∵β是x²-3x-5=0的根
∴β²-3β-5=0
β^4+2β^3-85β+1
=(β²)²+2β³-85β+1
=9β²+30β+25+2β³-85β+1
=2β³+9β²-55β+26
=2β(3β+5)+9β²-55β+26
=6β²+10β+9β²-55β+26
=15β²-45β+26
=15(β²-3β)+26
=75+26=101
∵β是x²-3x-5=0的根
∴β²-3β-5=0
β^4+2β^3-85β+1
=(β²)²+2β³-85β+1
=9β²+30β+25+2β³-85β+1
=2β³+9β²-55β+26
=2β(3β+5)+9β²-55β+26
=6β²+10β+9β²-55β+26
=15β²-45β+26
=15(β²-3β)+26
=75+26=101