若x^2+4x-1=0 求2x^4+8x^3-4x^2-8x+1的值

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若x^2+4x-1=0 求2x^4+8x^3-4x^2-8x+1的值
若x^2+4x-1=0 求2x^4+8x^3-4x^2-8x+1的值

若x^2+4x-1=0 求2x^4+8x^3-4x^2-8x+1的值
2x^4+8x^3-4x^2-8x+1
=2x^4+8x^3-2x^2-2x^2-8x+1
=2x^2(x^2+4x-1)-2x^2-8x+1
=2x^2(x^2+4x-1)-2(x^2+4x-1)-1 [x^2+4x-1=0]
=-1

2x^4+8x^3-4x^2-8x+1
=2x^4+8x³-2x²-2x²-8x+2-3
=2x²(x²+4x-1)-2(x²+4x-1)-3
=-3

x^2+4x-1=0
2x^4+8x^3-2x^2=0
2x^4+8x^3-4x^2-8x+1
=2x^4+8x^3-2x^2-2x^2-8x+2-2+1
=0-2(x^2+4x-1)-1
=0-0-1
=-1

2x^4+8x^3-4x^2-8x+1=(x^2+4x-1)(2x^2-2)-1=0-1=-1

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