We aim to find a basis and the dimension of the subspace $W$ of $P_3(\mathbb{R})$ defined by the conditions $p(2) = 0$ and $p(-1) = 0$. Here $P_3(\mathbb{R})$ is the space of all polynomials of degree at most 3 with real coefficients.

Step 1: General form of a polynomial in $P_3(\mathbb{R})$

A general polynomial $p(x) \in P_3(\mathbb{R})$ has the form: $p(x) = a_0 + a_1x + a_2x^2 + a_3x^3,$ where $a_0, a_1, a_2, a_3 \in \mathbb{R}$.

Step 2: Apply the given conditions

1. Condition $p(2) = 0$: Substituting $x = 2$ into $p(x)$, we get: $p(2) = a_0 + 2a_1 + 4a_2 + 8a_3 = 0.$

2. Condition $p(-1) = 0$: Substituting $x = -1$ into $p(x)$, we get: $p(-1) = a_0 - a_1 + a_2 - a_3 = 0.$

These conditions give two linear equations: $\text{(1)} \quad a_0 + 2a_1 + 4a_2 + 8a_3 = 0,$ $\text{(2)} \quad a_0 - a_1 + a_2 - a_3 = 0.$

Step 3: Solve for the subspace

The subspace is determined by the set of all $(a_0, a_1, a_2, a_3) \in \mathbb{R}^4$ that satisfy these equations. Represent this system as a matrix equation:

1 & 2 & 4 & 8 \\ 1 & -1 & 1 & -1 \end{bmatrix} \begin{bmatrix} a_0 \\ a_1 \\ a_2 \\ a_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}.$$ #### Row reduce the matrix: $$\begin{bmatrix} 1 & 2 & 4 & 8 \\ 1 & -1 & 1 & -1 \end{bmatrix} \to \begin{bmatrix} 1 & 2 & 4 & 8 \\ 0 & -3 & -3 & -9 \end{bmatrix} \to \begin{bmatrix} 1 & 2 & 4 & 8 \\ 0 & 1 & 1 & 3 \end{bmatrix}.$$ Rewriting the equations from the row-reduced form: 1. $$a_0 + 2a_1 + 4a_2 + 8a_3 = 0,$$ 2. $$a_1 + a_2 + 3a_3 = 0.$$ From the second equation: $$a_1 = -a_2 - 3a_3.$$ Substitute $$a_1$$ into the first equation: $$a_0 + 2(-a_2 - 3a_3) + 4a_2 + 8a_3 = 0,$$ $$a_0 - 2a_2 - 6a_3 + 4a_2 + 8a_3 = 0,$$ $$a_0 + 2a_2 + 2a_3 = 0,$$ $$a_0 = -2a_2 - 2a_3.$$ ### Step 4: Parametrize the solution Let $$a_2 = t$$ and $$a_3 = s$$, where $$t, s \in \mathbb{R}$$. Then: $$a_1 = -t - 3s, \quad a_0 = -2t - 2s.$$ The polynomial $$p(x)$$ becomes: $$p(x) = a_0 + a_1x + a_2x^2 + a_3x^3,$$ $$p(x) = (-2t - 2s) + (-t - 3s)x + tx^2 + sx^3.$$ Factorize $$t$$ and $$s$$: $$p(x) = t(-2 - x + x^2) + s(-2 - 3x + x^3).$$ The basis for $$W$$ is: $$\{-2 - x + x^2, \, -2 - 3x + x^3\}.$$ ### Step 5: Dimension of the subspace The dimension of $$W$$ is the number of basis elements: $$\dim(W) = 2.$$ ### Final Answer: - **Basis:** $$\{-2 - x + x^2, \, -2 - 3x + x^3\}$$, - **Dimension:** $$2$$. Would you like further clarifications or additional examples? --- ### Related Questions: 1. What is the dimension of $$P_3(\mathbb{R})$$ itself? 2. How do the conditions $$p(2) = 0$$ and $$p(-1) = 0$$ reduce the degrees of freedom in $$P_3(\mathbb{R})$$? 3. Can this subspace $$W$$ be visualized geometrically? How? 4. How would the basis change if $$p(1) = 0$$ were added as another condition? 5. What if the conditions were $$p'(2) = 0$$ and $$p(-1) = 0$$? --- ### Tip: When finding a basis, always focus on simplifying the system of equations to identify the free variables, which directly correspond to the dimension of the subspace.