Answer
$\frac{8 x^3-16x^2-42 x+15}{4 x-5}= 8x^2+6x-3$
Work Step by Step
In order to perform a synthetic division on this problem, the leading coefficient of the $x$ term must be $1$. However, Warrick did not consider that and he simply took the constant $c= 5$ to work with. The correct procedure is shown below. First divide the numerator and denominator by 4 to get an equivalent expression of the problem on the right hand side as shown. Do the synthetic division.
\begin{equation}
\begin{aligned}
\frac{8 x^3-16x^2-42 x+15}{4 x-5}&= \frac{8 x^3-4 x^2-10.5x+3.75}{x-1.25}.
\end{aligned}
\end{equation} \begin{equation}
\begin{array}{r|rrrr}
1.25& 8 & -4 & -10.5 & 3.75 \\
& & 10 & 7.5 & -3.75\\
\hline & 8 & 6 & -3&0
\end{array}
\end{equation} The solution is
\begin{equation}
\begin{aligned}
\frac{8 x^3-16x^2-42 x+15}{4 x-5}&= 8x^2+6x-3.
\end{aligned}
\end{equation}
