Answer:
[tex]\large\boxed{\dfrac{2s-6}{s+3}}[/tex]
Step-by-step explanation:
[tex]s^2-9=s^2-3^2\qquad\text{use}\ a^2-b^2=(a-b)(a+b)\\=(s-3)(s+3)\\\\s^2+6s+9=s^2+2(s)(3)+3^2\qquad\text{use}\ (a+b)^2=a^2+2ab+b^2\\=(s+3)^2[/tex]
[tex]\dfrac{s^2-9}{2s}\div\dfrac{s^2+6s+9}{4s}=\dfrac{s^2-9}{2s\!\!\!\!\diagup_1}\cdot\dfrac{4s}{s^2+6s+9}\\\\=\dfrac{(s-3)(s+3)}{2s\!\!\!\!\!\diagup_{_1}}\cdot\dfrac{4s\!\!\!\!\!\diagup^{^2}}{(s+3)^2}=\dfrac{2(s-3)(s+3)}{(s+3)(s+3)}\qquad\text{cancel}\ (s+3)\\\\=\dfrac{2(s-3)}{s+3}=\dfrac{2s-6}{s+3}[/tex]
