Is $P$ a subspace of $\mathbb{R_{≤2}[x]}$?

Question

I have a polynomial function defined as :$$P:= (p \in \mathbb{R_{≤2}[x] \ | \ p(0)=0})$$ and I want to prove that $P$ is a subset of $\mathbb{R_{≤2}[x]}$. I use the $3$ criteria that need to be fullfiled for it to be a subspace.

---

1. $P ≠ ∅ $ $$p_{zero}(x)= 0x^2+0x+0=0 \in P$$

---

2.$\vec v, \vec u \in P \ | \ \vec v+ \vec u \in P$ $$p(x):=p_2x^2+p_1x^1+p_0x^0$$ $$p(x):=q_2x^2+q_1x^1+q_0x^0$$ $$p(x)+q(x)= (p_2+q_2)x^2+(p_1+q_1)x^1+(p_0+q_0)$$ $$x=0, \ (p+q)(0)=0 \in P$$

---

3. $$α \in \mathbb{R} \ , \ α(p(x))=(αp_2)x^2+(αp_1)x^1+(αp_0)$$ $$p(0)=0 \ , \ α(p(0))=0+0+(αp_0)=αp_0∉P$$

---

Since the $3rd$ condition was not fullfiled therefore $P$ isn't a subspace of $\mathbb{R_{≤2}[x]}$. But in the answer sheet it says that the $3rd$ condition holds true, thus $P$ is a subset.

Can anyone explain why? I've been scratching my head over this for a while now.

Answer 1

The set $P$ is a subspace of $\mathbb{R}_{\leqslant 2}[x]$:

1. The null polynomial belongs to $P$; in particular, $P\neq\emptyset$.
2. If $p,q\in P$, then$$(p+q)(0)=p(0)+q(0)=0+0=0,$$and therefore $p+q\in P$.
3. If $p\in P$ and $\alpha\in\mathbb R$, then$$(\alpha p)(0)=\alpha\times p(0)=\alpha\times0=0,$$and therefore $\alpha p\in P$.

Answer 2

If $p \in P$, $p_0 = p(0)=0$, so your objection to the third condition is not true.