myalphonseelric
myalphonseelric myalphonseelric
  • 21-09-2018
  • Mathematics
contestada

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

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gmany
gmany gmany
  • 21-09-2018

[tex]\dfrac{c^2-4}{c+3}\div\dfrac{c+2}{3(c^2-9)}\\\\=\dfrac{c^2-2^2}{c+3}\cdot\dfrac{3(c^2-3^2)}{c+2}\ \ \ \ |\text{use}\ a^2-b^2=(a-b)(a+b)\\\\=\dfrac{(c-2)(c+2)}{c+3}\cdot\dfrac{3(c-3)(c+3)}{c+2}\\\\=\dfrac{c-2}{1}\cdot\dfrac{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}[/tex]

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