Polynomials - Complex Conjugate Root Theorem

(finding zeros - solving polynomial equations)


In this section we learn the complex conjugate root theorem for polynomials. This will allow us to find the zero(s) of a polynomial function in pairs, so long as the zeros are complex numbers.

Complex Conjugate Root Theorem

Given a polynomial functions: \[f(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_2x^2 + a_1x + a_0\] if it has a complex root (a zero that is a complex number), \(z\): \[f(z) = 0\] then its complex conjugate, \(z^*\), is also a root: \[f(z^*) = 0\]

What this means

The complex conjugate root theorem tells us that complex roots are always found in pairs. In other words if we find, or are given, one complex root, then we can state that its complex conjugate is also a root.

Tutorial 1

In the following tutorial we further explain the complex conjugate root theorem. We also work through an exercise, in which we use it. Indeed we look at the polynomial: \[f(x) = x^3 - 5x^2 + 17x- 13\] and are told \(2+3i\) is one of its roots. We then need to find all of its remaining roots and write this polynomial in its root-factored form.


Exercise 1

Using the complex conjugate root theorem, find all of the remaining zeros (the roots) of each of the following polynomial functions and write each polynomial in root factored form:

  1. 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.

  2. Given \(1-i\) is one of the zeros of \(f(x) = x^3 - 2x+4\), find its remaining roots and write \(f(x)\) in root factored form.

  3. Given \(2+3i\) is a root of \(f(x) = -2x^3 + 10x^2 -34x+26\), find the remaining roots and write \(f(x)\) in root factored form.

  4. Given \(3i\) is a root of \(f(x) = x^4 - 2x^3 + 6x^2 - 18x - 27\), find its remaining roots and write \(f(x)\) in its root-factord form.

  5. Given \(2- i \) is a root of \(f(x) = 2x^4 - 14x^3 + 38x^2 - 46x +20\), find this polynomial function's remaining roots and write \(f(x)\) in its root-factored form.

  6. Given \(i\) is a root of \(f(x) = x^5 + 2x^4 - 4x^3 - 4x^2 - 5x - 6\), find this polynomial function's remaining roots and write \(f(x)\) in its root-factored form.

Note: this exercise can be downloaded as a worksheet to practice with: Worksheet 1

Solution Without Working

  1. 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: \[\left \{ -2i,\ 2i, \ 3 \right \}\] in root-factored form we therefore have: \[f(x) = \begin{pmatrix}x - 3 \end{pmatrix}.\begin{pmatrix}x - 2i \end{pmatrix}.\begin{pmatrix}x + 2i \end{pmatrix}\]

  2. Given \(1-i\) is one of the zeros of \(f(x) = x^3 - 2x+4\), so is \(1+i\). We find its remaining roots are: \[\left \{ 1- i,\ 1+ i, \ -2 \right \}\] in root-factored form we therefore have: \[f(x) = \begin{pmatrix}x + 2 \end{pmatrix}.\begin{pmatrix}x - (1-i) \end{pmatrix}.\begin{pmatrix}x - (1+i) \end{pmatrix} \]

  3. Given \(2+3i\) is a root of \(f(x) = -2x^3 + 10x^2 -34x+26\), so is \(2 - 3i\). We find its remaining roots are: \[\left \{ 2 - 3i,\ 2+ 3i, \ 1 \right \}\] in root-factored form we therefore have: \[f(x) = -2.\begin{pmatrix}x - 1 \end{pmatrix}.\begin{pmatrix}x - (2 + 3i) \end{pmatrix}.\begin{pmatrix}x - (2 - 3i) \end{pmatrix} \]

  4. Given \(3i\) is a root of \(f(x) = x^4 - 2x^3 + 6x^2 - 18x - 27\), so is \(-3i\). We find the remaining roots are: \[\left \{ - 3i,\ 3i, \ -1, \ 3 \right \}\] in root-factored form we therefore have: \[f(x) = \begin{pmatrix}x - 3 \end{pmatrix}.\begin{pmatrix}x + 1 \end{pmatrix}.\begin{pmatrix}x - 3i \end{pmatrix}.\begin{pmatrix}x + 3i \end{pmatrix} \]

  5. Given \(2- i \) is a root of \(f(x) = 2x^4 - 14x^3 + 38x^2 - 46x +20\), so is \(2 + i\). We find the remaining roots are: \[\left \{ 2 - i,\ 2 + i, \ 1, \ 2 \right \}\] in root-factored form we therefore have: \[f(x) = 2.\begin{pmatrix}x - 2 \end{pmatrix}.\begin{pmatrix}x - 1 \end{pmatrix}.\begin{pmatrix}x - (2 - i) \end{pmatrix}.\begin{pmatrix}x - (2 + i) \end{pmatrix} \]
  6. Given \(i\) is a root of \(f(x) = x^5 + 2x^4 - 4x^3 - 4x^2 - 5x - 6\), so is \(-i\). We find the remaining roots are: \[\left \{ - i,\ i,\ -3, \ - 1, \ 2 \right \}\] in root-factored form we therefore have: \[f(x) = \begin{pmatrix}x - 2 \end{pmatrix}.\begin{pmatrix}x + 3 \end{pmatrix}.\begin{pmatrix}x + 1 \end{pmatrix}.\begin{pmatrix}x - i \end{pmatrix}.\begin{pmatrix}x + i \end{pmatrix} \]

Exercise 2

Using the method shown in the tutorial above, answer each of the questions below.

