<p>How many real solutions does the equation \(x^7+14x^5+16x^3+30x-560=0\) have?</p>
Step-by-Step Solution
Key Concept: Recognize that this polynomial can be factored by grouping or substitution to reveal it as a product of simpler factors. Rewrite the equation strategically to expose a quadratic-like structure in odd powers of x.
<p><strong>Step 1:</strong> Rearrange and attempt to factor. Notice the equation can be written as:<br/>(x² + 10)(x⁵ + 4x³ + 6x - 56) = 0</p><p><strong>Step 2:</strong> Analyze x² + 10 = 0 → x² = -10. This gives no real solutions (only complex roots ±i√10).</p><p><strong>Step 3:</strong> Analyze f(x) = x⁵ + 4x³ + 6x - 56 = 0. Find f'(x) = 5x⁴ + 12x² + 6 > 0 for all real x. This means f(x) is strictly increasing.</p><p><strong>Step 4:</strong> Check sign change: f(0) = -56 < 0 and f(2) = 32 + 32 + 12 - 56 = 20 > 0. By IVT, exactly one real root exists in (0,2).</p><p><strong>Step 5:</strong> Since f(x) is strictly monotonic, it has exactly one real solution.</p><p>∴ Answer: <strong>B (1 real solution)</strong></p>
Correct Answer: B