# Use the given zero to find the remaining zeros of the function `f(x) = x^3 - 9x^2 + 4x - 36` ; zero:2i Enter the remaining zeros of f

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If 2*i* is one zero of this function, then its conjugate, - 2*i*, is also zero of this function (this is because the function is a polynomial with real coefficients.)

This means, (*x - *2*i*) and (*x + *2*i*) are both factors of *f*(*x*).

`(x - 2i)(x+2i) = x^2+ 4`

Divide *f*(*x*) by `x^2 +4` to find the remaining factor:

See attached image for the long division problem that results in

*x* - 9.

Then, the last zero of the function is given by *x* - 9 = 0. It is 9.

**The three zeroes of f(x) are 2i, - 2i and 9.**

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