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Let P(x) = x¹⁰ + a*x⁸ + b*x⁶ + c*x⁴ + d*x² - 1 be a polynomial with real coefficients. Given P(1) = 1 and P(2) = -1, find the minimum number of real zeroes of P(x).

1. 2
2. 4
3. 6
4. 8

Correct answer: 4

Solution

P(x) has only even-degree terms, so it is an even function: P(x) = P(-x). Real roots come in symmetric pairs. Let Q(t) = t⁵ + a*t⁴ + b*t³ + c*t² + d*t - 1 where t = x² (so t >= 0 for real x). P(0) = -1 < 0. P(1) = 1+a+b+c+d-1 = 1 > 0 so a+b+c+d = 1. P(2) = 1024 + 256a+64b+16c+4d - 1 = -1 so 256a+64b+16c+4d = -1024. In terms of Q: Q(0) = -1 < 0, Q(1) = 1 > 0, Q(4) = 4⁵ + 256a+64b+16c+4d - 1 = 1024 - 1024 - 1 = -1 < 0. Sign changes: Q goes from -1 at t=0 to +1 at t=1 (root in (0,1)) and from +1 at t=1 to -1 at t=4 (root in (1,4)). Each positive root t₀ of Q gives x = +-sqrt(t₀), contributing 2 real roots of P. So P has at least 4 real roots.

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