Answer
$ x^3-2x^2 -4x-12-\frac{42}{x-3}$.
Work Step by Step
The given expression is
$(x^4-5x^3+2x^2-6)\div(x-3)$
Rewrite the expression as
$(x^4-5x^3+2x^2+0x-6)\div(x-3)$
$\begin{matrix}
&x^3& -2x^2 & -4x &-12 & & \leftarrow &Quotient\\
&--&-- &-- &--&--& \\
x-3) &x^4&-5x^3&+2x^2&+0x&-6 & \\
& x^4 & -3x^3 & & && \leftarrow &x^3(x-3) \\
& -- & -- & & && \leftarrow &subtract \\
& 0 & -2x^3 & +2x^2 & & \\
& & -2x^3 & +6x^2 & & & \leftarrow & -2x^2(x-3) \\
& & -- & -- & && \leftarrow & subtract \\
& & 0&-4x^2 &+0x & \\
& & & -4x^2& +12x && \leftarrow & -4x(x-3) \\
& & & -- & --& & \leftarrow & subtract \\
& & & 0 & -12x &-6& \\
& & & & -12x &+36& \leftarrow & -12(x-3) \\
& & & & --& --& \leftarrow & subtract \\
& & & & 0 &-42& \leftarrow & Remainder
\end{matrix}$
The answer is
$\Rightarrow Quotient + \frac{Remainder}{Divisor}$
$\Rightarrow x^3-2x^2 -4x-12-\frac{42}{x-3}$.
