Two difficult questions

用戶85377

2009-12-31 4:51 am

1a) Prove that A is k- multiple root (k > 1) of the polynomial equation f(x) = 0, then A must be a single root of the equation f ^(k - 1) (x) = 0
1b) Using a), show that if the equation x^5 - 10a^3x^2 + b^4 x + c^5 = 0, where a, b, c are real constants, has a 3- multiple root, then ab^4 - 9a^5 + c^5 = 0.

2. Theorem a): Let f(x) = a0x^n + a1x^(n - 1) + ... + an-1 x + an = 0 be an equation with integral coefficients. If the equation f(x) = 0 has a rational root p/q, where p and q are two relatively prime integers, then p is a factor of an and q is a factor of a0.
The converse of theorem a) may not be true. Try to find a counter-example to verify this.

回答 (1)

myisland8132

2009-12-31 5:51 am

✔ 最佳答案

(a) when n=1, if A is a root of f(x), f(x)=(x-A)Q(x), f'(x)=Q(x)+(x-A)Q'(x)
f(A)=Q(A) not equal to 0. So when n=1, the statement is true

Assume that when n=k, the statement is true. for n=k+1, Let f(x)=(x-A)^k+1 Q(x), f'(x) =(k+1)(x-A)^k Q(x)+ (x-A)^k+1 Q'(x) =(x-A)^k [(k+1)Q(x)+(x-A)Q'(x)] So A is a k-multiple root of f'(x). According to the hypothesis, A is a single root of [f'(x)]^(k-1)=f^(k) (x). So when n=k+1, the statement is true.

By M.I. for all integer values of n >=1, the statement is true.

(b) Let f(x)= x^5 - 10a^3x^2 + b^4 x + c^5 = 0
f'(x)=5x^4-20a^3 x +b^4, f''(x)=20x^3-20a^3 =>x=a is the required 3-multiple root. So a^5-10a^5+ab^4+c^5=0=>ab^4 - 9a^5 + c^5 = 0

2 The converse of theorem : if p is a factor of an and q is a factor of a0, then p/q is a rational root of f(x) = a0x^n + a1x^(n - 1) + ... + an-1 x + an = 0.

To show that it is wrong, consider g(x)=x^5-2,for g(x)=0, x=2^(1/5) which is an irrational number and so it cannot be written as the form p/q.

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