The Factorization of Polynomials with Complex Coefficients

# The Factorization of Polynomials with Complex Coefficients

The following theorem will tell us that if we have a polynomial with complex coefficients, then is can be uniquely factorized as a product of linear terms, each of which corresponds to a specific root of the polynomial.

 Theorem 1: Let $p(x) \in \wp ( \mathbb{C} )$ be a non-constant polynomial. Then $p(x)$ has a unique factorization $p(x) = c(x - \lambda_1)(x - \lambda_2)...(x - \lambda_m)$ where $\lambda_1, \lambda_2, ..., \lambda_m \in \mathbb{C}$ are the roots of $p$ and $c \in \mathbb{C}$.
• Proof: We will carry out this proof by induction to first show that such a factorization exists. Let $p(x) \in \wp (\mathbb{C})$ be a nonconstant polynomial. Then $\mathrm{deg} (p) = m ≥ 1$. For $m ≥ 1$, let $S(m)$ be the statement that every polynomial $p(x) \in \wp (\mathbb{C})$ with degree $m$ has a factorization in the form $c(x - \lambda_1)(x - \lambda_2)...(x - \lambda_m)$ where $\lambda_1, \lambda_2, ..., \lambda_m \in \mathbb{C}$ are the roots of $p(x)$ and $c \in \mathbb{C}$.
• Suppose that $m = 1$. Then $p(x) = a_0 + a_1x$ where $a_0, a_1 \in \mathbb{C}$ and $a_1 \neq 0$. Clearly $p(x)$ can be written in the form $p(x) = a_1 \left ( x - \left ( \frac{-a_0}{a_1} \right ) \right )$, and so $S(1)$ is true.
• Now suppose that $m > 1$ and that all polynomials of degree $m - 1$ can be factored in the form above. We know by The Fundamental Theorem of Algebra that since $\mathrm{deg} (p) = m ≥ 1$ then $p(x)$ has a root, say $\lambda_1$. Therefore for some polynomial $q(x) \in \wp ( \mathbb{C} )$ where $\mathrm{deg} (q) = m - 1$ we have that:
(1)
\begin{align} \quad p(x) = (x - \lambda_1) q(x) \end{align}
• Since $\mathrm{deg} (q) = m - 1$, then by our induction hypothesis it follows that $q(x)$ can be factored in the form specified earlier, so let $q(x) = c(x - \lambda_2)...(x - \lambda_m)$ and substitute this into the equation above so that we get:
(2)
\begin{align} \quad p(x) = c(x - \lambda_1)(x - \lambda_2)...(x - \lambda_m) \end{align}
• So $S(m)$ is true. Thus by the Principle of Mathematical, $S(m)$ is true for all $m ≥ 1$. We now only need to show that this factorization is unique.
• Suppose that $p(x)$ can be factorized in this form in two different ways, that is for $d, \tau_1, \tau_2, ..., \tau_m \in \mathbb{C}$ we have that:
(3)
\begin{align} \quad c(x - \lambda_1)(x - \lambda_2)...(x - \lambda_m) = d(x - \tau_1)(x - \tau_2)...(x - \tau_m) \end{align}
• We note that $c = d$ as when we expand both sides of these polynomials we get that $c$ and $d$ will both be the coefficient of the term $x^n$ and hence must be equal, therefore:
(4)
\begin{align} \quad (x - \lambda_1)(x - \lambda_2)...(x - \lambda_m) = (x - \tau_1)(x - \tau_2)...(x - \tau_m) \end{align}
• Now since $\lambda_1$ makes the lefthand side of the equation equal to zero, then it must also make the righthand side of this equation equal to zero, and so $\lambda_1$ is equal to one of the $\tau$'s. Without loss of generality, suppose that $\lambda_1 = \tau_1$. Thus the term $(x - \lambda_1)$ and $(x - \tau_1)$ are the same, and dividing both sides by this term yields:
(5)
\begin{align} \quad (x - \lambda_2)...(x - \lambda_m) = (x - \tau_2)...(x - \tau_m) \end{align}
• We can continue this process with pairing $\lambda_2, \lambda_3, ..., \lambda_m$ with $\tau_2, \tau_3, ..., \tau_m$ which shows that the factorization is unique. $\blacksquare$

## Example 1

Find the unique factorization of $p(x) = x^2 + x + 1$.

We should note that the polynomial above has strictly real coefficients, however Theorem 1 still holds since all real numbers are complex numbers.

This polynomial cannot be factored nicely, so we must use the quadratic formula to find these roots, which is $\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a$ is the coefficient on the $x^2$ term, $b$ is the coefficient on the $x$ term, and $c$ is the coefficient on the constant term. Therefore we have that:

(6)
\begin{align} \quad \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1}{2} \pm \frac{\sqrt{3}}{2} i \end{align}

Thus the roots are $\lambda_1 = \frac{-1}{2} + \frac{\sqrt{3}}{2} i$ and $\lambda_2 = \frac{-1}{2} - \frac{\sqrt{3}}{2} i$. Now we only need to find the value of $c \in \mathbb{C}$:

(7)
\begin{align} \quad p(x) = c \left (x - \left (\frac{-1}{2} + \frac{\sqrt{3}}{2}i \right ) \right ) \left (x - \left (\frac{-1}{2} - \frac{\sqrt{3}}{2}i \right ) \right ) \\ \quad p(x) = c \left (x + \frac{1}{2} - \frac{\sqrt{3}}{2}i \right ) \left (x + \frac{1}{2} + \frac{\sqrt{3}}{2}i \right ) \\ \quad \quad p(x) = c \left [x^2 + \frac{1}{2}x + \frac{\sqrt{3}i}{2}x + \frac{1}{2}x + \frac{1}{4} + \frac{\sqrt{3}i}{4} - \frac{\sqrt{3}i}{2}x - \frac{\sqrt{3}i}{4} + \frac{3}{4} \right] \\ \quad p(x) = c[x^2 + x + 1] \end{align}

Therefore $c = 1$, and so our unique factorization of $p(x)$ is $p(x) = \left (x - \left (\frac{-1}{2} + \frac{\sqrt{3}}{2}i \right ) \right) \left (x - \left (\frac{-1}{2} - \frac{\sqrt{3}}{2}i \right ) \right)$.

## Example 2

Let $p(x) \in \wp (\mathbb{C})$. Show that $p(x)$ has $m$ distinct roots if $p'(x)$ has no roots in common with $p(x)$.

Let $p$ be a polynomial with complex coefficients and suppose that $p(x)$ has $m$ distinct roots, $\lambda_1, \lambda_2, ..., \lambda_m$. By Theorem 1, we can factor $p(x)$ as:

(8)
\begin{align} \quad p(x) = c(x - \lambda_1)(x - \lambda_2) ...(x - \lambda_m) \end{align}

Now for any $\lambda_j$ where $j = 1, 2, ..., m$ we can write $p(x) = (x - \lambda_j) q(x)$. If we differentiate both sides of this equation and apply the product rule we get that:

(9)
\begin{align} \quad p'(x) = q(x) + (x - \lambda_j)q'(x) \end{align}

Therefore $p'(\lambda_j) = q(\lambda_j) \neq 0$. Hence $p(x)$ and $p'(x)$ have no roots in common.