Given the function:
[tex]f(x)=x^3-5x^2+36x-180[/tex]
The given function has 3 zeros
one of the zeros = -6i
so, the second one will be a conjugate to the first zero
So, the second one = 6i
The product of the three zeros = -180
let the third zero = a
So,
[tex]\begin{gathered} -6i\cdot6i\cdot b=-180 \\ -36i^2\cdot b=180 \\ -36\cdot-1\cdot b=180 \\ 36b=180 \\ \\ b=\frac{180}{35}=5 \end{gathered}[/tex]
So, the third zero = 5
So, the answer will be:
The remaining zeros of the function are: 6i and 5
