# The Factorization of Polynomials with Real Coefficients

We are about to look at an important way to factor polynomials with real coefficients, but before we do, we must first look at the following proposition.

Proposition 1: A quadratic polynomial with real coefficients $\alpha, \beta \in \mathbb{R}$ such that $p(x) = x^2 + \alpha x + \beta$ can be factored as $p(x) = (x – \lambda_1)(x – \lambda_2)$ where $\lambda_1, \lambda_2 \in \mathbb{R}$ if and only if $\alpha^2 ≥ 4 \beta$. |

**Proof:**$\Rightarrow$ Suppose that $p(x) = (x – \lambda_1)(x – \lambda_2)$. Then when we expand $p(x)$ we get that:

(1)

\begin{align} \quad p(x) = x^2 – \lambda_2 x – \lambda_1 x + \lambda_1 \lambda_2 \\ \quad p(x) = x^2 + (-\lambda_1 – \lambda_2)x + \lambda_1 \lambda_2 \end{align}

- Therefore $\alpha = -\lambda_1 – \lambda_2$ and $\beta = \lambda_1 \lambda_2$. Now notice that $(-\lambda_1 – \lambda_2)^2 = \lambda_1^2 + 2 \lambda_1 \lambda_2 + \lambda_2^2 ≥ 0$ and so $\lambda_1^2 + \lambda_2^2 ≥ 2 \lambda_1 \lambda_2$, thus:

(2)

\begin{align} \quad \alpha^2 = (-\lambda_1 – \lambda_2)(-\lambda_1 – \lambda_2) = \lambda_1^2 + \lambda_2^2 + 2 \lambda_1 \lambda_2 ≥ 2 \lambda_1 \lambda_2 + 2 \lambda_1 \lambda_2 = 4 \lambda_1 \lambda_2 = 4 \beta \end{align}

- $\Leftarrow$ Now suppose that $\alpha^2 ≥ 4 \beta$. Then there exists a real number $c \in \mathbb{R}$ such that $c^2 = \frac{\alpha^2}{4} – \beta$. Notice that we can rewrite the polynomial $x^2 + \alpha x + \beta$ as:

(3)

\begin{align} \quad x^2 + \alpha x + \beta = \left ( x + \frac{\alpha}{2} \right )^2 – \left ( \frac{\alpha^2}{4} – \beta \right ) = \left ( x + \frac{\alpha}{2} \right )^2 – c^2 = \left ( \left ( x + \frac{\alpha}{2} \right) + c \right ) \left ( \left ( x + \frac{\alpha}{2} \right ) – c \right ) \\ = \left ( x – \left ( – \frac{\alpha}{2} – c \right ) \right) \left ( x – \left ( – \frac{\alpha}{2} + c \right ) \right ) \end{align}

- Thus we have the desired factorization with $\lambda_1 = – \frac{\alpha}{2} – c$ and $\lambda_2 = – \frac{\alpha}{2} + c$. $\blacksquare$

We are now ready to look at the following theorem which gives us a second way to fac

Theorem 1: Let $p(x) \in \wp ( \mathbb{R} )$ be a non-constant polynomial. Then $p(x)$ has a unique factorization $p(x) = c(x – \lambda_1)(x – \lambda_2)…(x – \lambda_m)(x^2 + \alpha_1x + \beta_1)(x^2 + \alpha_2x + \beta_2)…(x^2 + \alpha_nx + \beta_n)$ where $\lambda_1, \lambda_2, …, \lambda_m \in \mathbb{R}$ are the real roots of $p(x)$ and $c, \alpha_1, \alpha_2, …, \alpha_n, \beta_1, \beta_2, …, \beta_n \in \mathbb{R}$. |

