Answer
The quotient is $x^4+x^3+x^2+x+1$
and the remainder is $0$.
Work Step by Step
The given expression is:-
$(x^5-1)\div (x-1)$
Rewrite as descending powers of $x$.
$(x^5+0x^4+0x^3+0x^2+0x-1)\div (x-1)$
The divisor is $x-1$, so the value of $c=1$.
and on the right side the coefficients of dividend in descending powers of $x$.
Perform synthetic division to obtain:
$\begin{matrix}
&-- &-- &--&--&--&-- \\
1) &1&0&0&0&0&-1& \\
& &1 &1 & 1 &1 &1\\
& -- & -- & --&-- &--&--& \\
& 1 & 1 & 1 & 1 &1 &0&\\
\end{matrix}$
The divisor is $x-1$
The dividend is $x^5-1$
The Quotient is $x^4+x^3+x^2+x+1$
The remainder is $0$.
Check:-
$\text{(Divisor)(Quotient)+Remainder}$
$=(x-1)(x^4+x^3+x^2+x+1)+0$
$=x^5+x^4+x^3+x^2+x-x^4-x^3-x^2-x-1+0$
$=x^5-1$
Hence, the quotient is $x^4+x^3+x^2+x+1$ and the remainder is $0$.
