Answer
$x-2$
Work Step by Step
The expression in the parentheses is a difference of rational expressions with equal denominators,
... dividing with $\displaystyle \frac{A}{B}$ equals multiplying with $\displaystyle \frac{B}{A}$
$...=\displaystyle \frac{3x^{2}-4x+4-(10x+9)}{3x^{2}+7x+2}\times\frac{x^{2}-4}{x-5}$
$=\displaystyle \frac{3x^{2}-4x+4-10x-9}{3x^{2}+7x+2}\times\frac{x^{2}-4}{x-5}$
... recognize the difference of squares
$=\displaystyle \frac{3x^{2}-14x-5}{3x^{2}+7x+2}\times\frac{(x+2)(x-2)}{x-5}$
... Factor $ax^{2}+bx+c$ by searching for factors of $ac$ whose sum is $b$
... if sucessful, rewrite $bx$ and factor in pairs.
$3x^{2}-14x-5$=
$...$factors of $3\cdot(-5)=-15$ ... whose sum is $-14 ..$. are -15 and +1
$3x^{2}-14x-5$=$3x^{2}-15x+x-5$=
$=3x(x-5)+(x-5)=(x-5)(3x+1)$
$3x^{2}+7x+2=$
$...$factors of $3\cdot(2)=6$ ... whose sum is $7 ..$. are $+6$ and $+1$
$3x^{2}+7x+2=3x^{2}+6x+x+2=$
$=3x(x+2)+(x+2)=(x+2)(3x+1)$
Problem, continued $=\displaystyle \frac{(x-5)(3x+1)}{(x+2)(3x+1)}\times\frac{(x+2)(x-2)}{(x-5)}$
$... $ common factors cancel
=$\displaystyle \frac{x-2}{1}$
= $x-2$
