Answer
$\frac{2(x-3)}{x-1}$.
Work Step by Step
The given expression is
$=\frac{3x^2-7x-6}{3x^2-13x-10}\div \frac{2x^2-x-1}{4x^2-18x-10}$
Factor $3x^2-7x-6$.
Rewrite the middle term $-7x$ as $-9x+2x$.
$=3x^2-9x+2x-6$
Group terms.
$=(3x^2-9x)+(2x-6)$
Factor each term.
$=3x(x-3)+2(x-3)$
Factor out $(x-3)$.
$=(x-3)(3x+2)$.
Factor $3x^2-13x-10$.
Rewrite the middle term $-13x$ as $-15x+2x$.
$=3x^2-15x+2x-10$
Group terms.
$=(3x^2-15x)+(2x-10)$
Factor each term.
$=3x(x-5)+2(x-5)$
Factor out $(x-5)$.
$=(x-5)(3x+2)$.
Factor $2x^2-x-1$.
Rewrite the middle term $-x$ as $-2x+x$.
$=2x^2-2x+x-1$
Group terms.
$=(2x^2-2x)+(x-1)$
Factor each term.
$=2x(x-1)+1(x-1)$
Factor out $(x-1)$.
$=(x-1)(2x+1)$.
Factor $4x^2-18x-10$.
Rewrite the middle term $-18x$ as $-20x+2x$.
$=4x^2-20x+2x-10$
Group terms.
$=(4x^2-20x)+(2x-10)$
Factor each term.
$=4x(x-5)+2(x-5)$
Factor out $(x-5)$.
$=(x-5)(4x+2)$.
$=2(x-5)(2x+1)$.
Substitute all factors into the given expression.
$=\frac{(x-3)(3x+2)}{(x-5)(3x+2)}\div \frac{(x-1)(2x+1)}{(x-5)(4x+2)}$
$=\frac{(x-3)(3x+2)}{(x-5)(3x+2)}\times \frac{2(x-5)(2x+1)}{(x-1)(2x+1)}$
Cancel common terms.
$=\frac{2(x-3)}{x-1}$.
