How to Factorize Complex Polynomials

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Encyclopedia>Numbers and Quantities>Numbers>Complex Numbers>Equations>How to Factorize Complex Polynomials

In the entry about complex quadratic equations, you can see that all quadratic equations have two complex solutions. This means that all quadratic polynomials can be factorized into a product of two complex linear polynomials.

Rule

Factorization of Quadratic Polynomials

Any quadratic polynomial

f (z) = a z^{2} + b z + c

can be written as

f (z) = (z - r_{1}) (z - r_{2}),

where $r_{1}, r_{2} \in ℂ$ are complex solutions to $f (z) = 0$ .

You factorize the quadratic polynomial

f (z)

by solving the equation

f (z) = 0

using the quadratic formula. The solutions to

f (z) = 0

are called the zeros of

f (z)

, or the roots of

f (z)

. Here, the word “roots” of

f (z)

—in the context of the zeros of

f (z)

—should not be confused with the

n

th roots of complex numbers.

If solving the equation $f (z) = 0$ yields one solution $r$ , you say that $r$ is a root of $f (z)$ with a multiplicity of 2. The multiplicity of $r$ is a measure of how many times $f (z)$ can be divided by $(z - r)$ . Factorization of algebraic expressions takes two forms: Real factorization and complex factorization. When performing real factorization, the expression is written as a product of factors with only real coefficients. When performing complex factorization, the expression is written as a product of factors involving both real and complex coefficients.

Rule

Complex and Real Factorization

In complex factorization, all the factors are of degree 1. The factors may have both real and complex coefficients.
In real factorization, the factors are either of degree 1 or degree 2. The quadratic factors have only complex roots. The factors have only real coefficients.

Example 1

Perform a complex factorization of the expression
$f (z) = z^{2} + i z + 2$

You want to write the expression in the form $f (z) = (z - r_{1}) (z - r_{2})$ , where $r_{1}$ and $r_{2}$ are solutions to the equation $f (z) = 0$ . You solve this equation by using the quadratic formula, where $a = 1$ , $b = i$ and $c = 2$ are the coefficients:

\begin{array}{l} z & = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ = \frac{- i \pm \sqrt{- 1 - 8}}{2} \\ = \frac{- i \pm 3 i}{2} . \end{array}

Thus, the solutions to the equation $f (z) = 0$ are $z_{1} = - 2 i$ and $z_{2} = i$ . The complex factorization of $f (z)$ is

f (z) = (z + 2 i) (z - i) .

By using complex numbers, you’re not only able to factorize quadratic polynomials into two linear factors. According to the fundamental theorem of algebra, you’re also able to factorize expressions of degree $n$ into $n$ linear factors, counted with multiplicity. Factors “counted with multiplicity” means the factors may appear more than once.

In order to find all the factors of complex polynomials of higher degree, you can for instance use polynomial long division.

Example 2

Show that $z = 1$ is a root of the polynomial

f (z) = z^{3} - 7 z^{2} + 31 z - 25

and perform both real and complex factorization of $f (z)$

If $z = 1$ is a root of $f (z)$ , then $f (1) = 0$ :

\begin{array}{l} f (1) & = 1^{3} - 7 \cdot 1^{2} + 31 \cdot 1 - 25 \\ = 0 . \end{array}

f (1) = 1^{3} - 7 \cdot 1^{2} + 31 \cdot 1 - 25 = 0 .

Since

z = 1

is a root of the polynomial, you know that the polynomial long division

f (z) \div (z - 1)

has a remainder of 0:

Polynomdial long division of f(z) by z - 1

The result of the polynomial long division is $z^{2} - 6 z + 25$ . This quadratic polynomial can be further factorized by solving the equation

z^{2} - 6 z + 25 = 0

The quadratic formula yields

\begin{array}{l} z & = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ = \frac{6 \pm \sqrt{36 - 100}}{2} \\ = \frac{6 \pm 8 i}{2} \\ = 3 \pm 4 i . \end{array}

You now have a total of three solutions to the equation $f (z) = 0$ . Since $f (z)$ is a cubic polynomial, you know that there can’t be more than three roots. Since the solutions to $z^{2} - 6 z + 25 = 0$ are complex numbers, it’s impossible to split $z^{2} - 6 z + 25$ into factors with real coefficients. Thus, the real factorization of $f (z)$ is

f (z) = (z - 1) (z^{2} - 6 z + 25) .

Furthermore, you can use all the roots to write the complex factorization of $f (z)$ :

(z - 1) (z - (3 - 4 i)) (z - (3 + 4 i)) .

f (z) = (z - 1) (z - (3 - 4 i)) (z - (3 + 4 i)) .

In Example 2 you saw that the non-real roots of the polynomial occurred as a complex conjugate pair. This is no coincidence. In all real polynomials the roots appear as conjugate pairs. Real polynomials are polynomials with only real coefficients.

Rule

Let $r \in ℂ ∖ ℝ$ be a non-real root of a real polynomial $f (z)$ . Then the conjugate $\overset{}{r} \in ℂ$ is also a root of $f (z)$ .

Example 3

Show that $r = 1 + i$ is a root of the polynomial
$f (z) = z^{2} - 2 z + 2$

and perform a complex factorization of $f (z)$

You can show that $r = 1 + i$ is a root of $f (z)$ by confirming that $f (r) = 0$ : $\begin{array}{l} f (r) & = {(1 + i)}^{2} - 2 (1 + i) + 2 \\ = 1 + 2 i + i^{2} - 2 - 2 i + 2 \\ = 1 + 2 i - 1 - 2 - 2 i + 2 \\ = 0 . \end{array}$

Since $f (z)$ is a real polynomial, you also know that $\overset{}{r} = 1 - i$ is a root of $f (z)$ . Thus, the complex factorization of $f (z)$ is

\begin{array}{l} f (z) & = (z - r) (z - \overset{}{r}) \\ = (z - (1 + i)) (z - (1 + i)) . \end{array}

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