Let V = P5(Q) the vector space consisting of polynomials in the variable x of degree at most 5. Let w E L(V) be a linear map defined by o(f) = x²f" – 6xf' + 12f for any f E V. (a) Prove that p is a linear transformation. T:V → W is a linear transformation if the following conditions hold: 1) T(x + у) %3D T(х) + T(),х, у € V. 2) T(ax) = aT(x); x E V & a scalar. Pf: 1) p(f + g) = x²f + g)" – 6x(f + g)' + 12(f + g) = x²(f" + g") – 6x(f' + g') + 12(f + g) = (x²f" – 6xf' + 12f) + (x²g" – 6xg' + 12g) = p(f) + «(g) d? (f + g) : d²f) + d?(g)\ dx2 dx? dx2 2) p(af) = x²(af)" – 6x(af)' + 12(af) = x²(af") – 6x(af") + 12(af) = ax²f" – a6xf' + a12f = ap(f) ) φ(αf) - αφ f) %3D Hence, p is a linear transformation. (b) Determine a basis for the image and nullspace of p.

1) f) Solve for x

I need to check my answers for a practice question in my precalculus class:   Find all the zeros of y=x^3 +27 and factor into linear factors.

Let u = = X₁ = -4 4 V= 5 x2 = (Simplify your answers.) and w= operations) to find x₁ and x₂ that satisfy the equation 0 - 41 -8 It can be shown that -5u-4v-w=0. Use this fact (and no row -4 5 5 4 4 - 3 X2 0 - 41 -8