Question

当x=2时,多项式ax^5+bx^4+cx^3+dx^2+ex+f,bx^4+dx^2+f的值分别为4和3,则当x=-2

Answer 1

x=2时,多项式ax^5+bx^4+cx^3+dx^2+ex+f=4,bx^4+dx^2+f=3

ax^5+bx^4+cx^3+dx^2+ex+f=4

32a+16b+8c+4d+2e+f=4

x=2时

bx^4+dx^2+f=3

16b+4d+f=3

32a+16b+8c+4d+2e+f=4

32a+8c+2e+3=4

32a+8c+2e=1

当x=-2时

ax^5+bx^4+cx^3+dx^2+ex+f

=-32a+16b-8c+4d-2e+f

=-(32a+8c+2e)+(16b+4d+f)

=-1+3

=2