Proving that a set of polynomials P2 is a subspace of P3 The questions specifically says:
Show that the set $W$ of all polynomials in $P_2$ (polynomials of degree $2$ or less) such that $P(1) = 0$ is a subspace of $P_3$.
To prove this I am assuming it must follow two axioms:


*

*closed under scalar multiplication

*closed under addition
The first one is simple to prove, if $k$ = scalar then :

$k*P(1) = 0$ , always, since $P(1)=0$   (is this logic right?)

But for addition, im not sure how to approach it?

$P(1) + P(1) = 0$ ?

 A: You're more or less correct. Your "proof" just needs a little polishing and a slight adjustment.
First, suppose $k \in \mathbb{R}$ (a scalar) and let $f,g \in W$ ($f$ and $g$ are two members of your set $W$). 
You need to show that $k\cdot f \in W$ and $f+g \in W$.
First, closure under scalar multiplication: Yep, $kf(1)=0$ because $f(1)=0$. Also, since $f \in P_2$ certainly $kf \in P_2$ (scaling doesn't magically raise the degree of the polynomial). Thus $kf \in W$.
Next, closure under addition: This needs to be adjusted a bit. You need $2$ arbitrary members of $W$ where your notation could be interpreted as the same element added to itself. So instead... $(f+g)(1)=f(1)+g(1)=0+0=0$ since $f,g \in W$. Also, $f+g \in P_2$ (since adding polynomials doesn't raise their degree). Thus $f+g \in W$.
One last thing, technically you also need to make sure $W$ isn't empty in order to conclude it is a subspace, but I suppose that's "obvious" for this one. :)  If not, notice that ${\bf 0} \in P_2$ (the zero polynomial) and ${\bf 0}(1)=0$ so ${\bf 0} \in W$. 
