Question 1

**Setting up.** f(x) − g(x) = (a₁ − b₁) + 7x + (a₂ − b₂)x² + (a₃ − b₃)x³. Since f(x) ≠ g(x) for all real x, this polynomial has no real root. A polynomial of odd degree with real coefficients always has a real root, so the cubic coefficient must vanish: a₃ = b₃. (The remaining quadratic can indeed have no real roots, e.g. if a₂ ≠ b₂ and 49 − 4(a₂−b₂)(a₁−b₁) < 0, so the condition is consistent.) **Coefficient of x³ in h(x) = f(x+1) − g(x+2).** - In f(x+1): the x⁴ term (x+1)⁴ contributes 4x³; the a₃(x+1)³ term contributes a₃x³. Total: 4 + a₃. - In g(x+2): (x+2)⁴ contributes 4·2 = 8 to x³; b₃(x+2)³ contributes b₃. Total: 8 + b₃. Coefficient of x³ in h(x) = (4 + a₃) − (8 + b₃) = −4 + (a₃ − b₃) = −4.

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