All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
Polynomial Ponder II (Posted on 2014-08-22) Difficulty: 3 of 5
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).

No Solution Yet Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution 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
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information