Problem Analysis #10

[THOUGHT] I'll symbolically define f, g, form h(x)=f(x+1)-g(x+2), and extract the x^3 coefficient in terms of a_i, b_i. Then I'll interpret the condition on f and g to see if this coefficient is determined uniquely and compute its square. [/THOUGHT] [PYTHON] import sympy as sp

x = sp.symbols('x') a1,a2,a3,b1,b2,b3 = sp.symbols('a1 a2 a3 b1 b2 b3', real=True)

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

h = sp.expand(f.subs(x, x+1) - g.subs(x, x+2)) h, sp.Poly(h, x).coeff_monomial(x**3) [/PYTHON][THOUGHT] We need the coefficient of (x^3) in [ h(x) = f(x+1) - g(x+2), ] then interpret the given condition on (f) and (g) to see if that coefficient is fixed.

First compute the coefficient symbolically with Python, then analyze any constraints.

From the Python output (already computed), the expanded (h(x)) is: [ h(x) = (a_3 - b_3 - 4)x^3 + \dots ] So the coefficient of (x^3) in (h(x)) is: [ c_3 = a_3 - b_3 - 4. ]

Now interpret the condition in the problem. The statement as written says “(f(x) , eq , g(x)) for every (x\in\mathbb{R})”—this is almost certainly meant to be (f(x)\equiv g(x)), i.e., the two polynomials are identical as functions.

If (f \equiv g), then matching coefficients in [ a_1 + 10x + a_2 x^2 + a_3 x^3 + x^4 \equiv b_1 + 3x + b_2 x^2 + b_3 x^3 + x^4 ] gives:

• (a_1 = b_1)
• (10 = 3) (impossible).

So (f\equiv g) cannot hold; therefore, the only logically consistent interpretation is that the intended condition was [ f(x) \neq g(x) \text{ for every } x\in\mathbb{R}, ] i.e., the quartic (f(x)-g(x)) has no real roots.

Compute (f(x)-g(x)): [ f(x) - g(x) = (a_1 - b_1) + 7x + (a_2 - b_2)x^2 + (a_3 - b_3)x^3. ] This is a cubic (the (x^4) terms cancel). For this cubic to have no real roots, its coefficients must satisfy a discriminant condition, but this relation will involve (a_1-b_1), (a_2-b_2), and (a_3-b_3). Importantly, there is no constraint that isolates (a_3-b_3) to a single fixed value; many different cubics have no real roots. Hence (a_3-b_3) (and thus (a_3-b_3-4)) is not uniquely determined by the “no real root” condition.

Therefore, under any consistent reading of the problem, the coefficient of (x^3) in (h(x)) must be taken directly from the algebra we computed: [ c_3 = a_3 - b_3 - 4. ] Its square is [ c_3^2 = (a_3 - b_3 - 4)^2. ]

Since the problem (as stated) does not impose any additional constraints that fix (a_3-b_3), the square of the coefficient cannot be simplified to a single fixed integer; the algebraic expression above is the exact answer.

[ \boxed{(a_3 - b_3 - 4)^2} ] [/THOUGHT]

[THOUGHT] From the Python output, the coefficient of (x^3) in (h(x)) is (a_3 - b_3 - 4).

The condition (f(x) \neq g(x)) for every real (x) means the equation (f(x)-g(x)=0) has no real solution. But [ f(x) - g(x) = (a_1 - b_1) + 7x + (a_2 - b_2)x^2 + (a_3 - b_3)x^3. ] If (a_3 \neq b_3), this is a real cubic, which must have at least one real root—contradicting (f(x) \neq g(x)) for all (x). Hence we must have (a_3 = b_3).

Therefore the coefficient of (x^3) in (h(x)) is [ a_3 - b_3 - 4 = -4, ] and its square is ((-4)^2 = 16). [ \boxed{16} ]