Find the reminder of x^8 + x + 1 when divided by x^2 + 2x + 1.?

Sol
設x^8+x+1=p(x)(x^2+2x+1)+ax+b
8x^7+1=p(x)(2x+2)+p’(x)(x^2+2x+1)+a
8*(－1)^7+1=a
a=－7
x^8+x+1=p(x)(x^2+2x+1)－7x+b
(－1)^8+(－1)+1=7+b
1=7+b
b=－6
餘式=－7x－6
or
x^2+2x+1=0
x^2=－2x－1
x^8+x+1
=>(－2x－1)^4+x+1
=>(4x^2+4x+1)(4x^2+4x+1)+x+1
=>(－8x－4+4x+1)(－8x+4x+1)+x+1
=(－4x－3)(－4x－3)+x+1
=>(16x^2+24x+9)+x+1
=>(－32x－16+24x+9)+x+1
=>－7x－6
or
y=x+1
x=y－1
x^8+x+1
=(y－1)^8+x+1
=y^8－8y^7+…－8y+1+x+1
餘式=－8y+1+x+1=－8(x+1)+1+x+1=－7x－6