Question:

<p><strong>Step 1:</strong> Let f be a quadratic polynomial. Then f(x) = 0 has at most 2 roots. Let's call them α and β.</p><p><strong>Step 2:</strong> For f(f(f(x))) = 0, we need f(f(x)) ∈ {α, β}. This gives us at most 2 equations to solve.</p><p><strong>Step 3:</strong> Each equation f(f(x)) = α and f(f(x)) = β is itself a quadratic-type composition. For each, we need f(x) = γ where γ is a root of f(f(x)) = α.</p><p><strong>Step 4:</strong> The equation f(f(x)) = α has at most 2 roots (say γ₁ and γ₂). Similarly, f(f(x)) = β has at most 2 roots (say δ₁ and δ₂).</p><p><strong>Step 5:</strong> So we have at most 4 values: {γ₁, γ₂, δ₁, δ₂}. For each, solve f(x) = γᵢ, which yields at most 2 solutions.</p><p><strong>Step 6:</strong> Maximum total: 4 × 2 = 8... but we need to reconsider. Actually, f(f(x)) = α or β gives at most 2+2=4 intermediate values, but solving f(x) equals each of these 4 values gives at most 2 roots each.</p><p><strong>Correction:</strong> f(f(f(x))) = 0 means f(f(x)) ∈ {at most 2 roots of f}. For each root, f(f(x)) = root₁ has at most 2 solutions in f(x) values. Each of those f(x) values, when solved via f(x) = c, gives at most 2 x-values. But more carefully: f(f(x)) = α has degree 4, so ≤4 solutions. Similarly for β. But actually f∘f∘f has degree 8, so at most 8 roots, but the constraint structure limits this to 6 roots maximum when f is quadratic.</p><p><strong>Refined Step:</strong> For a quadratic f: f(f(f(x))) = 0 has degree 2³ = 8, but due to the composition structure and f having exactly 2 roots, the maximum is actually <strong>6 roots</strong>.</p><p>∴ Answer: <strong>6</strong></p>