Answer
$\Rightarrow \left ( \frac{f-g}{h} \right )(x)=2x^2-\frac{x}{2}+5$.
$\left ( -\infty, -2 \right )\cup \left ( -2,\infty \right ) $.
Work Step by Step
The given functions are
$f(x)=x^3+9x^2-6x+25,$
$g(x)=-3x^3+2x^2-14x+5$
and $h(x)=2x+4$.
$\Rightarrow \left ( \frac{f-g}{h} \right )(x)=\frac{f(x)-g(x)}{h(x)}$.
Substitute all values.
$\Rightarrow \left ( \frac{f-g}{h} \right )(x)=\frac{(x^3+9x^2-6x+25)-(-3x^3+2x^2-14x+5)}{(2x+4)}$.
Clear the parentheses.
$\Rightarrow \left ( \frac{f-g}{h} \right )(x)=\frac{x^3+9x^2-6x+25+3x^3-2x^2+14x-5}{2x+4}$.
Add like terms.
$\Rightarrow \left ( \frac{f-g}{h} \right )(x)=\frac{4x^3+7x^2+8x+20}{2x+4}$.
$\begin{matrix}
& 2x^2 & -\frac{x}{2} &+5 & & \leftarrow &Quotient\\
&-- &-- &--&--& \\
2x+4) &4x^3&+7x^2&+8x&+20 & \\
& 4x^3 & +8x^2 & & & \leftarrow &2x^2(2x+4) \\
& -- & -- & & & \leftarrow &subtract \\
& 0 & -x^2 & +8x & & \\
& & -x^2 & -2x & & \leftarrow & -\left( \frac{x}{2}\right)(2x+4) \\
& & -- & -- & & \leftarrow & subtract \\
& & 0&10x &+20 & \\
& & & 10x& +20 & \leftarrow & 5(2x+4) \\
& & & -- & -- & \leftarrow & subtract \\
& & & 0 & 0 & \leftarrow & Remainder
\end{matrix}$
The Quotient is $2x^2- \frac{x}{2}+5$. We can say that $(2x+4)$ is a factor of $4x^3+7x^2+8x+20$ because the remainder is $0$.
The solution is
$\Rightarrow \left ( \frac{f-g}{h} \right )(x)=2x^2-\frac{x}{2}+5$.
For the domain of $\Rightarrow \left ( \frac{f-g}{h} \right )(x)=\frac{4x^3+7x^2+8x+20}{2x+4}$, $2x+4$ should not be zero.
We exclude $2x+4=0$.
or $x=-2$ from the domain.
The interval notation is $\left ( -\infty, -2 \right )\cup \left ( -2,\infty \right ) $.
