3.5 Complex Zeros and the Fundamental Theorem of Algebra

3.5.1 Learning Objectives

By the end of this section, you should be able to:

  • apply the Fundamental Theorem of Algebra;
  • count polynomial zeros with multiplicity;
  • use the Conjugate Zeros Theorem;
  • find all real and complex zeros of a polynomial; and
  • construct and factor polynomials from specified zeros.

3.5.2 The Fundamental Theorem of Algebra

The Fundamental Theorem of Algebra states that every polynomial of positive degree with complex coefficients has at least one complex zero.

Together with the Factor Theorem, this means that a polynomial of degree \(n\) can be written as

\[ P(x)=a_n(x-c_1)(x-c_2)\cdots(x-c_n), \]

where the \(c_i\) are complex zeros, repeated according to multiplicity. Therefore, a degree-\(n\) polynomial has exactly \(n\) complex zeros when multiplicity is counted.

Example 3.22 Counting Missing Zeros

A degree-\(6\) polynomial has known zeros \(2\), \(-1\), and \(4i\), each with multiplicity \(1\). How many zeros remain to be found?

Solution

A degree-\(6\) polynomial has \(6\) complex zeros counted with multiplicity. Three have been listed, so

\[ \boxed{3\text{ zeros remain}}. \]

Exercise 3.22 A degree-\(9\) polynomial has five known zeros counted with multiplicity. How many remain?

Show answer

Answer: \(\boxed{4}\)

3.5.3 Multiplicity and Complete Factorization

If a zero \(c\) has multiplicity \(m\), then the factor \((x-c)\) occurs \(m\) times. A polynomial is completely factored over the complex numbers when it is written as a product of linear factors.

Example 3.23 Factoring Completely over the Complex Numbers

Factor \(P(x)=x^4-16\) completely over the complex numbers.

Solution

First factor over the real numbers:

\[ x^4-16=(x^2-4)(x^2+4). \]

Then

\[ x^2-4=(x-2)(x+2) \]

and

\[ x^2+4=(x-2i)(x+2i). \]

\[ \boxed{P(x)=(x-2)(x+2)(x-2i)(x+2i)} \]

Exercise 3.23 Factor \(x^4+5x^2+4\) completely over the complex numbers.

Show answer

Answer: \(x^4+5x^2+4=(x^2+1)(x^2+4)=\boxed{(x-i)(x+i)(x-2i)(x+2i)}\).

3.5.4 The Conjugate Zeros Theorem

If a polynomial has real coefficients and \(a+bi\) is a zero, then \(a-bi\) is also a zero. Non-real zeros of real-coefficient polynomials therefore occur in conjugate pairs.

The product of the corresponding factors is a quadratic with real coefficients:

\[ \begin{aligned} [x-(a+bi)][x-(a-bi)] &=(x-a)^2+b^2. \end{aligned} \]

Example 3.24 Using a Known Complex Zero

A polynomial with real coefficients has zeros \(3\) and \(1+2i\). Write a least-degree monic polynomial with these zeros.

Solution

The conjugate \(1-2i\) must also be a zero. Therefore,

\[ \begin{aligned} P(x)&=(x-3)[x-(1+2i)][x-(1-2i)]\\ &=(x-3)\bigl((x-1)^2+4\bigr)\\ &=(x-3)(x^2-2x+5). \end{aligned} \]

\[ \boxed{P(x)=(x-3)(x^2-2x+5)} \]

Exercise 3.24 A real-coefficient polynomial has zero \(-2+5i\). State another required zero.

Show answer

Answer: \(\boxed{-2-5i}\)

3.5.5 Finding All Zeros

To find all zeros:

  1. use rational-zero techniques or graphing to find a real zero;
  2. divide out its factor;
  3. repeat until a quadratic or another manageable factor remains;
  4. solve remaining factors, using the quadratic formula when needed; and
  5. check that the total multiplicity equals the polynomial’s degree.

Example 3.25 Finding Real and Complex Zeros

Find all zeros of

\[ P(x)=x^3-4x^2+9x-36. \]

Solution

Grouping gives

\[ \begin{aligned} P(x)&=x^2(x-4)+9(x-4)\\ &=(x-4)(x^2+9). \end{aligned} \]

The linear factor gives \(x=4\). The quadratic gives \(x^2=-9\), so \(x=\pm3i\).

\[ \boxed{x=4,\quad x=3i,\quad x=-3i} \]

Exercise 3.25 Find all zeros of \(P(x)=x^3+2x^2+4x+8\).

Show answer

Answer: \(P(x)=(x+2)(x^2+4)\), so \(\boxed{x=-2,\ 2i,\ -2i}\).

3.5.6 Constructing Polynomials from Zeros

Specified zeros determine factors. Include every required conjugate when the coefficients must be real, repeat factors according to multiplicity, and multiply by the desired leading coefficient.

