Answer
$2x^3-5x^2-3x+6$.
Work Step by Step
The given expression is
$\Rightarrow (2x^4-13x^3+17x^2+18x-24)\div (x-4)$
$\begin{matrix}
& 2x^3 & -5x^2 &-3x & +6 & & \leftarrow &Quotient\\
&-- &-- &--&--& \\
x-4) &2x^4&-13x^3&+17x^2&+18x&-24 & \\
& 2x^4 & -8x^3 & & & & \leftarrow &2x^3(x-4) \\
& -- & -- & & & & \leftarrow &subtract \\
& 0 & -5x^3 & +17x^2 & & & \\
& & -5x^3 & +20x^2 & & & \leftarrow & -5x^2(x-4) \\
& & -- & -- & & & \leftarrow & subtract \\
& & 0&-3x^2 &+18x & \\
& & & -3x^2& +12x & & \leftarrow & -3x(x-4) \\
& & & -- & -- && \leftarrow & subtract \\
& & & 0 & 6x &-24& \\
& & & & 6x & -24 & \leftarrow & 6(x-4) \\
& & && -- & -- & \leftarrow & subtract \\
& & & & 0 &0& \leftarrow & Remainder
\end{matrix}$
The answer is $2x^3-5x^2-3x+6$.
