多項式-多次函數的問題

Question

若多項式p(x)滿足p(x 1) p(x-1)=2x^2-4x，求p(x)=?

已更新項目:

p(x +1) p(x-1)=2x^2-4x，求p(x)=?

2 個已更新項目:

sorry 搞錯了題目是 p(x 1) +p(x-1)=2x^2-4x

Answer 1

最佳解答

p(x+1) +p(x-1)=2x^2-4x p(x)次數=p(x+1)次數 = p(x-1)次數   且 p(x+1) +p(x-1)=2x2-4x    ---> 2 次 所以令 p(x) = x2+ax+b p(x+1) +p(x-1)= (x+1)2+a(x+1)+b + (x-1)2+a(x-1)+b = 2x2+2ax+ 2b+2比較係數,得 a= -2, b= -1 ; 故 p(x)= x2-2x-1

Answer 2

p(x)=x^2-2x-1

Answer 3

deg[p(x)]=deg[p(x+1)]=deg[p(x-1)]

2=deg[p(x+1)p(x-1)]=deg[p(x+1)]+deg[p(x-1)]

==> deg[p(x)]=1

p(x)=ax+b

p(x+1)p(x-1)

=(ax+b+a)(ax+b-a)

=aaxx+2abx+bb-aa

=2xx-4x

(1) a=sqrt2, b=-sqrt2

p(x)=sqrt2*(x-1)

(2) a=-sqrt2, b=sqrt2

p(x)=-sqrt2*(x-1)