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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.

John757
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$$f(x)=x^6+x^3+1= x^6 +x^5 -x^5 +x^3+1\\ = x^5(x+1) -x^5-x^4+x^4+x^3+1 \\=x^5(x+1)-x^4(x+1)+x^3(x+1)+1\\= (x+1)(x^5-x^4+x^3)+1$$

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

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

You may apply so-call long division method.

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mengdie1982
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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 $q(x)=x^3(x^2-x+1)$

Benoit Gaudeul
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  • Sorry, but $x^3$ is "an obvious factor" of what? Certainly not of $f(x)$ in this problem. Likely you have in mind that $x^3$ is a factor of $q(x)$ relatively prime to the "divisor" $x+1$, and can thus be removed to simplify the calculation. – hardmath Jul 16 '18 at 05:04
  • an obvious factor of $f(x)-1$: The OP as very smartly proved that the remainder is $1$, so you just have to divide $f(x)-1$ by $x+1$. – Benoit Gaudeul Jul 16 '18 at 07:48