Question

5x^3-6x^2+10=a(x-1)^3+b(x-1)^2+c(x-1)+d，求a，b，c，d的值

Answer 1

原式=a（x^3-3x^2+3x-1）+b（x^2-2x+1）+c（x-1）+d
=ax^3-（3a-b）x^2+（3a-2b+c）x-（a-b+c+d）
解得：a=5
b=-9
c=-33
d=29

Answer 2

5x^3-6x^2+10=a(x-1)^3+b(x-1)^2+c(x-1)+d=3ax^3-3ax^2+3ax-1+bx^2-2bx+b+cx-c+d所以得出a=5；b-3a=-6；3a-2b+c=0；-1+b-c+d=10。求解a=5,b=9,c=3,d=5