數學題（保證勁難）1

As follows:

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餘式定理，
(x + 1)²=0
x=-1
reminder=f(-1) 設f(x)=x¹ºº
= (-1)¹ºº
=1

Let f(x)=x^100

Let consider the reminder of f(x)/(x+1)

Sub x=-1 into f(x)

f(-1)=1

The reminder of f(x)/(x+1) is 1.

f(x)-1=x^100-1=(x^99-x^98+x^97-.....+x-1)(x+1)

Let g(x)=(x^100-1)/(x+1)

sub x=-1, g(x)=-1-1-1-1...-1-1=-100

(x+1) is a factor of g(x)+100

h(x)=[g(x)+100](x+1)=(x^100-1)+100(x+1)

h(x) can be divided by (x+1)^2

Expanding h(x)

x^100-1+100x+100=x^100+100x+99

That tell us that we need to add 100x+99 to x^100 , then it can be divided by (x+1)^2

Implied the reminder is -100x-99

Let x be 0,

The equation is:

0¹ºº ÷ (0² + 2x0 +3)
= 0 ÷ 3
= 0