Answer
Domain: $(-\infty,-7)\cup(-7,2)\cup(2,\infty)$;
$f(x)=\frac{5}{x+7}$.
Work Step by Step
The given function is
$f(x)=\frac{5x-10}{x^2+5x-14}$
$\Rightarrow 5x-10$
Factor out $5$ from both terms.
$\Rightarrow 5(x-2)$
$\Rightarrow x^2+5x-14$
Rewrite the middle term $5x$ as $7x-2x$.
$\Rightarrow x^2+7x-2x-14$
Group terms.
$\Rightarrow (x^2+7x)+(-2x-14)$
Factor each group.
$\Rightarrow x(x+7)-2(x+7)$
Factor out $(x+7)$.
$\Rightarrow (x+7)(x-2)$
Back substitute all factors into the fraction.
$f(x)=\frac{5(x-2)}{(x+7)(x-2)}$
The denominator should not be zero.
We exclude the value $(x+7)(x-2)=0$
Set each factor equal to zero.
$x+7=0$ or $x-2=0$
Simplify.
$x=-7$ or $x=2$
The interval notation is
$(-\infty,-7)\cup(-7,2)\cup(2,\infty)$
The factored form of the given function is
$f(x)=\frac{5(x-2)}{(x+7)(x-2)}$
Cancel common terms.
$f(x)=\frac{5}{x+7}$.
