Question

question
perform the following operation and express in simplest form.
$\frac{x^{2}-2x-3}{x^{3}} div \frac{x^{2}-6x-7}{x}$

Explanation:

Step1: Rewrite division as multiplication

$\frac{x^2 - 2x - 3}{x^3} \times \frac{x}{x^2 - 6x - 7}$

Step2: Factor quadratic numerators

Factor $x^2-2x-3=(x-3)(x+1)$; Factor $x^2-6x-7=(x-7)(x+1)$
$\frac{(x-3)(x+1)}{x^3} \times \frac{x}{(x-7)(x+1)}$

Step3: Cancel common factors

Cancel $(x+1)$ and $x$: $\frac{(x-3)}{x^2} \times \frac{1}{(x-7)}$

Step4: Multiply remaining terms

$\frac{(x-3)}{x^2(x-7)}$

Answer:

$\frac{x-3}{x^2(x-7)}$ (or expanded form $\frac{x-3}{x^3-7x^2}$)