The options have not been given but three correct alternatives have been presented as the solution.
Answer:
[tex]3(c-2)(c-3)[/tex] or[tex]3(c^2-5c+6)[/tex] or [tex]3c^2 - 15c + 18[/tex]
Step-by-step explanation:
[tex]\dfrac{c^2-4}{c+3}\div\dfrac{c+2}{3(c^2-9)}[/tex]
Applying the rule [tex]a^2 - b^2 = (a+b)(a-b)[/tex] to the numerator of the first term and the denominator of the second term,
[tex]\dfrac{c^2-2^2}{c+3}\div\dfrac{c+2}{3(c^2-3^2)}[/tex]
[tex]\dfrac{(c+2)(c-2)}{c+3}\div\dfrac{c+2}{3(c+3)(c-3)}[/tex]
By changing ÷ to ×, we invert the second term
[tex]\dfrac{(c+2)(c-2)}{c+3}\times\dfrac{3(c+3)(c-3)}{c+2}[/tex]
[tex](c-2)\times3(c-3)[/tex]
[tex]3(c-2)(c-3) = 3(c^2-5c+6) = 3c^2 - 15c + 18[/tex]
