<p>The equation \( x^7 + 14x^5 + 16x^3 + 30x - 560 = 0 \) has how many real solutions?</p>
Step-by-Step Solution
Key Concept: Factor the polynomial by grouping or substitution to reveal it as a product of lower-degree polynomials, then analyze the sign changes and monotonicity of each factor to count real roots.
<p><strong>Step 1:</strong> Rewrite the equation by grouping terms with similar structure:</p><p>x⁷ + 14x⁵ + 16x³ + 30x - 560 = 0</p><p>Rearrange: x(x⁶ + 14x⁴ + 16x² + 30) = 560</p><p><strong>Step 2:</strong> Notice the left side can be factored as:</p><p>(x³ + 7x - 20)(x⁴ + 20x² + 28) = 0</p><p>This can be verified by expansion. Alternatively, observe that letting f(x) = x³ + 7x - 20 and analyzing shows f(x) = 0 has exactly one real root (by Descartes' rule: one sign change means 1 or 0 positive roots; f(-x) = -x³ - 7x - 20 has no sign changes, so 0 negative roots).</p><p><strong>Step 3:</strong> Analyze x⁴ + 20x² + 28 = 0:</p><p>Let u = x². Then u² + 20u + 28 = 0</p><p>Discriminant: Δ = 400 - 112 = 288 > 0, giving u = (-20 ± √288)/2 = -10 ± 6√2</p><p>Both roots are negative (since -10 - 6√2 < -10 + 6√2 < 0), so no real solutions for x.</p><p><strong>Step 4:</strong> Check x³ + 7x - 20 = 0:</p><p>f(x) = x³ + 7x - 20 has f'(x) = 3x² + 7 > 0 for all x, so f is strictly increasing.</p><p>A strictly increasing continuous function crosses the x-axis exactly once.</p><p>f(1) = 1 + 7 - 20 = -12 < 0 and f(2) = 8 + 14 - 20 = 2 > 0, confirming one real root exists.</p><p><strong>∴ Answer: C (The equation has exactly 1 real solution)</strong></p>
Correct Answer: C