Find the quotient and remainder

Question

Find the quotient and remainder when $x^6+x^3+1$ is divided by $x+1$

Let $f(x)=x^6+x^3+1$
Now $f(x)=(x+1).q(x) +R $ where r is remainder Now putting $x=-1$ we get $R=f(-1)$ i.e $R=1-1+1=1$ Now $q(x)=(x^6+x^3)/(x+1)$ But what I want to know if there is another way to get the quotient except simple division.

Answer 1

$$f(x)=x^6+x^3+1= x^6 +x^5 -x^5 +x^3+1\\ = x^5(x+1) -x^5-x^4+x^4+x^3+1 \\=x^5(x+1)-x^4(x+1)+x^3(x+1)+1\\= (x+1)(x^5-x^4+x^3)+1$$

Answer 2

You're on the right track: $$ x^6+x^3=x^3(x^3+1)=x^3(x+1)(x^2-x+1) $$ Therefore $$ q(x)=(x^6+x^3)/(x+1)=x^3(x^2-x+1)=x^5-x^4+x^3 $$

Answer 3

$$x^6+x^3+1=(x+1)(x^5-x^4+x^3)+1.$$

You may apply so-call long division method.

Answer 4

In this precise case, you can notice that $x^3$ is an obvious factor. You are left with: $(x^3+1)/(x+1)$ Using the remarkable identity $A^n-B^n=(A-B)(A^{n-1}+A^{n-2}B+...+B^{n-1})$ with $A=x$ and $B=-1$ gives $q(x)=x^3(x^2-x+1)$