$\displaystyle y'' - 3y' + 2y = 5e^{2x}$

Homogenous solution is:

$\displaystyle Ae^{2x} + Be^x$

Particular solution guess:

$\displaystyle y = axe^{2x}$

$\displaystyle y' = \alpha e^{2x}(2x + 1)$

$\displaystyle y'' = 4\alpha e^{2x}(x + 1)$

Subbing into ODE:

$\displaystyle 4\alpha e^{2x}(x + 1) - 3\alpha e^{2x}(2x + 1) + 2\alpha xe^{2x} = 5e^{2x}$

My text says this simplifies to:

$\displaystyle 4\alpha e^{2x} - 3\alpha e^{2x} = 5e^{2x}$

How is this so? I understand that you can make this true by making x = 0, but wouldn't this make the RHS = 5?

I'm sure I'm confusing myself, any help would be great.