x^4+ax^3+bx^2+cx−1=0 \quad Wielomian 4-go stopnia podzielny przez (x-1)^3
W(x)=(x-1)^3\cdot (x+d)\\ =(x^3-3x^2+3x-1)(x+d)\\ =x^4+dx^3-3x^3-3dx^2+3x^2+3dx-x-d\\ =x^4+(d-3)x^3+(3x^2-3dx^2)+(3d-1)x-d\\ =x^4+(d-3)x^3+(3-3d)x^2+(3d-1)x-d\\ =x^4 \ + \ \qquad ax^3 \ + \ \quad bx^2 \ \quad \ + \ \qquad cx \quad - \ 1 \ \Rightarrow \ d=1
a=d-3=1-3=-2\\ b=3-3d=3-3\cdot 1=0\\ c=3d-1=3\cdot 1-1=2
Odpowiedź:
a = -2
b = 0
c = 2