The Conjugate Zeros Theorem

If f(x) defines a polynomial function having only real coefficients and if z = a + bi is a zero of f(x), where a and b are real numbers, then the conjugate of z, $LaTeX: a-bi,\:$ $a-bi,\:$ is also a zero of f(x).

Note:

Caution When the conjugate zeros theorem is applied, it is essential that the polynomial have only real coefficients.

For example,

f(x) = x − (1 + i)

has 1 + i as a zero, but the conjugate 1 − i is not a zero.

Example

Find a polynomial function f(x) of least degree having only real coefficients and zeros 3 and 2 + i.

Solution

The complex number 2 − i must also be a zero, so the polynomial has at least three zeros: 3, 2 + i, and 2 − i. For the polynomial to be of least degree, these must be the only zeros. By the factor theorem there must be three factors, x − 3, x − (2 + i), and x – (2 − i). So the factored form of the polynomial must be:

$LaTeX: f\left(x\right)=\left(x-3\right)\left(x-\left(2+i\right)\right)\left(x-\left(2-i\right)\right)$ $f\left(x\right)=\left(x-3\right)\left(x-\left(2+i\right)\right)\left(x-\left(2-i\right)\right)$

Simplifying, we get:

$LaTeX: f\left(x\right)=\left(x-3\right)\left(x-2-i\right)\left(x-2+i\right)=\left(x-3\right)\left(\left(x-2\right)-1\right)\left(\left(x-2\right)+i\right)=\left(x-3\right)\left(\left(x-2\right)^2-i^2\right)=\left(x-3\right)\left(x^2-4x+4-\left(-1\right)\right)=\left(x-3\right)\left(x^2-4x+5\right)=x^3-7x^2+17x-15$ $f\left(x\right)=\left(x-3\right)\left(x-2-i\right)\left(x-2+i\right)=\left(x-3\right)\left(\left(x-2\right)-1\right)\left(\left(x-2\right)+i\right)=\left(x-3\right)\left(\left(x-2\right)^2-i^2\right)=\left(x-3\right)\left(x^2-4x+4-\left(-1\right)\right)=\left(x-3\right)\left(x^2-4x+5\right)=x^3-7x^2+17x-15$

Any nonzero multiple of $LaTeX: x^3-7x^2+17x-15\:$ $x^3-7x^2+17x-15\:$ also satisfies the given conditions on zeros. The information on zeros given in the problem is
not sufficient to give a specific value for the leading coefficient.

Example

Find all zeros of $LaTeX: f\left(x\right)=x^4-7x^3+18x^2-22x+12,\:$ $f\left(x\right)=x^4-7x^3+18x^2-22x+12,\:$ given that 1 − i is a zero.

Solution

Since the polynomial function has only real coefficients and since 1 − i is a zero, by the conjugate zeros theorem 1 + i is also a zero. To find the remaining zeros, first use synthetic division to divide the original polynomial by x − (1 − i).

Using Synthetic Division to Divide by x-(1-i).PNG

By the factor theorem, since x = 1 − i is a zero of f(x), x − (1 − i) is a factor, and f(x) can be written as follows.

$LaTeX: f\left(x\right)=\left[x-\left(1-i\right)\right]\left[x^3+\left(-6-i\right)x^2+\left(11+5i\right)x+\left(-6-6i\right)\right]$ $f\left(x\right)=\left[x-\left(1-i\right)\right]\left[x^3+\left(-6-i\right)x^2+\left(11+5i\right)x+\left(-6-6i\right)\right]$

We know that x = 1 + i is also a zero of f(x). Continue to use synthetic division and divide the quotient polynomial above by x − (1 + i).

Using Synthetic Division to Divide by x-(1+i).PNG

Now f(x) can be written in the following factored form.

$LaTeX: f\left(x\right)=\left[x-\left(1-i\right)\right]\left[x-\left(1+i\right)\right]\left[x^2-5x+6\right]=\left[x-\left(1-i\right)\right]\left[x-\left(1+i\right)\right]\left(x-2\right)\left(x-3\right)\:$ $f\left(x\right)=\left[x-\left(1-i\right)\right]\left[x-\left(1+i\right)\right]\left[x^2-5x+6\right]=\left[x-\left(1-i\right)\right]\left[x-\left(1+i\right)\right]\left(x-2\right)\left(x-3\right)\:$

The remaining zeros are 2 and 3. The four zeros are 1 − i, 1 + i, 2, and 3.