Answer
The quotient is $4x^5+4x^4+x^3+x^2+2x+2$
and the remainder is $7$.
Work Step by Step
The given expression is:-
$(4x^6-3x^4+x^2+5)\div (x-1)$
Rewrite as descending powers of $x$.
$(4x^6+0x^5-3x^4+0x^3+x^2+0x+5)\div (x-1)$
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 the syntehtic division to obtain:
$\begin{matrix}
&-- &-- &--&--& \\
1) &4&0&-3&0&1&0&5 \\
& &4 &4 &1 &1& 2&2\\
& -- & -- & --& -- &--&--&-- \\
& 4 & 4 & 1 &1 &2 &2&7\\
\end{matrix}$
The divisor is $x-1$
The dividend is $4x^6-3x^4+x^2+5$
The Quotient is $4x^5+4x^4+x^3+x^2+2x+2$
The remainder is $7$.
Check:-
$=\text{(Divisor)(Quotient)+Remainder}$
$=(x-1)(4x^5+4x^4+x^3+x^2+2x+2)+7$
$=4x^6+4x^5+x^4+x^3+2x^2+2x-4x^5-4x^4-x^3-x^2-2x-2+7$
$=4x^6-3x^4+x^2+5$
Hence, the quotient is $4x^5+4x^4+x^3+x^2+2x+2$ and the remainder is $7$.
