已知x²+4x-1=0,求x⁴+8x³-4x²-8x+1的值
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已知x²+4x-1=0,求x⁴+8x³-4x²-8x+1的值
已知x²+4x-1=0,求x⁴+8x³-4x²-8x+1的值
已知x²+4x-1=0,求x⁴+8x³-4x²-8x+1的值
多项式除法:得到x⁴+8x³-4x²-8x+1=( x²+4x-1)( x²+4x-19)+(72x-18)=72x-18,方程算一下:x=-2±根号5,则原式=-162±根号5
x^3=1-4x
2x^2+8x^3-4x^2-8x+12x^4+8x^3-4x^2-8x+1=2x(1-4x)+8(1-4x)-4x^2-8x+1=2x-8x^2+8-32x-4x^2-8x+1=-4x^2-38x+10