大學入學試題兩條(餘式定理)

1 f(x) = (2x^2 - 3x - 2)Q(x) + 2x + 3

g(x) = (2x^2 - 3x - 2)T(x) + 4x - 1

f(x) - g(x)

= (2x^2 - 3x - 2)[Q(x) - T(x)] - 2x + 4

= (2x + 1)(x - 2)[Q(x) - T(x)] - (2x + 1) + 5

因此餘式為 5

2 f(x) = ax^4 + bx^3 + cx^2 + dx

f(1) = a + b + c + d = 4

f(-1) = a - b + c - d = 4

f(2) = 16a + 8b + 4c + 2d = 4

f(-2) = 16a - 8b + 4c - 2d = 4

因此a + c = 4, b + d = 0

4a + c = 1, 4b + d = 0

即a = -1, c = 3, b = d = 0

f(x) = -x^4 + 3x^2

Q2.
可採用因式定理。依題意，f(x)-4均被x-1,x+1,x-2,x+2整除。