Question

C^2-4/c+3 divided by c+2/3(c^2-9)

Answer 1

`(c^2-4)/(c+3)/(c+2)/(3(c^2-9))\\\\=(c^2-2^2)/(c+3)\cdot(3(c^2-3^2))/(c+2)\ \ \ \ |\text{use}\ a^2-b^2=(a-b)(a+b)\\\\=((c-2)(c+2))/(c+3)\cdot(3(c-3)(c+3))/(c+2)\\\\=(c-2)/(1)\cdot(3(c-3))/(1)=(c-2)(3)(c-3)\ \ \ \ |\text{use distributive property}\\\\=(c-2)(3c-9)=(c)(3c)+(c)(-9)+(-2)(3c)+(-2)(-9)\\\\=3c^2-9c-6c+18=\boxed{3c^2-15c+18}`