數學題3題

(4)設f(x)=(x+2)(x-3)+3,當f(x)除以x+k時,餘式是-k,求k的值。
f(x) = (x+2)(x-3) + 3
根據餘式定理，f(x)除以x+k時,餘式 = f(-k)
所以 f(-k) = -k
(-k+2)(-k-3) + 3 = -k
k2 + 3k - 2k - 6 + 3 = -k
k2 + 2k - 3 = 0
(k + 3)(k - 1) = 0
k + 3 = 0 或 k - 1 = 0
k = -3 或 k = 1
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(7)已知g(x)=px3-x2+qx+6可被x+2整除,當g(x)除以x-3時,餘式是30。
(a) 求p和q的解。
根據餘式定理，g(x)除以x+2時,餘式 = g(-2)
所以 g(-2) = 0
p(-2)3 - (-2)2 + q(-2) + 6 = 0
-8p - 4 - 2q + 6 = 0
4p + q - 1 = 0
q = 1 - 4p ... (1)
根據餘式定理，g(x)除以x-3時,餘式 = g(3)
所以 g(3) = 30
p(3)3 - (3)2 + q(3) + 6 = 30
27p - 9 + 3q + 6 = 30
27p + 3q = 33
9p + q = 11
9p + (1 - 4p) = 11 【由 (1)】
9p + 1 - 4p = 11
5p = 10
p = 2
代 p = 2 至 (1)
q = 1 - 4(2)
q = -7
所以 p = 2, q = -7
(b)解方程g(x)=0
由 (a)
g(x) = 2x3 - x2 - 7x + 6
g(x) = 2x2(x+2) - 5x2 - 7x + 6
g(x) = 2x2(x+2) - 5x(x+2) + 3x + 6
g(x) = 2x2(x+2) - 5x(x+2) + 3(x+2)
g(x) = (x+2)(2x2 - 5x + 3)
g(x) = (x + 2)(2x - 3)(x - 1)
所以如 g(x) = 0
(x + 2)(2x - 3)(x - 1) = 0
x+2 = 0 或 2x-3 = 0 或 x-1 = 0
x = -2 或 x = 3/2 或 x = 1
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(8)設f(x)=4x3+px2+7x-2及g(x)=2x3-5x2+qx+8,當f(x)和g(x)除以x-3時,餘式分別是19和 -19。
(a)求p和q的值。
根據餘式定理，f(x)除以x-3時,餘式 = f(3)
所以 f(3) = 19
4(3)3 + p(3)2 + 7(3) - 2 = 19
108 + 9p + 21 - 2 = 19
9p = -108
p = -12
根據餘式定理，g(x)除以x-3時,餘式 = g(3)
所以 g(3) = -19
2(3)3 - 5(3)2 + q(3) + 8 = -19
54 - 45 + 3q + 8 = -19
3q = -36
q = -12
所以 p = -12, q = -12
(b)解方程f(x)+g(x)=0
由 (a)
f(x) = 4x3 - 12x2 + 7x - 2
g(x) = 2x3 - 5x2 - 12x + 8
根據餘式定理，f(x)+g(x)除以x-3時,餘式 = f(3)+g(3)
由於 f(3)+g(3) = 19+(-19) = 0
所以 f(x)+g(x) 可被x-3整除。
f(x) + g(x) = 0
4x3 - 12x2 + 7x - 2 + 2x3 - 5x2 - 12x + 8 = 0
6x3 - 17x2 - 5x + 6 = 0
6x2(x-3) + x2 - 5x + 6 = 0
6x2(x-3) + x(x-3) - 2x + 6 = 0
6x2(x-3) + x(x-3) - 2(x-3) = 0
(x-3)(6x2 + x - 2) = 0
(x-3)(3x-1)(2x+1) = 0
x-3 = 0 或 3x-1 = 0 或 2x+1 = 0
x = 3 或 x = 1/3 或 x = -1/2

(4)設f(x)=(x+2)(x-3)+3,當f(x)除以x+k時,餘式是-k,求k的值。
(7)已知g(x)=px^3-x^2+qx+6可被x+2整除,當g(x)除以x-3時,餘式是30。
(a) 求p和q的解。
(b)解方程g(x)=0
(8)設f(x)=4x^3+px^2+7x-2及g(x)=2x^3-5x^2+qx+8,當f(x)和g(x)除以x-3時,餘式分別是19和 -19。
(a)求p和q的值。
(b)解方程f(x)+g(x)=0

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