Respuesta :
Answer with explanation:
[tex]\rightarrow \frac{\frac{2y^2-6 y-20}{4 y+12}}{\frac{y^2+5 y+6}{3 y^2+28 y+27}}\\\\\rightarrow \frac{\frac{y^2-3y-10}{2 y+6}}{\frac{(y+2)(y+3)}{3 y^2+28 y+27}}\\\\\rightarrow \frac{\frac{(y-5)(y+2)}{2 (y+3)}}{\frac{(y+2)(y+3)}{3 y^2+28 y+27}}\\\\\rightarrow \frac{(y-5)(y+2)}{2 (y+3)}} \times {\frac{3 y^2+28 y+27}{(y+2)(y+3)}}\\\\ \rightarrow\frac{(y-5)\times(3 y^2+28 y+27)}{2 (y+3)^2}}[/tex]
→y²+5y+6
=y²+3 y+2 y+6
=y×(y+3)+2×(y+3)
=(y+2)(y+3)
→y² -3 y-10
=y² -5 y+2 y -10
=y×(y-5)+2×(y-5)
=(y+2)(y-5)