4th time posting this...........

Answer:D is the correct representation.[tex]\frac{\frac{4x^2 + 2x}{x^{2}+x-2} }{ \frac{8x^2 +4x}{3x^2 +10x+8} }[/tex] =  [tex]{\frac{(3x+4)}{2(x-1)}[/tex]Step-by-step explanation:Here, the given equation is:[tex]\frac{\frac{4x^2 + 2x}{x^{2}+x-2} }{ \frac{8x^2 +4x}{3x^2 +10x+8} }[/tex]here, numerator  = [tex]{\frac{4x^2 + 2x}{x^{2}+x-2} }[/tex]and denominator = [tex]\frac{8x^2 +4x}{3x^2 +10x+8}[/tex]Solving numerator and denominator separately, we getNUMERATOR: [tex]{\frac{4x^2 + 2x}{x^{2}+x-2} } \implies\frac{2x(2x +1)}{x^{2} + 2x-x-2 } \\\implies\frac{2x(2x +1)}{(x+2)(x-1) }[/tex]Denominator:[tex]\frac{8x^2 +4x}{3x^2 +10x+8}  = \frac{4x(2x+1)}{3x^2 +6x+ 4x+8}\\ \implies \frac{4x(2x+1)}{3x(x+2)+ 4(x+2)}  =  \frac{4x(2x+1)}{(3x+4)(x+2)}[/tex]Hence, the transformed  fraction is: [tex]\frac{\frac{2x(2x +1)}{(x+2)(x-1) }}{\frac{4x(2x+1)}{(3x+4)(x+2)}}  = {\frac{2x(2x +1)}{(x+2)(x-1) }} \times{\frac{(3x+4)(x+2)}{4x(2x+1)}[/tex]or, implied fraction is  [tex]{\frac{(3x+4)}{2(x-1)}[/tex]Hence, D is the correct representation.