已知x2+5x-4=0,求x3+6x2+x+2的值已知x2+x-1=0,求x3+2x2+3的值
问题描述:
已知x2+5x-4=0,求x3+6x2+x+2的值
已知x2+x-1=0,求x3+2x2+3的值
答
x^3+6x^2+x+2
=x^3+5x^2-4x+x^2+5x-4+6
=x(x^2+5x-4)+(x^2+5x-4)+6
=0+0+6
=6
x^3+2x^2+3
=x^3+x^2-x+x^2+x-1+4
=x(x^2+x-1)+(x^2+x-1)+4
=0+0+4
=4
答
x2+5x-4=0
所以
x2+5x=4
x3+6x2+x+2
=x(x2+5x)+x2+x+2
=4x+x2+x+2
=x2+5x+2
=4+2
=6
x3+2x2+3
=x(x2+x)+x2+3
=x+x^2+3
=1+3
=4