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Let(x) = a[n] x^n +a[n-1] x^(n-1) +...+a[1] x + a[0]be a polynomial with integer coefficients.Suppose a[n] is odd and that f(x)=0 has a rational root r/s ,where r and sare two integers.a) If s is even , show that r is also even.b) If r and s are relatively prime, show that s must be odd and hence show... 顯示更多 Let(x) = a[n] x^n +a[n-1] x^(n-1) +...+a[1] x + a[0] be a polynomial with integer coefficients. Suppose a[n] is odd and that f(x)=0 has a rational root r/s ,where r and s are two integers. a) If s is even , show that r is also even. b) If r and s are relatively prime, show that s must be odd and hence show that f(r) is even. c) Show that if a[0] and a[n] are both odd and f(x)=0 has a rational root, then f(1) is even.
最佳解答:
f(x) = an x? + an-1 x??1 + ... + a1 x + a0 a)f(r/s) = 0 an (r/s)? + an-1 (r/s)??1 + ... + a1 (r/s) + a0 = 0 an r? + an-1 sr??1 + ... + a1 s??1 r + s? a0 = 0 ...... (*) r? = - ( an-1 sr??1 + ... + a1 s??1 r + s? a0 ) / an- ( an-1 sr??1 + ... + a1 s??1 r + s? a0 ) is even since each term have an even factor 's' , by given that an is odd therefore r? = even / odd = even , and hence r is even. b)If s is even , by the result of part (a) , r is also even , then s and r have a common factor 2. (Contradiction) So s must be odd.By (*) , an r? + an-1 sr??1 + ... + a1 s??1 r + s? a0 = 0an r? + [an-1 (s-1)r??1 + an-1 r??1] + ... + [a1 (s??1 - 1) r + a1 r] + [a0 (s?-1) + a0] = 0 [an r? + an-1 r??1 + ... + a1 r + a0] + [an-1 (s-1)r??1 + ... + a1 (s??1 - 1) r + a0 (s?-1)] = 0f(r) = - [an-1 (s-1)r??1 + ... + a1 (s??1 - 1) r + a0 (s?-1)] - [an-1 (s-1)r??1 + ... + a1 (s??1 - 1) r + a0 (s?-1)] is even since each term have an even factor 's? - 1' since s is odd were 1 ≤ k ≤ n is an integer. Hence f(r) is even. c)Let r/s be a rational root of f(x) = 0 and r and s are relatively prime , by the result of b) , f(r) = an r? + an-1 r??1 + ... + a1 r + a0 is even , ==> an r? + an-1 r??1 + ... + a1 r = even - a0 = even - odd = odd==> r (an r??1 + an-1 r??2 + ... + a1) = odd==> r must be odd Then start from an r? + an-1 r??1 + ... + a1 r + a0 = even ==> [an (r?-1) + an-1 (r??1-1) + ... + a1 (r-1)] + [an + an-1 + ... + a1 + a0] = even==> an + an-1 + ... + a1 + a0 = even - [an (r?-1) + an-1 (r??1-1) + ... + a1 (r-1)] ==> f(1) = even - even = even since each term of [an (r?-1) + an-1 (r??1-1) + ... + a1 (r-1)] have an even factor 'r? - 1' since r is odd were 1 ≤ k ≤ n is an integer.
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