perplexus.info :: Just Math : Polynomial Ponder II

Polynomial Ponder II (Posted on 2014-08-22)

F(x) is a polynomial with real coefficients such that:

F(0) = 1, and:
F(2) + F(3) = 125, and:
F(x)*F(2x²) = F(2x³ + x)

Find the value of F(5).

See The Solution Submitted by K Sengupta

Rating: 4.5000 (2 votes)

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Solution

| Comment 2 of 3 |

Following Larry’s lead, and in the hope that F(x) is a quartic,

Let F(x) = 1 + a₁x + a₂x² + a₃x³ + a₄x⁴

F(x)*F(2x²) = F(2x³ + x) then gives the identity:

(1 + a₁x + a₂x² + a₃x³ + a₄x⁴)(1 + 2a₁x² + 4a₂x⁴ + 8a₃x⁶ + 16a₄x⁸)

    = 1 + a₁(2x³ + x) + a₂(2x³ + x)² + a₃(2x³ + x)³ + a₄(2x³ + x)⁴

Equating coefficients of selected powers of x..

x² :       2a₁ + a₂ = a₂                             =>        a₁ = 0

x⁵ :       4a₁a₂ + 2a₃a₁ = 6a₃                    =>        a₃ = 0

x¹² :      16a₄² = 16a₄                              =>        a₄ = 1 or 0*

x⁸ :       16a₄ + 8a₂a₃ + 4a₄a₂ = 24a₄        =>        a₂ = 2

* a₄ = 0 corresponds to solutions F(x) = 1+x² and F(x) = 1,

neither of which satisfies the criterion F(2) + F(3) = 125, so we

are left with F(x) = 1 + 2x² + x⁴ which satisfies the identity

(coefficients of all remaining powers checked) and also gives:

F(2) + F(3) = 25 + 100 = 125    and    F(5) = 676

Posted by Harry on 2014-08-24 21:51:31

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