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dimisalkan suku banyak berderajat 3 -> f(x) = ax³+ bx² + cx + 3
dibagi (x-1) sisa 4 --> f(1) = 4 --> a + b+ c + 3 = 4
a + b + c = 1 ....(i)
dibagi (x +1) sisa 2 --> f(-1) = 2 --> -a + b - c + 3 = 2
-a + b - c = - 1....(ii)
dibgai (x -2) sisa 17 --> f(2) = 17 --> 8a + 4b + 2c + 3= 17
8a + 4b + 2c = 14 atau 4a + 2b + c = 7 ...(iii)
eliminasikan (i - ii)
a+ b + c = 1
-a + b - c = -1
2a+ 2c = 2
a+ c = 1....(iv)
dari (2i - iii)
2a + 2b + 2c = 2
4a +2b + c = 7
-2a + c = - 5...(v)
dari (iv - v)
a+ c = 1
-2a + c = - 5
3a = 6
a= 2 --> c= -1
dari (i) --> a + b + c = 1
2 + b - 1 = 1 --> b = 0
a= 2, b = 0 , c = - 2
f(x) = a x³ + bx² + cx + 3
f(x) = 2 x³ - x + 3