多項式

[happy520]

(1) f(X)除以(2X+3)^2及(X-2)^2之餘式依次為X+6及78X-99,求
(a)以(2X+3)(X-2)^2除f(X)之餘式
(b)以(2X+3)^2 (X-2)^2除f(X)之餘式

(2) f(X)=ax^4+bx^3+cxx+dx+e,以(x-1)^2除f(x)餘式為9X-4,以(x+1)^3除f(x)餘式為8xx+9x+4,則(a,b,c,d,e)=

(3) f(x)=xxx+7xx+ax+6與g(x)=xxx+2xx+bx-9之H.C.F為二次式則a+b可能的值為?(A){0,1,2,3} (B){1,2,6,8} (C){2,4,6,8}
(D){1,5,7,9} (E){10,11,12,13,14}

(4)請看附檔

[thepiano]

第 1 題
(1)
令 f(x) = (2x + 3)(x - 2)^2 * q(x) + a(x - 2)^2 + 78x - 99
利用 f(-3/2) = 9/2
可求出 a = 18
所求之餘式 = 18x^2 + 6x - 27

(2)
利用 (1) 之方法可求出 f(x) 除以 (2x + 3)^2(x - 2) 之餘式為 4x^2 + 13x + 15
再令
f(x) = (2x + 3)^2(x - 2)^2 * q(x) + m(2x + 3)^2(x - 2) + (4x^2 + 13x + 15) ...... (i)
f(x) = (2x + 3)^2(x - 2)^2 * q(x) + n(2x + 3)(x - 2)^2 + (18x^2 + 6x - 27) ...... (ii)

由 (i) 和 (ii) 分別比較 x^3 項係數和常數項可知
4m = 2n
-18m + 15 = 12n - 27
......


第 2 題
令 f(x) = ax^4 + bx^3 + cx^2 + dx + e = (x - 1)^2 * q_1(x) + (9x - 4) = (x + 1)^3 * q_2(x) + (8x^2 + 9x + 4)
易知
f(1) = a + b + c + d + e = 5 ...... (1)
f(-1) = a - b + c - d + e = 3 ...... (2)

f'(x) = 4ax^3 + 3bx^2 + 2cx + d = [(x - 1)^2 * q_1(x)]' + 9 = [(x + 1)^3 * q_2(x)]' + (16x + 9)
f'(1) = 4a + 3b + 2c + d = 9 ...... (3)
f'(-1) = -4a + 3b - 2c + d = -7 ...... (4)

f''(x) = 12ax^2 + 6bx + 2c = [(x + 1)^3 * q_2(x)]'' + 16
f''(-1) = 12a - 6b + 2c = 16 ...... (5)

解 (1)，(2)，(3)，(4)，(5) 之聯立方程式可得 a，b，c，d，e


第 3 題
f(x) - g(x) = 5x^2 + (a - b)x + 15
3f(x) + 2g(x) = 5x^3 + 25x^2 + (3a + 2b)x = x[5x^2 + 25x + (3a + 2b)]
5x^2 + (a - b)x + 15 = 5x^2 + 25x + (3a + 2b)
.....


第 4 題
(1)
x = 10t * cos37 = 10t * (4/5) = 8t
t = x/8

y = 10t * sin37 - (1/2)gt^2 = 6t - 4.9t^2 = (3/4)x - (49/640)x^2

(2)
6t - 4.9t^2 = 0
解 t

(3)
求 y = (3/4)x - (49/640)x^2 的頂點坐標