Complete Polynomial Factorization

Factoring into Linear Factors over the Complex Numbers

We have learned to factor polynomials, for example using the Factor Theorem. However, sometimes the factorization result still leaves factors that are not linear (like quadratic factors) which cannot be factored further using real numbers.

Complete Factorization (or Complete Linear Factorization) is the process of factoring a polynomial into a product of linear factors, where these factors may involve complex numbers.

This concept is based on the Fundamental Theorem of Algebra: every polynomial of degree $n\ge1$ has exactly $n$ complex roots when multiplicities are counted. Real roots are included among those complex roots.

Complete Factorization Property of Polynomials

If $P(x)$ is a polynomial of degree $n \ge 1$ with leading coefficient $a_n \neq 0$, then there exist complex numbers $c_1, c_2, \dots, c_n$ (which are the roots of $P(x)$) such that:

This means that every polynomial of degree $n$ can be broken down into exactly $n$ linear factors $(x - \text{root})$ multiplied by its leading coefficient.

Steps for Complete Factorization

To perform a complete factorization of a polynomial $P(x)$:

1. Find All Complex Roots: Find all $n$ complex roots (zeros) of $P(x) = 0$. This might involve:

   • Factoring directly (grouping, etc.).
   • Using the Rational Zero Theorem to find rational roots.
   • Using division (Horner/long division) to reduce the degree of the polynomial after a root is found.
   • Solving quadratic equations (using the quadratic formula) which might yield complex roots $a \pm bi$.

2. Apply the Factor Theorem: For each root $c_i$ found, form its linear factor, which is $(x - c_i)$.

3. Write the Complete Factorization: Multiply all the obtained linear factors by the leading coefficient $a_n$ of $P(x)$.

   $$\begin{aligned} &P(x)= \\ &a_n(x - c_1)(x - c_2)\dots(x - c_n) \end{aligned}$$

Using Complete Factorization

Find all complex zeros of $P(x) = x^3 - x^2 + x - 1$ and factor the polynomial completely.

Solution:

1. Find Roots: Try to factor $P(x)$ first.

   • Factor by grouping:

     $$P(x) = (x^3 - x^2) + (x - 1)$$

     $$P(x) = x^2(x - 1) + 1(x - 1)$$

     $$P(x) = (x^2 + 1)(x - 1)$$

   • Now, find the roots by setting $P(x) = 0$:

     $$(x^2 + 1)(x - 1) = 0$$

   • This gives two possibilities:

     $$x - 1 = 0 \implies x = 1$$

     $$\begin{aligned} x^2 + 1 &= 0 \\ &\implies x^2 \\ &= -1 \\ &\implies x \\ &= \pm\sqrt{-1} \\ &\implies x \\ &= \pm i \end{aligned}$$

   So, the complex roots are $1, i, -i$.

2. Form Linear Factors:

   • From the root $1$, the factor is $(x - 1)$.
   • From the root $i$, the factor is $(x - i)$.
   • From the root $-i$, the factor is $(x - (-i)) = (x + i)$.

3. Write Complete Factorization:

   The leading coefficient of $P(x)$ is $a_3 = 1$.

   $$P(x) = 1 \cdot (x - 1)(x - i)(x + i)$$

   $$P(x) = (x - 1)(x - i)(x + i)$$

Exercise

Find all complex zeros of $P(x) = x^3 - 3x^2 + x + 5$, then factor $P(x)$ completely. Start from the possible rational roots, test them with substitution, and divide out each factor you find until only a quadratic remains.

Answer Key

1. Find Rational Roots (Rational Zero Theorem):

   • $a_0 = 5$, factors $p$: $\pm 1, \pm 5$.

   • $a_n = 1$, factors $q$: $\pm 1$.

   • Possible roots $p/q$: $\pm 1, \pm 5$.

   • Test $x = -1$:

     $$\begin{aligned} P(-1) &= (-1)^3-3(-1)^2 \\ &\quad{}+(-1)+5 \\ &= -1-3(1)-1+5 \\ &= -1-3-1+5 \\ &= 0 \end{aligned}$$

     So, $x = -1$ is a root, and $(x+1)$ is a factor.

2. Divide using Horner's Method ($c = -1$):

   $$\begin{array}{c|cccc} -1 & 1 & -3 & 1 & 5 \\ & & -1 & 4 & -5 \\ \hline & 1 & -4 & 5 & \boxed{0} \\ \end{array}$$

   Quotient $H(x) = x^2 - 4x + 5$.

   $P(x) = (x+1)(x^2 - 4x + 5)$.

3. Find Roots of the Quotient: Solve $x^2 - 4x + 5 = 0$ using the quadratic formula.

   $$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(5)}}{2(1)}$$

   $$x = \frac{4 \pm \sqrt{16 - 20}}{2}$$

   $$x = \frac{4 \pm \sqrt{-4}}{2}$$

   $$x = \frac{4 \pm 2i}{2}$$

   $$x = 2 \pm i$$

   The other roots are $2 + i$ and $2 - i$.

4. All Complex Roots: The roots are $-1, 2+i, 2-i$.

5. Complete Factorization ($a_n = 1$):

   $$\begin{aligned} &P(x) \\ &=1\cdot(x-(-1)) \\ &\quad{}\cdot(x-(2+i))(x-(2-i)) \end{aligned}$$

   $$\begin{aligned} &P(x) \\ &= (x + 1)(x - 2 - i)(x - 2 + i) \end{aligned}$$

Non Real Roots of Real Polynomials

Can a polynomial with real coefficients have exactly one non-real complex zero? No.

If $a+bi$ with $b\neq0$ is a root of a polynomial with real coefficients, then $a-bi$ is also a root. The two roots form a conjugate pair, so such a polynomial cannot have exactly one non-real root.