Answer:
The remaining zeros are -7i, 4
Explanations:
Given the polynomial function:
[tex]f(x)=x^3-4x^2+49x-196[/tex]
Group the functions:
[tex]f(x)=(x^3-4x^2)+(49x-196)[/tex]
Factor out the common term in both parenthesis
[tex]\begin{gathered} f(x)=x^2(x-4)+49(x-4) \\ f(x)=(x^2+49)(x-4) \end{gathered}[/tex]
Get the zeros of the function by making f(x) = 0
[tex]\begin{gathered} (x^2+49)(x-4)=0 \\ \end{gathered}[/tex]
Equate both factors to zero;
[tex]\begin{gathered} x^2+49=0 \\ x^2=-49 \\ x=\pm\sqrt[]{-49} \\ x=\pm\sqrt[]{49}\times\sqrt[]{-1} \\ x=\pm7i \end{gathered}[/tex]
For the other factor (x-4) = 0
[tex]\begin{gathered} x-4=0 \\ x=0+4 \\ x=4 \end{gathered}[/tex]
Given one of the zeros as 7i, the other zeros of the function will be -7i and 4
