如题所述
r=0或2或±√3+1
解析:
r^4-r³-r²-2r=0
r(r³-r²-r-2)=0
r=0或r³-r²-r-2=0
后者,
r³-r²-r-2
=r³-8+8-2r²+r²-r-2
=(r-2)(r²+r+1)+2(2-r)(2+r)+(r-2)(r+1)
=(r-2)(r²+r+1-4-4r+r+1)
=(r-2)(r²-2r-2)
=(r-2)(r²-2r+1-3)
=(r-2)[(r-1)²-3)]
=0
解得,r=2或±√3+1