The number of partial fraction of $\frac{3{\mathrm{x}}^2+70\mathrm{x}+93}{{\left(\mathrm{x}-1\right)}^5}$ is

$$\begin{gathered} \frac{3{\mathrm{x}}^2+70\mathrm{x}+93}{{\left(\mathrm{x}-1\right)}^5} \\ x-1=y\Rightarrow x=1+y \\ \frac{3{\mathrm{x}}^2+70\mathrm{x}+93}{{\left(\mathrm{x}-1\right)}^5}=\frac{3{(1+\mathrm{y})}^2+70(1+y)+93}{y^5} \\ =\frac{3y^2+76y+166}{y^5} \\ =\frac{3}{y^3}+\frac{76}{y^4}+\frac{166}{y^5} \\ No. of partial fractions are 3 \end{gathered}$$