In general, a polynomial of degree 3 is given by the next expression
[tex]f(x)=a_3x^3+a_2x^2+a_1x+a_0=c(x-b_1)(x-b_2)(x-b_3)[/tex]Where b_1, b_2, and b_3 are the roots of the polynomial.
Therefore, in our case,
[tex]\begin{gathered} x=2\to x-2=0,x=-2\to x+2=0,x=-3\to x+3=0 \\ f(x)=c(x-2)(x+2)(x+3) \end{gathered}[/tex]We need to find the value of c.
For this, notice that the f(0)=4
[tex]\begin{gathered} f(0)=4 \\ \Rightarrow c(-2)(2)(3)=4 \\ \Rightarrow c=\frac{4}{-2\cdot2\cdot3}=\frac{4}{-12}=-\frac{1}{3} \end{gathered}[/tex]Therefore,
[tex]\begin{gathered} f(x)=-\frac{1}{3}(x-2)(x+2)(x+3)=-\frac{1}{3}(x^2-4)(x+3)=-\frac{1}{3}(x^3+3x^2-4x-12)=-\frac{x^3}{3}-x^2+\frac{4x}{3}+4 \\ \end{gathered}[/tex]The answer is f(x)=(-1/3)(x-2)(x+2)(x+3), which is equivalent to -x^3/3-x^2+4x/3+4
