If x⁴-5x³+9x²-7x+2=0 has a multiple root of order 3, then the roots are

$\begin{array}{l}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^4-5{\mathrm{x}}^3+9{\mathrm{x}}^2-7\mathrm{x}+2=0\\ \Rightarrow {\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)=4{\mathrm{x}}^3-15{\mathrm{x}}^2+18\mathrm{x}-7\text{ and }{\mathrm{f}}^{\prime \prime }\left(\mathrm{x}\right)=12{\mathrm{x}}^2-30\mathrm{x}+18\end{array}$
Clearly  is root of  f'(x)=0, f'(x)=0 and f''(x)=0. Therefore x=1 is multiple root of order 3. The fourth root of f(x)=0 is 5-3=2.