We have $f,ginRR[X];f=(X-1)^n-X^n+1;g=X^2-3X+2$ .How to find the rest of dividing $f$ to $g$ ?

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Answer 1 · 2017-04-17T21:35:49

$r(x) = (2-2^n)x+2^n-2$

For degree consistence in $f(x) = q(x)g(x)+r(x)$

If $deg(f)=n-1$ and $deg(g)=2$ then

$deg(q)=n-1-2=n-3$ and $deg(r) = 1$

so

$r(x) = ax+b$ and

$g(x)=(x-1)(x-2)$

so

$f(x)=q(x)(x-1)(x-2)+a x + b$

Now we have

$f(1)=a cdot 1 + b = 0$
$f(2)= a cdot 2 + b =1-2^n+1 = 2-2^n$

Solving

${(a cdot 1 + b = 0),(a cdot 2 + b = 2-2^n):}$

for $a,b$ we have $a = 2-2^n$ and $b=2^n-2$

and finally

$r(x) = (2-2^n)x+2^n-2$

We have f,ginRR[X];f=(X-1)^n-X^n+1;g=X^2-3X+2f,ginRR[X];f=(X-1)^n-X^n+1;g=X^2-3X+2.How to find the rest of dividing ff to gg?