Answer
$\frac{x-7}{x^2}$.
Work Step by Step
The given expression is
$=\frac{x^2-9x+14}{x^3+2x^2}\cdot \frac{x^2-4}{x^2-4x+4}$
Factor $x^2-9x+14$.
Rewrite the middle term $-9x$ as $-7x-2x$.
$=x^2-7x-2x+14$
Group terms.
$=(x^2-7x)+(-2x+14)$
Factor each term.
$=x(x-7)-2(x-7)$
Factor out $(x-7)$.
$=(x-7)(x-2)$
Factor $x^3+2x^2$.
$=x^2(x+2)$
Factor $x^2-4$.
$=x^2-2^2$
Use the algebraic expression $a^2-b^2=(a+b)(a-b)$.
$=(x+2)(x-2)$
Factor $x^2-4x+4$
Rewrite the middle term $-4x$ as $-2x-2x$.
$=x^2-2x-2x+4$
Group terms.
$=(x^2-2x)+(-2x+4)$
Factor each term.
$=x(x-2)-2(x-2)$
Factor out $(x-2)$.
$=(x-2)(x-2)$
Substitute all factors into the given expression.
$=\frac{(x-7)(x-2)}{x^2(x+2)}\cdot \frac{(x+2)(x-2)}{(x-2)(x-2)}$
Cancel common terms.
$=\frac{x-7}{x^2}$.