**Proof:**Let $p(x) \in \wp ( \mathbb{R} )$ be a non-constant polynomial with $\mathrm{deg} (p) = r$. Since $p(x)$ is a polynomial of real coefficients, it is also a polynomial of complex coefficients since all real numbers are complex numbers, and from The Factorization of Polynomials with Complex Coefficients page we have that for $\tau_1, \tau_2, …, \tau_r \in \mathbb{C}$ (which are the roots of $p(x)$), $p(x)$ can be uniquely factored as:

(4)

\begin{align} \quad p(x) = (x – \tau_1)(x – \tau_2)…(x – \tau_r) \end{align}

- Now if some root $\tau_i$ where $1 ≤ i ≤ r$ is a complex number whose imaginary part is nonzero (that is $\tau_i$ is a non-real complex root), then we saw from the Pairs of Complex Roots for Polynomials with Real Coefficients page that the complex conjugate $\bar{\tau_i}$ is also a non-real complex root of $p(x)$ and so $\bar{\tau_i} = \tau_j$ for $i \neq j$ and $1 ≤ j ≤ r$. The product of $(x – \tau_i)(x – \tau_j) = (x – \tau_i)(x – \bar{\tau_i})$ will yield one of our quadratic factors, $x^2 + \alpha_k x + \beta_k$ where $1 ≤ k ≤ n$.

- To guarantee that such a factorization $c(x – \lambda_1)(x – \lambda_2)…(x – \lambda_m)(x^2 + \alpha_1x + \beta_1)(x^2 + \alpha_2x + \beta_2)…(x^2 + \alpha_nx + \beta_n)$ exists, we must show that for all non-real $\tau$ that $(x – \tau)$ and $(x – \bar{\tau})$ appear the same number of times in such a factorization so that we can group them to make our quadratic factors. We note that for $q(x) \in \wp ( \mathbb{C} )$ where $\mathrm{deg} (q) = r – 2$ we can write $p(x)$ as follows:

(5)

\begin{align} p(x) = (x – \tau)(x – \bar{\tau}) q(x) = (x^2 – 2 \Re (\tau) x + \mid \tau \mid^2 ) q(x) \end{align}

- If we can show that $q(x)$ has real coefficients then this will imply that $(x – \tau)$ appears as many times as $(x – \bar{\tau})$. Note that $(x^2 – 2 \Re (\tau) x + \mid \tau \mid^2 ) \neq 0$, and so:

(6)

\begin{align} \quad q(x) = \frac{p(x)}{x^2 – 2 \Re (\tau) x + \mid \tau \mid^2} \end{align}

- Now since $p(x) \in \wp (\mathbb{R})$ and $x^2 – 2 \Re (\tau) x + \mid \tau \mid^2 \in \mathbb{R}$ then for all $x \in \mathbb{R}$ we have that $q(x) \in \mathbb{R}$ and so $q(x)$ is a polynomial whose range contains only real numbers. Let’s write the polynomial $q(x)$ as:

(7)

\begin{equation} q(x) = a_0 + a_1x + … + a_{r-2} x^{r-2} \end{equation}

- If we take the imaginary parts of both sides of this equation (by using the properties of the imaginary part of numbers), then we have that:

(8)

\begin{align} \quad \Im (q(x)) = \Im (a_0 + a_1x + … + a_{r-2} x^{r-2}) \\ \quad 0 = \Im (a_0) + \Im (a_1)x + … + \Im (a_{r-2}) x^{r-2} \end{align}

- But this implies that $\Im (a_0) = \Im (a_1) = … = \Im (a_{r-2}) = 0$ otherwise the lefthand side of the above equation would not be equal to $0$. Thus $a_0, a_1, …, a_{r-2} \in \mathbb{R}$. Therefore, such a factorization exists.

- Now we only need to show that this factorization is unique. This is easy to show. Since $p(x)$ is a polynomial of real coefficients, then $p(x)$ is also a polynomial of complex coefficients and so two different factorizations would lead to a contradiction of the theorem we proved on The Factorization of Polynomials with Complex Coefficients page. $\blacksquare$

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