Question

解一道一元二次方程题 急需

已知x3-7x+1＝0
求x8-6x6+3x5-7x4-12x3+2x2-7x+6的值。

Answer 1

x^8-6x^6+3x^5-7x^4-12x^3+2x^2-7x+6
=x^5(x^3 -7x+1)+x^6 +2x^5-7x^4-12x^3+2x^2-7x+6
=x^6+2x^5-7x^4-12x^3+2x^2-7x+6
=x^3(x^3-7x+1) +2x^5-13x^3+2x^2-7x+6
= 2x^5-13x^3+2x^2-7x+6
=2x^2(x^3-7x+1)+x^3-7x+6
=x^3-7x+6
=(x^3-7x+1)+5
=5
或：原式=x^5(x^3 -7x+1)+x^3(x^3-7x+1) +2x^2(x^3-7x+1)+(x^3-7x+1)+5=5
或：原式=（x^5+x^3+2X^2+1)(x^3-7x+1)+5=5

Answer 2

x^8-6x^6+3x^5-7x^4-12x^3+2x^2-7x+6
=x^8-(7x^6-x^6)-7x^4+(x^5+2x^5)-7x^4-12x^3+2x^2-7x+6
=x^8-7x^6+x^6-7x^4+x^5+2x^5-7x^4-12x^3+2x^2-7x+6
=(x^8-7x^6+x^5)+x^6+2x^5-7x^4-12x^3+2x^2-7x+6
=x^5(x^3 -7x+1)+x^6 +2x^5-7x^4-12x^3+2x^2-7x+6
下同如上步骤凑出x^3-7x+1 消去
=x^3(x^3-7x+1) +2x^5-13x^3+2x^2-7x+6
= 2x^5-13x^3+2x^2-7x+6
=2x^2(x^3-7x+1)+x^3-7x+6
=x^3-7x+6
=(x3-7x+1)+5
=5