餘式定理(急!!)

(a)
設 f(x) = x^2 + 98x - 100
f(1) = (1)^2 + 98(1) -100 = -1
利用餘式定理，(x^2 + 98x - 100)除(x - 1) 的餘數是 -1

(b)
設 f(x) = x^3 - x^2 + 2x + 1
f(-2) = (-2)^3 - (-2)^2 + 2(-2) + 1 = -15
利用餘式定理，(x^3 - x^2 + 2x + 1)除(x + 2)的餘數是 -15

(c)
設 f(x) = x^3
f(5) = (5)^3 = 125
利用餘式定理，(x^3)除(x - 5)的餘數是 125

(d)
設 f(x) = x^4 + x^2 + 1
f(-1/2) = (-1/2)^4 + (-1/2)^2 + 1 = 21/16
利用餘式定理，(x^4 + x^2 + 1)除(2x + 1)的餘數是 21/16

(e)
設 f(x) = x^99 + 1
f(-1) = (-1)^99+1 = 0
利用餘式定理，(x^99 + 1)除(x + 1)的餘數是 0

(f)
設 f(x) = 1 - x^100 + x^200
f(1) = 1 - (1)^100 + (1)^200 = 1
利用餘式定理，(1 - x^100 + x^200)除(x - 1)的餘數是 1

(a)(x^2+98x-100)除(x-1)
餘數 = (1)^2 + 98(1) -100 = -1


(b)(x^3-x^2+2x+1)除(x+2)
餘數 = (-2)^3 - (-2)^2 + 2(-2) + 1 = -8 - 4 - 4 + 1 = -15


(c)x^3除(x-5)
餘數 = (5)^3 = 125


(d)(x^4+x^2+1)除(2x+1)
餘數 = (-1/2)^4 + (-1/2)^2 + 1 = 1/16 + 1/4 + 1 = 21/16

(e)(x^99+1)除(x+1)
餘數 = (-1)^99+1 = -1 + 1 = 0

(f)(1-x^100+x^200)除(x-1)
餘數 = 1 - (1)^100 + (1)^200 = 1 - 1 + 1 = 1