Assuming the problem comes from the happy world of textbook problems... a typically successful method is to assume integer factorizations: $$ (3x+a)(x+b) = 0$$ where $ab=-5$. In the world free of those complicated fractions, we have just $a= \pm 1 $ and $b= \mp 5$ to choose. So, our options are: $$ (3x+1)(x-5) \qquad \& \qquad (3x-5)(x+1)$$ and multiplication reveals $(3x+1)(x-5)=0$ is the winner. In summary, guided guess and check.

Assuming the problem comes from the happy world of textbook problems... a typically successful method is to assume integer factorizations: $$ (3x+a)(x+b) = 0$$ where $ab=-5$. In the world free of those complicated fractions, we have just $a= \pm 1 $ and $b= \mp 5$ to choose. So, our options are: $$ (3x+1)(x-5) \qquad \& \qquad (3x-5)(x+1)$$ $$ (3x-1)(x+5) \qquad \& \qquad (3x+5)(x-1)$$ and multiplication reveals $(3x+1)(x-5)=0$ is the winner. In summary, guided guess and check.