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## Homework Statement

Prove the following statement:

Let f be a polynomial, which can be written in the form

fix) = a(n)X^(n) + a(n-1)X^(n-1) + • • • + a0

and also in the form

fix) = b(n)X^(n) + b(n-1)X^(n-1) + • • • + b0

Prove that a(i)=b(i) for all i=0,1,2,...,n-1,n

## Homework Equations

3. The Attempt at a Solution [/B]

0 = f(x) - f(x) = (a(n)-b(n))X^(n) + (a(n)-b(n))X^(n) + ... + a0 - b0

Need to prove that Di = (a(i)-b(i)) = 0 for i = 0,1,2,3,4,...n

But I do not know how.

This is the answer from book, using the method of contradiction:

Suppose that there exists some index i such that Di does not equal 0.

Let m be the largest of these indices, so that we can write

0 = D(m)X^(m) + ... + D0 for all x and D(m) does not equal 0. This contradicts Theorem 2. Therefore we conclude that Di = 0 for all i = 1, . . . , n, thus proving the theorem.

Theorem 2 is as stated:

Let f be a polynomial. Let a0,a1,...,a(n-1),a(n) be numbers such that a(n) does not equal 0, and such that we have: f(x) = a0 + a1X + ... + a(n)X^(N) for all x. Then f has at most n roots.

I don't understand the contradiction between theorem 2 and the statement regarding D(m). Could someone explain?