If one of the zeroes of the quadratic polynomial is $\left(\mathrm{k}-1\right){\mathrm{x}}^2+\mathrm{kx}+1$ is -3, then k equal to

$$\begin{gathered} \mathrm{x}=-3 \\ \begin{array}{l}\mathrm{p}\left(\mathrm{x}\right)=(\mathrm{K}-1){\mathrm{x}}^2+\mathrm{Kx}+1\\ \mathrm{p}(-3)=0\\ (\mathrm{K}-1)(-3{)}^2+\mathrm{K}(-3)+1=0\\ (\mathrm{K}-1)9-3\mathrm{K}+1=0\\ 9\mathrm{K}-9-3\mathrm{K}+1=0\\ 6\mathrm{K}-8=0\\ 6\mathrm{K}=8\\ \mathrm{K}=\frac{8}{6}=\frac{4}{3} \\ \mathrm{K}=\frac{4}{3}\end{array} \end{gathered}$$