H(x)=x^{3}-2x^{2}+16x-32

Find all zeros of this function.

4i, -4i, 2

2, -2, -4i

4i+1, 4i-1, 1

4i-1, -2, 2

Write the polynomial as a product of linear factors

(x-(4i-1))(x+2)(x-2)

(x-(4i+1)(x-(4i-1))(x-1)

(x-4i)(x+4i)(x-2)

(x-2)(x+2)(x+4i)

The possible rational zeros (according to Rational Zero test) are ±1, ±2, ±4, ±8, ±16, ±32.

To help determine which of those choices is actually a zero, take a look at the graph of h(x).

It looks like there is a rational zero at x=2, so test the factor (x-2) with synthetic division.

H(x)= x^{3}-2x^{2}+16x-32=(x-2)(x^{2}+16)=(x-2)(x-4i)(x+4i)
[written as a product of linear factors]

Therefore all the zeros of this function are x=2, 4i, and -4i.