6.3 Roots of Polynomials and the Fundamental Theorem of Algebra

Fundamental Theorem of Algebra

more resources to help you study

practice questions

Key Concepts

The Fundamental Theorem of Algebra states that every non-constant single-variable polynomial with complex coefficients has at least one complex root. From this, a powerful consequence follows: a polynomial of degree $n$ has exactly $n$ complex roots when you count each root according to its multiplicity.

This means any polynomial can be factored completely into linear factors over the complex numbers. For example, a degree-4 polynomial will always break down into four linear factors (though some may repeat, and some may involve complex numbers).

One crucial rule for polynomials with real coefficients: complex roots always come in conjugate pairs. If $2 + 3i$ is a root, then $2 - 3i$ must also be a root. This is why real-coefficient polynomials of odd degree are guaranteed to have at least one real root.

Implications and Limitations

• The theorem guarantees roots exist but gives you no method for actually finding them. You still need techniques like the Rational Root Theorem or synthetic division.
• "Complex coefficients" includes real numbers as a subset, so the theorem covers every polynomial you'll encounter in this course.
• Multiplicity refers to how many times a root appears as a factor. For instance, $(x - 2)^3$ has a root of $2$ with multiplicity 3, which counts as three of the polynomial's roots.
• The theorem does not guarantee any roots will be real. A polynomial like $x^2 + 1$ has two complex roots ($i$ and $-i$) and no real roots at all.

Complex Zeros of Polynomials

Finding Complex Zeros

The Factor Theorem is your starting point: a polynomial $f(x)$ has a factor $(x - c)$ if and only if $f(c) = 0$. So finding zeros and finding factors are the same problem.

Here's a typical process for finding all zeros of a polynomial:

1. Use the Rational Root Theorem (covered below) to generate a list of possible rational zeros.
2. Test candidates using synthetic division or direct substitution.
3. When you find a zero, factor it out. This reduces the polynomial's degree by one.
4. Repeat on the reduced polynomial until you reach a quadratic.
5. Solve the remaining quadratic using factoring or the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

If the discriminant ($b^2 - 4ac$) is negative, the quadratic has two complex conjugate roots. That's completely normal and expected.

Higher-Degree Polynomials

For cubics and quartics, more advanced methods exist (Cardano's formula for cubics, Ferrari's method for quartics), but these are rarely used by hand in this course. What matters is the general strategy: reduce the degree step by step using known roots until you reach a quadratic you can solve with the quadratic formula.

Coefficients and Roots

Vieta's Formulas

Vieta's formulas create a direct link between a polynomial's coefficients and its roots, without requiring you to solve the polynomial. They're especially useful when a problem gives you information about the roots and asks for a coefficient (or vice versa).

For a quadratic $ax^2 + bx + c$ with roots $r$ and $s$:

• Sum of roots: $r + s = -\frac{b}{a}$
• Product of roots: $r \cdot s = \frac{c}{a}$

For a cubic $ax^3 + bx^2 + cx + d$ with roots $r$, $s$, and $t$:

• $r + s + t = -\frac{b}{a}$
• $rs + rt + st = \frac{c}{a}$
• $r \cdot s \cdot t = -\frac{d}{a}$

The General Pattern

Notice the pattern that extends to any degree:

• The sum of the roots always equals the negative of the second-highest degree coefficient divided by the leading coefficient.
• The product of all roots equals the constant term divided by the leading coefficient, with a sign that depends on the degree: positive for even degree, negative for odd degree.

For example, if you know a cubic with leading coefficient 1 has roots $2$, $-1$, and $3$, you can immediately find: sum = $2 + (-1) + 3 = 4$, so the $x^2$ coefficient is $-4$. Product = $2 \cdot (-1) \cdot 3 = -6$, so the constant term is $-(-6) = 6$.

Rational Root Theorem

Applying the Theorem

The Rational Root Theorem narrows down which rational numbers could be zeros of a polynomial with integer coefficients. It states:

Example: For $2x^3 - 5x^2 - 11x + 6$:

1. Factors of the constant term (6): $\pm 1, \pm 2, \pm 3, \pm 6$
2. Factors of the leading coefficient (2): $\pm 1, \pm 2$
3. All possible rational roots ($\frac{p}{q}$): $\pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2}$

Note that the original list in step 3 should include fractions. The candidates $\pm \frac{1}{2}$ and $\pm \frac{3}{2}$ come from dividing factors of 6 by 2.

Testing and Factoring

Once you have your list of candidates, test them systematically:

1. Plug a candidate into the polynomial (or use synthetic division). If the result is 0, you've found a root.

2. Use the Factor Theorem to factor out $(x - c)$ where $c$ is the confirmed root.

3. Work with the reduced (lower-degree) polynomial and repeat.

Continuing the example: Testing $x = 3$ in $2x^3 - 5x^2 - 11x + 6$:

Using synthetic division with 3:

• $2x^3 - 5x^2 - 11x + 6 = (x - 3)(2x^2 + x - 2)$

Now solve $2x^2 + x - 2 = 0$ with the quadratic formula to find the remaining roots. If the quadratic doesn't factor neatly, the remaining roots may be irrational or complex, and that's fine. The Rational Root Theorem only finds the rational ones.