數學二題：指數方程式、餘式定理

1. 設 t=10^x, 原方程2根x1,x2>0,則
t^2 - 10t+k =0, 兩根t=10^x>1,
so, (1) 100-4k >=0, then k<= 25
(2) (t1-1)(t2-1)>0 , then 1-10+k>0, k>9
(3) (t1-1)+(t2-1)>0, then 10-2>0
hence 9<k<=25

(note: x1, x2>0 , then t1, t2> 1 )

2.
Let t=x+1, then
Problem : f=(t-1)^10-2(t-1)^5+3(t-1)^2-1 divided by t^3
f=(t-1)^10-2(t-1)^5+3(t-1)^2-1
=(1-10t+45t^2-120t^3+...)+2(1-5t+10t^2-10t^3+...)+3(t-1)^2-1
Divided by t^3, then the remainder= 1-10t+45t^2+2(1-5t+10t^2)+3(t-1)^2-1
=1-10(x+1)+45(x+1)^2+2-10(x+1)+20(x+1)^2+3x^2 -1
=68x^2+110x+47

第二題如果不用微分就只能二項式定理硬做了

第二題不能用微分直接來嗎?