Subspace of P_2? Is the set of the polynomials in the form $cx^2+dx+e$ with $c+d+e=0$  a subspace of $P_2$? Why? 
Is there a zero component in this if $c=d=e=0$, then $0x^2+0x+0$ is not a part of $P_2$?
Or is 
$cx^2+dx+e$$-cx^2-dx-e$ = $0x^2+0x+0$? It's not an element of $P_2$ is it?
 A: $a)$: Yes
$b)$: Yes, $0 = 0x^2+0x+ 0$
$c)$: No.
A: As $P_2 = \{ax^2 + bx + c:~~a\in \mathbb{R}, b\in\mathbb{R}, c\in\mathbb{R}\}$, the set of all parabolas, lines, and constant functions, even if $a=b=c=0$, you still have $0x^2+0x+0 = 0\in P_2$
As for if the set $\{ax^2 + bx + c:~~a\in \mathbb{R}, b\in\mathbb{R}, c\in\mathbb{R}, a+b+c = 0\}$ is a subspace (that is, the set of all polynomials of degree at most 2 whose sum of coefficients equals zero), note that if you have two elements of this form, $ax^2 + bx + c$ and $a_2x^2 + b_2 x + c_2$
$(ax^2 + bx + c) + (a_2x^2 + b_2 x + c_2) = (a+a_2)x^2 + (b+b_2)x + (c+c_2)$
Note that $a+b+c=0$ and $a_2+b_2+c_2 = 0$.  It follows that $(a+a_2) + (b+b_2) + (c+c_2) = 0$ so it is closed under addition
As for if it is closed under scalar multiplication:
$\alpha(ax^2 + bx + c) = \alpha ax^2 + \alpha bx + \alpha c$
Note that $\alpha a + \alpha b + \alpha c = \alpha (a+b+c) = \alpha (0) = 0$
Showing that it is closed under scalar multiplication.  Therefore, yes it is indeed a subspace.
