Explanation
Since -3 and 1 are zeros of the functions, it implies that
[tex](x+3)\text{ }and\text{ }(x-1)[/tex]are factors of the equation.
Therefore we can find the remaining factors below
[tex](x+3)(x-1)=x^2+2x-3[/tex]By long division
[tex]remaining\text{ expression =}\frac{x^4+6x^3+7x^2-8x-6}{x^2+2x-3}=x^2+4x+2[/tex]By quadratic formula
[tex]\begin{gathered} x_{1,2}=\frac{-4\pm\sqrt{4^2-4\times1\times2}}{2\times1} \\ x_1=\frac{-4+2\sqrt{2}}{2},x_2=\frac{-4-2\sqrt{2}}{2} \\ x=-2+\sqrt{2},x=-2-\sqrt{2} \\ therefore \\ (x+2-\sqrt{2})(x+2+\sqrt{2}) \end{gathered}[/tex]The linear factor are
Answer:
[tex]f(x)=(x+3)(x-1)(x+2-\sqrt{2})(x+2+\sqrt{2})[/tex]