[tex]\bf \cfrac{x+3}{4x^2-8}\cdot \cfrac{1x-2}{1x^2+5x+6}\implies \cfrac{x+3}{4x^2-8}\cdot \cfrac{x-2}{x^2+5x+6}
\\\\\\
\cfrac{\underline{x+3}}{4(x^2-2)}\cdot \cfrac{x-2}{\underline{(x+3)}(x+2)}\implies \cfrac{1}{4(x^2-2)}\cdot \cfrac{x-2}{x+2}
\\\\\\
\cfrac{x-2}{4(x^2-2)(x+2)}\implies \cfrac{x-2}{4(x^3+2x^2-2x-4)}
\\\\\\
\cfrac{x-2}{4x^3+8x^2-8x-16}[/tex]
