Q: Show evaluation ev₂: ℝ[x] → ℝ, ev₂(p) = p(2), is a homomorphism. Kernel? Image? ev₂(p+q) = (p+q)(2) = p(2)+q(2) ✓. ev₂(pq) = p(2)q(2) ✓: HOMOMORPHISM! Kernel: p(2) = 0: (x-2) (polynomials vanishing at 2!)! Image: ℝ (surjective: constants!). ℝ[x]/(x-2) ≅ ℝ!