Polynomial chain.

Algebra Level 2

Given a n t h n^{th} degree polynomial f ( x ) = a n x n + a n − 1 x n − 1 + . . . + a 2 x 2 + a 1 x 1 + a 0 f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+...+a_{2}x^{2}+a_{1}x^{1}+a_{0} satisfying a n + a n − 1 + . . . + a 2 + a 1 + a 0 = 0 a_{n}+a_{n-1}+...+a_{2}+a_{1} + a_{0} = 0 ( a n a_{n} must not equal to 0).

Can 1 be a solution to this polynomial? Prove it.

Yes. This will create a paradox. No.

Tin Le by Tin Le , 15, Vietnam

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2 solutions

Jordan Cahn
Apr 9, 2019

Not only can 1 1 be a solution, 1 1 must be a solution! f ( 1 ) = a n + a n − 1 + ⋯ + a 2 + a 1 + a 0 = 0 f(1) = a_n+a_{n-1} + \cdots+a_2+a_1+a_0 = 0

0 Helpful 1 Interesting 0 Brilliant 0 Confused

f(1)=a(n)+a(n-1)+a(n-2)+...+a(0)=0 (given). This implies 1 is a root of f(x)

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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