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Problem 69: $V= P_2$, and S is the subset of $P_2$ consisting of all polynomials of the form $p(x) = ax^2 + b$

Problem 69: $V= P_2$, and S is the subset of $P_2$ consisting of all polynomials of the form $p(x) = ax^2 + b$. Determine whether it is a subspace of the given vector space $\V$.

Solution:
$S = \{p \in P_2 : p(x) = ax^2 + b, a, b \in \R\}$.
Note that S$ \ne \varnothing$ ; since $p(x) = 0$ belongs to $S$.
Closure under Addition:
Let $p, q \in S$. Then for some $a_1, a_2, b_1, b_2 \in \R$,
$p(x) = a_1x^2 + b_1$ and $q(x) = a_2x^2 + b_2$.
Hence,
$(p + q)(x) = p(x) + q(x) = (a_1 + a_2)x^2 + (b_1 + b_2) = ax^2 + b$,
where $a = a_1 + a_2$ and $b = b_1 + b_2$, so that S is closed under addition.
Closure under Scalar Multiplication:
If $k \in \R$, then
$(kp)(x) = kp(x) = ka_1x^2 + kb_1 = cx^2 + d$,
where $c = ka_1$ and $d = kb_1$, so that S is also closed under scalar multiplication.
Therefore S is a subspace of $P_2$

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