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The problem asks:

Find the basis and the subspaces defined by the following sets of conditions:

$$\{p \in P_3(\mathbb{R}) \mid p(2) = p(-1) = 0\}$$

How would I go about solving this problem since it doesn't explicitly specify any of the objects in the set?

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    related http://math.stackexchange.com/questions/2138354/a-basis-for-px-in-mathbb-r-nxp1-p2-0 – 2017-02-17

2 Answers 2

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I would guess that $P_3(\mathbb{R})$ is the vector space of polynomials with real coefficients whose degree is $\leq 3$. Namely,

$$ P_3(\mathbb{R}) := \{ p(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 \, | \, a_0,a_1,a_2,a_3 \in \mathbb{R} \}. $$

This is a four dimensional vector space and you are asked to find a basis for the subspace

$$ S = \{ p \in P_3(\mathbb{R}) \, | \, p(2) = p(-1) = 0 \}. $$

This will be a two-dimensional subspace of $P_3(\mathbb{R})$. Write down the conditions on the coefficients of $p$ that will make $p$ an element of $S$ and find two linearly independent polynomials that satisfy those conditions.

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    For example, $p(2) = 0$ iff $p(2) = a_0 + 2a_1 + 4a_2 + 8a_3 = 0$ and $p(-1) = 0$ iff $a_0 - a_1 + a_2 - a_3 = 0$. – 2017-02-16
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A polynomial $p$ has $2$ as a root if and only if it is divisible by $x-2$. Similarly it has root $-1$ if and only if it is divisible by $x+1$. These divisors are coprime, hence if $p(2)=p(-1)=0$, it is divisible by $(x-2)(x+1)$.

If further, $p$ has degree $\le 3 $, the quotinet must have degree $\le 3-2=1$, hence $p(x)=(x-2)(x+1)(ax+b$ for some $a, b$. Thus a basis is $$\{(x-2)(x+1),\;x(x-2)(x+1)\}.$$