  1. \(z_1 = 3\) and \(z_2 = 1+2i\) are roots of the equation: \[x^3 + bx^2 + cx + d = 0\]
    1. What is \(z_3\)?
    2. Find \(b\), \(c\) and \(d\).

  2. \(z_1 = -2\) and \(z_2 = 3 + i\) are roots of the equation: \[2x^3 + bx^2 + cx + d = 0\]
    1. What is \(z_3\)?
    2. Find \(b\), \(c\) and \(d\).

  3. \(z_1 = 2i\) and \(z_2 = 3+i\) are both roots of the equation: \[x^4 + bx^3 + cx^2 + dx + e = 0\]
    1. What are the other roots?
    2. Find \(b\), \(c\), \(d\) and \(e\)

  4. \(z_1 = 1+\sqrt{2}i\) and \(z_2 = 2-3i\) are roots of the equation: \[-2x^4 + bx^3 + cx^2 + dx + e = 0 \]
    1. What are this equation's remaining roots?
    2. Find \(b\), \(c\), \(d\) and \(e\).

  5. \(z_1 = 3\), \(z_2 = i\) and \(z_3 = 2-3i\) are roots of the equation: \[x^5 +bx^4 + cx^3 + dx^2 + ex + f = 0 \]
    1. What are this equation's remaining roots?
    2. Find \(b\), \(c\), \(d\), \(e\) and \(f\).

Note: this exercise can be downloaded as a worksheet to practice with: Worksheet 2

Solution Without Working

  1. Using the fact that \(z_1 = 3\) and \(z_2 = 1+2i\) are roots of the equation \(x^3 + bx^2 + cx + d = 0\), we find the following:
    1. \(z_3 = 1 - 2i\)
    2. Using the fact that: \[x^3 + bx^2 + cx + d = \begin{pmatrix}x - 3 \end{pmatrix}.\begin{pmatrix}x - (1+2i)\end{pmatrix}.\begin{pmatrix}x - (1-2i)\end{pmatrix}\] expanding the right hand side, simplifying as much as possible, and equating the coefficients to those on the left hand side we find: \[b = -5, \ c = 11, \ d = -15\]

  2. Using the fact that \(z_1 = -2\) and \(z_2 = 3 + i\) are roots of the equation \(2x^3 + bx^2 + cx + d = 0\), we find:
    1. \(z_3 = 3 - i\)
    2. Using the fact that: \[2x^3 + bx^2 + cx + d = 2.\begin{pmatrix}x + 2 \end{pmatrix}.\begin{pmatrix}x - (3 + i)\end{pmatrix}.\begin{pmatrix}x - (3 - i)\end{pmatrix}\] expanding the right hand side, simplifying as much as possible, and equating the coefficients to those on the left hand side we find: \[b = -8, \ c = -4, \ d = 40\]

  3. Using the fact that \(z_1 = 2i\) and \(z_2 = 3+i\) are both roots of the equation \(x^4 + bx^3 + cx^2 + dx + e = 0\), we find:
    1. The other roots are: \(z_3 = -2i\) and \(z_4 = 3 - i\).
    2. Using the fact that: \[x^4 + bx^3 + cx^2 + dx + e = \begin{pmatrix}x - 2i \end{pmatrix}.\begin{pmatrix}x + 2i\end{pmatrix}.\begin{pmatrix}x - (3 + i)\end{pmatrix}.\begin{pmatrix}x - (3 - i)\end{pmatrix}\] expanding the right hand side, simplifying as much as possible, and equating the coefficients to those on the left hand side we find: \[b = -6, \ c = 14, \ d = -24, \ e = 40 \]

  4. Using the fact that \(z_1 = 1+\sqrt{2}i\) and \(z_2 = 2-3i\) are roots of the equation \(-2x^4 + bx^3 + cx^2 + dx + e = 0 \), we find:
    1. The remaining roots are \(z_3 = 1 - \sqrt{2}i\) and \(z_4 = 2 + 3i\).
    2. Using the fact that: \[-2x^4 + bx^3 + cx^2 + dx + e = -2.\begin{pmatrix}x - (1 - \sqrt{2}i) \end{pmatrix}.\begin{pmatrix}x + (1 + \sqrt{2}i)\end{pmatrix}.\begin{pmatrix}x - (2 - 3i)\end{pmatrix}.\begin{pmatrix}x - (2 + 3i)\end{pmatrix}\] expanding the right hand side, simplifying as much as possible, and equating the coefficients to those on the left hand side we find: \[b = -12, \ c = 48, \ d = -76, \ e = 78 \]

  5. Using the fact that \(z_1 = 3\), \(z_2 = i\) and \(z_3 = 2-3i\) are roots of the equation \(x^5 +bx^4 + cx^3 + dx^2 + ex + f = 0 \), we find:
    1. \(z_4 = -i\) and \(z_5 = 2 + 3i\)
    2. Using the fact that: \[x^5 + bx^4 + cx^3 + dx^2 + e = \begin{pmatrix} x - 3\end{pmatrix}.\begin{pmatrix}x - i \end{pmatrix}.\begin{pmatrix}x + i\end{pmatrix}.\begin{pmatrix}x - (2 - 3i)\end{pmatrix}.\begin{pmatrix}x - (2 + 3i)\end{pmatrix}\] expanding the right hand side, simplifying as much as possible, and equating the coefficients to those on the left hand side we find: \[b = -7, \ c = 26, \ d = -46, \ e = 25, \ f = -39 \]


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