g(x)=\frac{(x+5)(x+3)}{3x^2-27}
dziedzina
3x^2-27\ne 0
3(x^2-9)\ne 0
x^2-9\ne0
(x-3)(x+3)\ne0
x\ne 3 i x\ne-3
D=\mathbb R \backslash \{-3,3\}
miejsca zerowe
\frac{(x+5)(x+3)}{3x^2-27}=0
\frac{(x+5)(x+3)}{3(x^2-9)}=0|*3
\frac{(x+5)(x+3)}{x^2-9}=0
(x+5)(x+3)(x^2-9)=0
(x+5)(x+3)(x+3)(x-3)=0
(x+5)(x+3)^2*(x-3)=0
(x=-5\vee x=-3 \not \in D \vee x=3\not \in D …\vee - lub
x=-5