From the question,
The zeros of the polynomial function are
[tex]\frac{2}{3},-4,-\frac{4}{3}[/tex]
This implies that
[tex]x=\frac{2}{3},x=-4,x=-\frac{4}{3}[/tex]
Hence we have
[tex]x-\frac{2}{3}=0,x+4=0,x+\frac{4}{3}=0[/tex]
Hemce, the function f(x) will be
[tex]f(x)=(x-\frac{2}{3})(x+4)(x+\frac{4}{3})[/tex]
Next, we are to expand the polynomial
This gives
[tex]\begin{gathered} f(x)=(x^2+4x-\frac{2}{3}x-\frac{8}{3})(x+\frac{4}{3}) \\ f(x)=(x^2+\frac{10}{3}x-\frac{8}{3})(x+\frac{4}{3}) \end{gathered}[/tex]
Next we have
[tex]\begin{gathered} f(x)=x^3+\frac{4}{3}x^2+\frac{10}{3}x^2+\frac{40}{9}x-\frac{8}{3}x-\frac{32}{9} \\ f(x)=x^3+\frac{14}{3}x^2+\frac{16}{9}x-\frac{32}{9} \\ f(x)=9x^3+42x^2+16x-32 \end{gathered}[/tex]
Therefore, the polynomial function is
[tex]f(x)=9x^3+42x^2+16x-32[/tex]
