I once again have a disagreement with my textbook

If anyone can find a mistake I would be grateful.

$\displaystyle y'+\frac{1}{x-2}y = 3x$; $\displaystyle y(3)=4$

integration factor $\displaystyle I(x)=e^{\int{\frac{1}{x-2}dx}}=k(x-2)$ where k is non zero.

$\displaystyle k(x-2)y' +\frac{k(x-2)}{x-2}y = 3kx^2-6kx$

$\displaystyle k\frac{d}{dx}y(x-2) = 3kx^2-6kx$

$\displaystyle \int{(k\frac{d}{dx}y(x-2))dx} = \int{(3kx^2-6kx)dx}$

$\displaystyle ky(x-2) = kx^3-3kx^2+C$

$\displaystyle

y=\frac{x^3}{x-2}-\frac{3x^2}{x-2}+\frac{A}{x-2}$ where $\displaystyle A=\frac{C}{k}$

This is about where I stopped, I tried expanding the numerators and factoring to reduce my highest power of x by one which worked, but I still have x's in the denominators of my solution which the book does not have, so I guess I have made a mistake already.

The final solution in the book is $\displaystyle y=x^2-x-2$

Thanks for any help.