Answer
It makes sense
Work Step by Step
Suppose we have to determine:
$$E=\dfrac{f(x)}{h(x)}-\dfrac{g(x)}{h(x)}=\dfrac{f(x)-g(x)}{h(x)},$$
where $g(x)=a_nx^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0$.
As usually the coefficients $a_k$ have different signs, in order to prevent mistakes caused by wrong signs when subtracting $g(x)$, it is a good idea, instead of subtracting each term directly, to write:
$$\begin{align*}
E&=\dfrac{f(x)-(a_nx^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0)}{h(x)}\\
&=\dfrac{f(x)-a_nx^n-a_{n-1}x^{n-1}-\dots-a_1x-a_0}{h(x)}
\end{align*}$$
because this way we can distribute the negative sign to each term without the risk of making mistakes.
For example:
$$\begin{align*}
E&=\dfrac{10x^4}{2x+3}-\dfrac{7x^3-3x^2+4x-1}{2x+3}\\
&=\dfrac{10x^4-(7x^3-3x^2+4x-1)}{2x+3}\\
&=\dfrac{10x^4-7x^3+3x^2-4x+1}{2x+3}
\end{align*}$$
So the statement makes sense.
