Find a basis and dimension of V When finding a basis for a vector space defined by $V = \{p(x) \in P_3(\Bbb R) \mid p(3)=0\text{ and }p(2)=0\}$, I started by showing $p(x) = (x-2)(x-3)q(x)$ where $q(x)\in P_1(\Bbb{R})$.
So, $p(2) = 8a+4b+2c+d = 0 $
And, $p(3) = 27a+9b+3c+d = 0$
I am wondering do I set these equal and solve for each independent variable a,b,c,d and with that answer is my basis?
Any help would be appreciated!

Go for $$\{(x-2)(x-3), x(x-2)(x-3)\}$$ for your basis.
These two polynomials  will span your space and they are linearly independent. 
