{x-b_3} = 0$ has exactly two solutions in $\mathbb{R}$.
First, I showed that the solutions must be bounded in $[b_1,b_3]$.
Since there are 3 points that the function is undefined at, I couldn't use any theorem that requires the function to be continuous. I know that it's continuous between $(b_1,b_2)$ and $(b_2,b_3)$, so I was trying to use IVT, but it also didn't work.
Could you please give me a hint?