How do I find the subspace of $P_{2}$ from the following subsets? One of the following is a subspace of $P_{2}$.
A. $W= \{ax^2+bx+c\mid a+b+c=1\}$
B. $W= \{ax^2+bx+c\mid a+b+c=0\}$
C. $W= \{ax^2+bx+c\mid a \geq 0\}$
D. $W= \{ax^2+bx+c \mid b,c \geq 0\}$
So it's either $B, C$, or $D$ because the zero vector condition won't work on $A$. What I don't understand is the difference between $C$ and $D$. How does having 'a' greater than or equal zero differ from having '$b$' and '$c$' greater than or equal zero? Can someone please explain how it works? This concept is a bit new to me and I couldn't find any examples of this particular case online.
How does having 'a' greater than or equal zero differ from having 'b' and 'c' greater than or equal zero? Can someone please explain how it works?

So, for example, consider the polynomials
$$
p_1(x) = 2x^2 + 3x + 4,\\
p_2(x) = -3x^2 + 4x + 5,\\
p_3(x) = 0x^2 - 5x + 6,\\
p_4(x) = -5x^2 + 6x - 7.
$$
The polynomials $p_1$ and $p_3$ are elements of the set described in C because the $x^2$ coefficient (the &quot;$a$&quot; coefficient) is non-negative. The polynomials $p_1$ and $p_2$ are in the set described in D because the other coefficients (the &quot;$b$&quot; and &quot;$c$&quot; coefficients) are non-negative. $p_4$ is in neither set because it is not true that $a \geq 0$, and it is also not true that we have both $b \geq 0$ and $c \geq 0$.
Hint to answer the title question:
A subspace must be closed under multiplication by a scalar such as $−1$.