AOS3 Topic 7: Solutions to Complex Polynomials

Complex Polynomials

Complex polynomials are polynomials in which the coefficients and/or the variables can take complex values. A complex polynomial is an algebraic expression consisting of terms involving complex numbers. These polynomials can have both real and imaginary parts.

For example, a complex polynomial might look like this:

\[ P(z) = a_n z^n + a_{n-1} z^{n-1} + \ldots + a_1 z + a_0 \]

where \( z \) is a complex variable, and \( a_n, a_{n-1}, \ldots, a_1, a_0 \) are complex coefficients.

Polynomial Equations

Polynomial equations are equations in which a polynomial expression is set equal to a constant or to another polynomial expression. A polynomial equation typically involves variables raised to non-negative integer powers and can have one or more solutions.

For example, \(2x^2 + 3x - 5 = 0\) and \(x^3 - 4x^2 + x + 6 = 0\) are polynomial equations. In these equations, the variables \(x\) are raised to non-negative integer powers (2, 1, and 0 in the first equation; 3, 2, 1, and 0 in the second equation), and the coefficients are constants.

Completing the Square:

For a quadratic equation \( ax^2 + bx + c = 0 \), you can complete the square by following these steps:

  1. Move the constant term \( c \) to the other side of the equation.
  2. Divide all terms by the coefficient of \( x^2 \) if it's not already 1.
  3. Take half of the coefficient of \( x \), square it, and add it to both sides of the equation.
  4. Factor the left side of the equation as a perfect square trinomial.
  5. Solve for \( x \) using square root.

    6. Factoring Quadratic Expressions

    There are several methods for factoring quadratic expressions (ax2 + bx + c) where a, b, and c are constants and a ≠ 0. Here are three common approaches:

    1. Factoring by Grouping (Rearranging):

      This method is applicable when you can rewrite the middle term (bx) by grouping the terms and finding two values that add up to b and multiply to c.

    2. Using the Quadratic Formula (Direct Formula):

      This method provides a direct formula to find the roots (solutions) of the quadratic equation, and the factors can be obtained from the roots.

      Formula:

      x = (-b ± √(b2 - 4ac)) / (2a)

      where a, b, and c are the coefficients of the quadratic expression (ax2 + bx + c).

      Conjugate Root Theorem

      Let P(z) be a polynomial with real coefficients. If a + bi is a solution of the equation P(z) = 0, with a and b real numbers, then the complex conjugate a − bi is also a solution.

Example 1

By completing square
Use completing the square formula to solve the quadratic equation: x^2 - 4x - 8 = 0.

Solution
Given the quadratic equation: \(x^2 - 4x - 8 = 0\)
To solve this equation using completing the square method, follow these steps:
  • Move the constant term to the other side of the equation: \(x^2 - 4x = 8\)
  • Take half of the coefficient of x, square it, and add it to both sides: \(x^2 - 4x + (-4/2)^2 = 8 + (-4/2)^2\)
  • Factor the left side of the equation as a perfect square trinomial: \((x - 2)^2 = 8 + 4\)
  • Solve for x using square root: \(x - 2) = \pm\sqrt(12)\)
    • Therefore, the solutions for the equation are:
    x = 2 ± \sqrt{12}

    Method 2:
    Let's transpose the constant term to the other side of the equation: \(x^2 - 4x = 8\). Take half of the coefficient of the \(x\)-term, which is -4, including the sign, which gives -2. Square -2 to get +4, and add this squared value to both sides of the equation:
    \[ x^2 - 4x + 4 = 8 + 4 \]
    ⇒ \(x^2 - 4x + 4 = 12\)
    This process creates a quadratic expression that is a perfect square on the left-hand side of the equation. Simply, we can replace the quadratic with the squared-binomial form: \((x - 2)^2 = 12\)
    Now, we've completed the expression to create a perfect-square binomial, let's solve:
    \((x - 2)^2 = 12\)
    ⇒ \(x - 2 = \pm \sqrt{12}\)
    ⇒ \(x - 2 = \pm 2\sqrt{3}\)
    ⇒ \(x = 2 \pm 2\sqrt{3}\)

    Example 2

    Factoring by Grouping (Rearranging):
    Factor the quadratic expression \( x^2 + 7x + 10 \).

    Solution: We can rewrite the expression by grouping:

    \[ x^2 + 7x + 10 = (x^2 + 2x) + (5x + 10) = x(x + 2) + 5(x + 2) = (x + 2)(x + 5) \]

    Therefore, the factored form is \( (x + 2)(x + 5) \).

    Example 3

    Factoring Polynomial by Grouping
    Factor the polynomial \(3x^3 + 5x^2 - 6x - 10\).

    Solution: We can group the terms in pairs:
    \[ (3x^3 + 5x^2) + (-6x - 10) \]
    Now, factor out the greatest common factor from each pair:
    \[ x^2(3x + 5) - 2(3x + 5) \]
    Notice that both terms have a common factor of \(3x + 5\). Factor out this common factor:
    \[ (3x + 5)(x^2 - 2) \]
    Therefore, the factored form is \((3x + 5)(x^2 - 2)\).

    Example 4

    Complex polynomial by completing square.
    Solve the following complex polynomial by completing the square:
    \(2z^2 − 2(3 − i)z + 4 − 3i\)

    Solution:
    Let \( P(z) = 2z^2 − 2(3 − i)z + 4 − 3i \). Then
    \( P(z) = 2z^2 − (3 − i)z + \frac{4 − 3i}{2} \)
    \( = 2(z^2 − (3 − i)z + \frac{(3 − i)^2}{2}) + \frac{4 − 3i}{2} - \frac{(3 − i)^2}{2} \)
    \( = 2(z - \frac{3 − i}{2})^2 + 4 − 3i - \frac{(3 − i)^2}{2} \)
    \( = 2(z - \frac{3 − i}{2})^2 + 8 − 6i − 9 + 6i + 1 \)
    \( = 2(z - \frac{3 − i}{2})^2 \)

    Example 5

    Using quadratic formula
    solve this complex polynomial by using quadratic formula:\(z^2 = 2z-5 \).

    Solution
    Rearrange the equation into the form \( z^2 - 2z + 5 = 0 \).
    Now apply the quadratic formula:
    \( z = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} = \frac{{2 \pm \sqrt{{-16}}}}{2} = \frac{{2 \pm 4i}}{2} = 1 \pm 2i \)
    The solutions are \( 1 + 2i \) and \( 1 - 2i \).

    Example 6

    Using Conjugate root theorem
    Given \(2i\) is one of the roots of \(f(x) = x^3 - 3x^2 + 4x - 12\), find its remaining roots and write \(f(x)\) in root factored form.

    Solution:
    Given \( 2i \) is one of the roots of \( f(x) = x^3 - 3x^2 + 4x - 12 \), so is \( -2i \). We find its remaining roots are: { \( -2i, 2i, 3 \) }
    In root-factored form we therefore have: \( f(x) = (x - 3)(x - 2i)(x + 2i) \)

    Exercise 1

    Exercise 2

    Exercise 3

    Exercise 4

    Back: AOS3 Topic 5: Complex Polynomials Next: AOS4 Topic 1: Differentiation