Question

Using the given zero to find the remaining zeros of the function f(x)=x^3-4^2+49x-196; zero: -7i

Answer 1

Because $f$ has real coefficients, you know the complex root occurs along with its conjugate. That is, both $x=\pm7i$ are roots to $f(x)$.

This means that dividing through by both factors yields another polynomial with no remainder:

$\dfrac{x^3-4x^2+49x-196}{(x-7i)(x+7i)}=\dfrac{x^3-4x^2+49x-196}{x^2+49}=x-4$

This means the last root (there are only three according to the fundamental theorem of algebra) is $x=4$.