find a basis for the subspace $S$ = {$p \in \mathbb{P}_3$ |$ p$(5) = 0} This is as far as I've gotten:
$p$($x$) = $a$ + $bx$ + $c$$x^2$ + $dx^3$ then $p$($5$) = $a$ + $5b$ + $25c$ + $125d$ = $0$
which gets the vector  $(1, 5, 25, 125)$ but then I'm not sure how to a basis from this vector. I tried setting some other $p$($n$)$= 0$ and solving a system of linear equations  but that didn't get me anywhere.
What do? 
 A: Assuming that $P_3$  is the set of degree $\le 3$ polynomials over $\mathbb{R}$, note that the space has a basis of four elements. But $S$ is a proper subspace, and so has a basis with at most three elements; hence it is sufficient to find three linearly independent polynomials of degree at most $3$.
Such a set could be $\{x - 5, x^2 - 5x, x^3 - 5x^2\}$.
A: A starting point would be to note that it should have dimension 3 because you can factor any $P$ into linear factors over $\mathbb{C}$ and so you can choose $P$ by choosing either two real roots, or else one complex root (i.e. two choices of a real number $x,y$ giving $x+iy$), and then the other root must be the complex conjugate. Finally, you can multiply the polynomial by a scalar, so it should have three dimensions. 
A: An straightforward algebraic approach is to note that "evaluation at 5" is a linear transformation from $\mathbb{P}_3$ to $\mathbb{R}$ (relative to suitable bases for $\mathbb{P}_3$ and $\mathbb{R}$, the coordinate matrix of this transformation is the vector you discovered).
Thus, your goal is to find a basis for the kernel (aka null space) of this transformation, which is one of the algebraic techniques you've been learning!
Although spaces of polynomials do offer a number of useful tricks to use as short-cuts, as the other answers show, but they are not actually necessary to do the problem. Even if you do use the shortcuts, parts of the above analysis are still useful, because it lets you know immediately the dimension of $S$, which is useful to know.
A: Look at the map $T \colon \{ p \in {\mathbb P}_ 3 \mid p(0) = 0 \} \to \{ p \in {\mathbb P}_3 \mid p(5) = 0 \}$ given by $p(X) \mapsto p(X-5)$. This is an isomorphism of vector spaces. Since $X, X^2, X^3$ form a basis of the former one, $X-5, (X-5)^2, (X-5)^3$ form a basis of the latter one.
