Jawaban:
cara kerja teorema sisa: cari nilai x yg membuat H(x) × pembagi = 0
f(x) = A(x)(x-1) + 3 ==> (cari x yg bs membuat A(x) × (x-1) = 0, yaitu x = 1, karena A(1) × (1-1) = A(1) × (0) = 0
f(1) = 0+3
f(1) = 3
f(x) = B(x)(x+1) - 1 ==> (cari x yg bs membuat B(x) × (x+1) = 0, yaitu x = -1, karena B(-1) × (-1+1) = B(1) × (0) = 0
f(-1) = 0-1
f(-1) = -1
f(x) = C(x)(x²-1)(x-1) + (ax²+bx+c) ==> sisa berupa polinom tingkat 2, karena pembaginya tingakat 3
f(x) = C(x)(x+1)(x-1)(x-1) + (ax²+bx+c)
supaya C(x)(x+1)(x-1)(x-1) = 0 ==> x = 1 atau -1
f(1) = C(1)(1+1)(1-1)(1-1) + (a(1)²+b(1)+c)
3 = 0 + a+b+c
a+b+c = 3 … [1]
f(-1) = C(-1)(-1+1)(-1-1)(-1-1) + (a(-1)²+b(-1)+c)
-1 = 0 + a-b+c
a-b+c = -1 [2]
[1] & [2]
a+b+c = 3
a-b+c = -1
--------------- (-)
2b = 4
b = 2
a+c = 1 ==> a & c terdapat byk kemungkinan, bisa 0 & 1, -1 & 2 dsb
sisa = ax²+2x+c dimana a+c=1
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