To solve this problem, you find the value of x that will make the function to be = 0 by substituting the likely values from the option into the eqaution and checking if after the simplification the value is 0
so checking
[tex]\begin{gathered} \text{The factors betwe}en\text{ }3\text{ and 45 are } \\ 1,3,5,9,15,45 \\ \text{factors of 3 are 1,3} \end{gathered}[/tex]we have
[tex]\begin{gathered} =3x^{^3}-13x^2-3x\text{ +45} \\ \pm1,\text{ 3, 5,9, 15,45} \\ \pm\frac{1}{3},\text{ 1, 5/3, 3, 5 , 15} \\ \text{values that apply are +3 twice and -5/3} \end{gathered}[/tex]