Answer:
[tex]\dfrac{3x+4}{2x-2}[/tex]
Step-by-step explanation:
Invert and multiply the way you would any fraction division problem. Then cancel common factors from numerator and denominator.
[tex]\displaystyle\frac{\left(\frac{4x^2+2x}{x^2+x-2}\right)}{\left(\frac{8x^2+4x}{3x^2+10x+8}\right)}=\frac{4x^2+2x}{x^2+x-2}\cdot\frac{3x^2+10x+8}{8x^2+4x}\\\\=\frac{2x(2x+1)(3x+4)(x+2)}{4x(2x+1)(x-1)(x+2)}=\frac{3x+4}{2(x-1)}\\\\=\frac{3x+4}{2x-2}[/tex]
