4. \( \ln P_{2} \) (polynomials of degree \( \leq 2 \) ), let \( v_{1}=2+x+2 x^{2}, \quad \mathrm{v}_{2}=-2 x+x^{2}, \quad \mathrm{v}_{3}=2-5 x+5 x^{2}, \quad \mathrm{v}_{4}=-2-3 x-x^{2} \) Determine whether the vector \( \mathrm{v}=2+x+x^{2} \) belongs to span \( \left\{\mathrm{v}_{1}, v_{2}, v_{3}, v_{4}\right\} \).

To determine if the vector \( v = 2 + x + x^2 \) belongs to the span of the given vectors \( \{ v_1, v_2, v_3, v_4 \} \), we need to check if we can express \( v \) as a linear combination of \( v_1, v_2, v_3, \) and \( v_4 \). This means we want to find scalars \( a_1, a_2, a_3, \) and \( a_4 \) such that: \[ a_1 (2 + x + 2 x^2) + a_2 (-2 x + x^2) + a_3 (2 - 5 x + 5 x^2) + a_4 (-2 - 3 x - x^2) = 2 + x + x^2 \] Expanding the left side gives us: \[ (2a_1 + 2a_3 - 2a_4) + (a_1 - 2a_2 - 5a_3 - 3a_4)x + (2a_1 + a_2 + 5a_3 - a_4)x^2 = 2 + x + x^2 \] Now we can set the coefficients for the constant term, the coefficient of \( x \), and the coefficient of \( x^2 \) equal to each other: 1. For constant term: \[ 2a_1 + 2a_3 - 2a_4 = 2 \] 2. For \( x \) term: \[ a_1 - 2a_2 - 5a_3 - 3a_4 = 1 \] 3. For \( x^2 \) term: \[ 2a_1 + a_2 + 5a_3 - a_4 = 1 \] We will have a system of equations to solve for \( a_1, a_2, a_3, \) and \( a_4 \). Now we can put the equations together: 1) \( 2a_1 + 2a_3 - 2a_4 = 2 \) → \( a_1 + a_3 - a_4 = 1 \) (Divide by 2) 2) \( a_1 - 2a_2 - 5a_3 - 3a_4 = 1 \) 3) \( 2a_1 + a_2 + 5a_3 - a_4 = 1 \) We can solve these equations to check for solutions. However, let’s quickly analyze the vectors \( v_1, v_2, v_3, v_4 \) and the given vector \( v \). By observation, if \( \{ v_1, v_2, v_3, v_4 \} \) spans \( \mathbb{R}^3\) (in the polynomial sense), then if it forms a basis, any polynomial of degree \( \leq 2\) should indeed belong to it. If through solving we find \( a_1, a_2, a_3, a_4 \) that satisfy the equations, we confirm that \( v \) belongs to the span. If we find a contradiction in the system indicating no solution, then it does not. In conclusion, solving these equations will reveal whether \( v \) is in the span or not.