Answer:
Step-by-step explanation:
We have to solve the given expression,
[tex]\frac{9y-33y^2-3y^4}{100-49y^2}.\frac{7y^2+17y+10}{14y^2+28y}[/tex]
[tex]\frac{9y-33y^2-3y^4}{100-49y^2}.\frac{7y^2+17y+10}{14y^2+28y}[/tex] = [tex]\frac{-y(-9+33y+3y^3)}{100-49y^2}.\frac{7y^2+17y+10}{14y(y+2)}[/tex]
= [tex]\frac{-y(-9+33y+3y^3)}{(10-7y)(10+7y)}.\frac{7y^2+10y+7y+10}{14y(y+2)}[/tex]
= [tex]\frac{-y(-9+33y+3y^3)}{(10-7y)(10+7y)}.\frac{y(7y+10)+1(7y+10)}{14y(y+2)}[/tex]
= [tex]\frac{-y(-9+33y+3y^3)}{(10-7y)(10+7y)}.\frac{(y+1)(7y+10)}{14y(y+2)}[/tex]
= [tex]\frac{-3y(-3+11y+y^3)}{(10-7y)}.\frac{(y+1)}{14y(y+2)}[/tex]
= [tex]\frac{-3(-3+11y+y^3)}{(10-7y)}.\frac{(y+1)}{14(y+2)}[/tex]
= [tex]\frac{3(3-11y-y^3)(y+1)}{(10-7y)(14(y+2)}[/tex]
