Problem Analysis #10

[THOUGHT] To find the coefficient of $x^3$ in $h(x) = f(x+1) - g(x+2)$, expand using SymPy. The coefficient will be in terms of $a_3$ and $b_3$. The condition $f(x) \neq g(x)$ for all real $x$ implies $f(x) - g(x)$ has no real roots. Since $f(x) - g(x) = (a_1 - b_1) + 7x + (a_2 - b_2)x^2 + (a_3 - b_3)x^3$, this cubic must have leading coefficient zero (otherwise it has a real root), so $a_3 = b_3$. Thus, the $x^3$ coefficient in $h(x)$ is independent of $a_i, b_i$ and equals -4. Its square is 16. Verify the coefficient with SymPy. [/THOUGHT] [PYTHON] from sympy import symbols, expand

a1, a2, a3, b1, b2, b3, x = symbols('a1 a2 a3 b1 b2 b3 x')

f = a1 + 10x + a2x2 + a3*x3 + x4 g = b1 + 3x + b2x2 + b3*x3 + x4

f_shifted = f.subs(x, x+1) g_shifted = g.subs(x, x+2)

h = expand(f_shifted - g_shifted) coeff_x3 = h.coeff(x**3)

print(coeff_x3) [/PYTHON]

[THOUGHT] The coefficient of $x^3$ in $h(x)$ is $a_3 - b_3 - 4$. Since $f(x) \neq g(x)$ for all real $x$, $f(x) - g(x)$ is a cubic with no real roots, requiring $a_3 - b_3 = 0$. Thus, the coefficient is $-4$, and its square is 16. [/THOUGHT] $\boxed{16}$

Explanation

The reasoning begins by expanding $h(x) = f(x+1) - g(x+2)$ using SymPy, which yields the coefficient of $x^3$ as $a_3 - b_3 - 4$. The condition $f(x) \neq g(x)$ for all real $x$ means $f(x) - g(x) = (a_1 - b_1) + 7x + (a_2 - b_2)x^2 + (a_3 - b_3)x^3$ has no real roots. A cubic polynomial always has at least one real root unless its leading coefficient is zero, so $a_3 - b_3 = 0$. Substituting this into the coefficient gives $-4$. The square of $-4$ is 16, matching the Python output and problem requirements.