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From the given problem :
[tex]\frac{x^2y^4}{x^2+11x+28}\times\frac{x^2+2x-8}{x^6y^3}[/tex]x^2 + 11x + 28 can be factored as, you need to think of two numbers that has a product of 28 and a sum of 11. In this case it's 7 and 4 :
(x + 7)(x + 4)
and x^2 + 2x - 8 can be factored as, you need to think of two numbers that has a product of -8 and a sum of 2. In this case it's 4 and -2 :
(x + 4)(x - 2)
So the expression will be :
[tex]\frac{x^2y^4}{(x+7)(x+4)}\times\frac{(x+4)(x-2)}{x^6y^3}[/tex]The term (x+4) will be cancelled out
[tex]\frac{x^2y^4(x-2)}{x^6y^3(x+7)}[/tex]Simplify further and the answer is :
[tex]\frac{y(x-2)}{x^4(x+7)}[/tex]