已知x2-2x-1=0，求2x5+4x4-3x3+3x2-x+4的值
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x²-2x-1=0
（x-1）²=2
x=1±√2
2x^5+4x^4-3x^3+3x²-x+4
=2x^3(x²-2x-1)+8x^4-x^3+3x²-x+4
=8x^4-x^3+3x²-x+4
=8x²(x²-2x-1)+15x^3+11x²-x+4
=15x^3+11x²-x+4
=15x(x²-2x-1)+41x²+14x+4
=41x²+14x+4
=41(x²-2x-1)+96x+45
=96x+45=96(1±√2)+45=141±96√2
（x-1）²=2
x=1±√2
2x^5+4x^4-3x^3+3x²-x+4
=2x^3(x²-2x-1)+8x^4-x^3+3x²-x+4
=8x^4-x^3+3x²-x+4
=8x²(x²-2x-1)+15x^3+11x²-x+4
=15x^3+11x²-x+4
=15x(x²-2x-1)+41x²+14x+4
=41x²+14x+4
=41(x²-2x-1)+96x+45
=96x+45=96(1±√2)+45=141±96√2
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