Finding bases for vector spaces with two constraints I'm having problems understanding this question. Compute the dimension and find bases for the vector space of polynomials $p(x)$ of degree 4 which have the property that $p(2) = 0$ and $p(3) = 0$. 
Right now, I have tried substituting x = 2 and x = 3 into the polynomial equation $a + bx + cx^2 + dx^3 + ex^4$, then  I obtained two equations. I put these two equations into a matrix and solved for $a$ and $b$ in terms of the other three variables, $c, d, e$. I then substituted these variables into the equation, thus obtaining an equation of $p(x)$ in terms of $c, d, e$. 
I ended up with the bases $(6-5x+x^2), (30-19x+x^3), (114-65x+x^4)$. However I'm not sure if this is correct - seems like they are not independent (could not row reduce to 1s on diagonal..)
Does anyone have an idea of how to solve this question correctly? Would really appreciate it!
 A: You have to use some arithmetic of polynomials: $p(2)=0$ means $p(x)$ is divisible by $x-2$. Similarly $p(x)$ is divisible by $x-3$. As $x-2$ and $x-3$ are coprime, $p(x) $ is divisiible by $(x-2)(x-3)$ by Gauß's lemma.
So we can write: $$p(x)=(x-2)(x-3)q(x),\quad \deg q=2,$$ so that the vector space of polynomials $p$ such that $p(2)=p(3)=0$ is isomorphic to the vector space of polynomials of degree at most $2$ through the mapping $p\longmapsto q(x)=\dfrac{p(x)}{(x-2)(x-3)}$.
This shows the vector space has dimension $3$. Since the reciprocal isomorphism is $q(x)\longmapsto (x-2)(x-3)q(x)$ and we know a basis for the polynomials of degree at most $2$, a basis for the vector space is, e.g.:
$$\{(x-2)(x-3), x(x-2)(x-3),x^2(x-2)(x-3)\}.$$
A: They are indeed linearly independent, since each has a term of a degree not shared by any other: only the first one has an $x^2$ term, only the second has an $x^3$ term, and only the third has an $x^4$ term.
Observe that if you arrange them into a matrix as
$$\begin{bmatrix}
1 & 0 & 0 & -65 & 114\\
0& 1 & 0 & -19 & 30 \\
0 & 0 & 1 & -5 & 6 \\
\end{bmatrix}$$
you will see that you already have row-reduced them. The fact that the right-hand columns are not zeros is just a consequence of the fact that you are in a 5-dimensional vector space but have only 3 vectors.
