Use synthetic division to rewrite the following fraction in the form $q(x)+\frac{r(x)}{d(x)}$, where $d(x)$ is the denominator of the original fraction, $q(x)$ is the quotient, and $r(x)$ is the remainder.
\[
\frac{5 x^{5}-25 x^{4}+27 x^{3}-20 x^{2}-32 x}{x-4}
\]

Step 1 ：Given the fraction \(\frac{5 x^{5}-25 x^{4}+27 x^{3}-20 x^{2}-32 x}{x-4}\), we are asked to perform synthetic division to rewrite the fraction in the form \(q(x)+\frac{r(x)}{d(x)}\), where \(d(x)\) is the denominator of the original fraction, \(q(x)\) is the quotient, and \(r(x)\) is the remainder.

Step 2 ：The divisor in this case is \(x-4\), so we will use 4 as the divisor in synthetic division.

Step 3 ：Performing synthetic division on the given polynomial, we get the quotient as \([5, -5, 7, 8, 0]\) and the remainder as 0.

Step 4 ：This means that the polynomial \(5x^{5}-25x^{4}+27x^{3}-20x^{2}-32x\) divided by \(x-4\) gives a quotient of \(5x^{4}-5x^{3}+7x^{2}+8x\) and a remainder of 0.

Step 5 ：Therefore, the fraction can be rewritten as \(5x^{4}-5x^{3}+7x^{2}+8x + \frac{0}{x-4}\).

Step 6 ：Final Answer: The fraction can be rewritten as \(\boxed{5x^{4}-5x^{3}+7x^{2}+8x + \frac{0}{x-4}}\)