Given,
\(f(x)=x^5-5 x^3-5 x-1=0\)
Now by inspection,
\(x=1 \text { is a zero of } f(x)\)
So, we can divide \(f(x)\) by \(x-1\) to get,
\(f(x)=(x-1)\left(x^4+x^3-4 x^2+x+1\right)\)
We similarly proceed to get
\(\begin{aligned}
& f(x)=(x-1)(x-1)\left(x^3+2 x^2-2 x-1\right) \\
& f(x)=(x-1)(x-1)(x-1)\left(x^2+3 x+1\right)
\end{aligned}\)
So, \(f(x)\) has 3 equal roots.