Question

已知x^2+4x-1=0,试求代数式2x^4+8x^3-4x^2-8x+1的值

Answer 1

x^2+4x=1

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

=-2x^2-8x+1

=-2（x^2+4x）+1

=-1