Example 3.26 Constructing a Real Polynomial

Find a degree-\(4\) polynomial with real coefficients, leading coefficient \(2\), and zeros \(1\), \(-3\), and \(2i\).

Solution

Because the coefficients are real, \(-2i\) is also a zero:

\[ \begin{aligned} P(x)&=2(x-1)(x+3)(x-2i)(x+2i)\\ &=2(x-1)(x+3)(x^2+4). \end{aligned} \]

\[ \boxed{P(x)=2(x-1)(x+3)(x^2+4)} \]

Exercise 3.26 Write a least-degree monic polynomial with real coefficients and zeros \(4\) and \(-1+3i\).

Show answer

Answer: Include \(-1-3i\). One form is \(\boxed{P(x)=(x-4)(x^2+2x+10)}\).

3.5.7 Real Factorization

A polynomial with real coefficients can be factored over the real numbers into linear factors and irreducible quadratic factors. An irreducible quadratic has no real zeros, so its discriminant is negative.

Example 3.27 Factoring over the Real Numbers

Factor \(P(x)=x^4+3x^2-4\) completely over the real numbers.

Solution

Treat the expression as a quadratic in \(x^2\):

\[ x^4+3x^2-4=(x^2+4)(x^2-1). \]

Then

\[ x^2-1=(x-1)(x+1). \]

The factor \(x^2+4\) is irreducible over the real numbers.

\[ \boxed{P(x)=(x-1)(x+1)(x^2+4)} \]

Exercise 3.27 Factor \(x^4-2x^2-3\) completely over the real numbers.

Show answer

Answer: \(\boxed{(x^2-3)(x^2+1)=(x-\sqrt3)(x+\sqrt3)(x^2+1)}\)

3.5.8 Conceptual Takeaways

  • A degree-\(n\) polynomial has exactly \(n\) complex zeros when multiplicity is counted.
  • Complete factorization over the complex numbers uses only linear factors.
  • Non-real zeros of real-coefficient polynomials occur in conjugate pairs.
  • A complete solution should account for every zero predicted by the degree.
  • Over the real numbers, irreducible quadratic factors represent conjugate non-real zeros.

3.5.9 Skills You Should Be Able to Do

  • Count zeros using degree and multiplicity.
  • Apply the Fundamental Theorem of Algebra.
  • Use conjugate pairs to identify missing zeros.
  • Find all real and complex zeros of a polynomial.
  • Construct and factor polynomials over the real and complex numbers.

3.5.10 Practice Problems with Solutions

  1. How many complex zeros does a degree-\(12\) polynomial have, counting multiplicity?

    Show Solution

    By the Fundamental Theorem of Algebra,

    \[ \boxed{12}. \]

  2. A real-coefficient polynomial has zero \(5-2i\). State another zero.

    Show Solution

    The conjugate must also be a zero:

    \[ \boxed{5+2i}. \]

  3. Factor \(x^3-9x\) completely over the complex numbers.

    Show Solution

    \[ x^3-9x=x(x^2-9)=x(x-3)(x+3). \]

    All factors are already linear, so

    \[ \boxed{x(x-3)(x+3)}. \]

  4. Factor \(x^4+4x^2\) completely over the complex numbers.

    Show Solution

    \[ x^4+4x^2=x^2(x^2+4)=x^2(x-2i)(x+2i). \]

    Therefore, \(\boxed{x^2(x-2i)(x+2i)}\).

  5. Find all zeros of \(P(x)=x^3-2x^2+9x-18\).

    Show Solution

    Grouping gives

    \[ P(x)=x^2(x-2)+9(x-2)=(x-2)(x^2+9). \]

    Thus, \(\boxed{x=2,\ 3i,\ -3i}\).

  6. Find all zeros of \(P(x)=x^4-5x^2+6\).

    Show Solution

    Let \(u=x^2\). Then

    \[ u^2-5u+6=(u-2)(u-3). \]

    Thus, \(x^2=2\) or \(x^2=3\), so

    \[ \boxed{x=\pm\sqrt2,\quad x=\pm\sqrt3}. \]

  7. Write a monic polynomial with real coefficients and zeros \(-2\), \(3i\), and \(-3i\).

    Show Solution

    \[ P(x)=(x+2)(x-3i)(x+3i)=(x+2)(x^2+9). \]

    Therefore,

    \[ \boxed{P(x)=x^3+2x^2+9x+18}. \]

  8. A degree-\(5\) polynomial with real coefficients has zeros \(1\) with multiplicity \(2\), and \(-2+i\). If \(3\) is another zero, list all five zeros.

    Show Solution

    The zero \(1\) counts twice. Since \(-2+i\) is non-real and the coefficients are real, \(-2-i\) must also be a zero. Including \(3\), the five zeros counted with multiplicity are

    \[ \boxed{1,\ 1,\ -2+i,\ -2-i,\ 3}. \]