Answer
The quotient is $x^4-3x^3+5x^2-15x+46$
and the remainder is $-138$.
Work Step by Step
The given expression is:-
$(x^5-4x^3+x)\div (x+3)$
Rewrite as descending powers of $x$.
$(x^5+0x^4-4x^3+0x^2+x+0)\div (x+3)$
The divisor is $x+3$, so the value of $c=-3$.
and on the right side the coefficients of dividend in descending powers of $x$.
Perform syntehtic division to obtain:
$\begin{matrix}
&-- &-- &--&--& \\
-3) &1&0&-4&0&1 &0 \\
& &-3 &9 &-15 &45&-138 \\
& -- & -- & --& -- &--& -- \\
& 1 & -3 & 5 &-15 &46 &-138\\
\end{matrix}$
The divisor is $x+3$
The dividend is $x^5-4x^3+x$
The Quotient is $x^4-3x^3+5x^2-15x+46$
The remainder is $-138$.
Check:-
$=\text{(Divisor)(Quotient)+Remainder}$
$=(x+3)(x^4-3x^3+5x^2-15x+46)-138$
$=x^5-3x^4+5x^3-15x^2+46x+3x^4-9x^3+15x^2-45x+138-138$
$=x^5-4x^3+x$
Hence, the quotient is $x^4-3x^3+5x^2-15x+46$ and the remainder is $-138$.
