Solve the following quadratic equations by factorization:

$\frac{1}{x-3}+\frac{2}{x-2}=\frac{8}{x}$

$\Rightarrow \frac{(x-2)+2(x-3)}{(x-3)(x-2)}=\frac{8}{x}$

$\Rightarrow \frac{x-2+2 x-6}{x^{2}-2 x-3 x+6}=\frac{8}{x}$

$\Rightarrow \frac{3 x-8}{x^{2}-5 x+6}=\frac{8}{x}$

$\Rightarrow x(3 x-8)=8\left(x^{2}-5 x+6\right)$

$\Rightarrow 3 x^{2}-8 x=8 x^{2}-40 x+48$

$\Rightarrow 5 x^{2}-32 x+48=0$

$\Rightarrow 5 x^{2}-20 x-12 x+48=0$

$\Rightarrow 5 x(x-4)-12(x-4)=0$

$\Rightarrow(5 x-12)(x-4)=0$

$\Rightarrow 5 x-12=0$ or $x-4=0$

$\Rightarrow x=\frac{12}{5}$ or $x=4$

Hence, the factors are 4 and $\frac{12}{5}$.