# Complex Roots of a Polynomial, Further

The complex roots page built introduced how to effectively approach working out the complex roots of a polynomial that was a cubic or a quartic equation.

This page will further demonstrate the type of examples that can be encountered involving polynomials with complex solutions, and how to solve them.

Examples

(1.1)

Verify that    a = 2 + i    is a solution of the polynomial equation    a^3 \space {\text{–}} \space 3a^2 + a + 5 = 0.

Solution

If we remember the factor theorem.

For a polynomial   f(x),  when   f(c) = 0.

Then   (x \space {\text{–}} \space c)   is a factor of   f(x).

For verification We just need to input a value into the equation and obtain a result of  0.
This is true for complex numbers as well as real numbers.

We will have to work out,    (2 + i)^3 \space {\text{–}} \space 3(2 + i)^2 + (2 + i) + 5.

(2 + i)^2 \space = \space (2 + i)(2 + i) \space = \space 4 + 4i \space {\text{–}} \space 1

(2 + i)^3 \space = \space (2 + i)(2 + i)^2 \space = \space (2 + i)(4 + 4i \space {\text{–}} \space 1) \space\space = \space\space 2 + 11i

So:

(2 + i)^3 \space {\text{–}} \space 3(2 + i)^2 + (2 + i) + 5 \space\space = \space\space 2 + 11i \space {\text{–}} \space 12 \space {\text{–}} \space 12i + 3 + 2 + i + 5 \space = \space 0

a = 2 + i   is a solution.

(1.2)

Solve the equation    x^3 \space {\text{–}} \space 4x^2 \space {\text{–}} \space 2x + 20 = 0    given that    x = {\text{-}}2    is a root.

Solution

With this example, as we already know that  -2  is a root.
We can begin with the use of synthetic division.

The result below now writes as   x^2 \space {\text{–}} \space 6x + 10.

Giving us    (x + 2)(x^2 \space {\text{–}} \space 6x + 10).

The quadratic can be factored further with the quadratic formula, giving us the other two roots of the original cubic polynomial.

Here:   a = 1 \space\space , \space\space b = {\text{-}}6 \space\space , \space\space c = 10

{\scriptsize{x = \space}} {\boldsymbol{\frac{{\text{-}}b \space \pm \space \sqrt{b^{2} \space – \space 4ac}}{2a}}}

{\scriptsize{x = \space}} {\boldsymbol{\frac{6 \space \pm \space \sqrt{({\text{-}}6^{2} \space\space – \space\space (4 \times 1 \times 10)}}{2 \times 1}}}    =    {\boldsymbol{\frac{6 \space \pm \space \sqrt{36 \space\space {\text{–}} \space\space 40}}{2}}}

{\scriptsize{x = \space}} {\boldsymbol{\frac{6 \space \pm \space \sqrt{{\text{-}}4}}{2}}}    =    {\boldsymbol{\frac{6 \space \pm \space \sqrt{4 \space \times \space {\text{-}}1}}{2}}}    =    {\boldsymbol{\frac{6 \space \pm \space \sqrt{4}\sqrt{{\text{-}}1}}{2}}}

{\scriptsize{x = \space}} {\boldsymbol{\frac{6 \space \pm \space 2i}{2}}} \space {\scriptsize{\space = \space}} \space {\boldsymbol{\frac{6}{2} {\scriptsize\pm} \frac{2i}{2}}} \space {\scriptsize{\space = \space}} \space {\footnotesize{\boldsymbol{3 \pm i}}}

Thus,     x = {\text{-}}2 \space , \space 3 + i \space , \space 3 \space {\text{–}} \space i.

## Complex Roots of a Polynomial,Further Examples

(2.1)

If    x_{\tiny{1}} = 1    and    x_{\tiny{2}} = 3 + 2i    are solutions to    x^3 + bx^2 + cx + d = 0.

Find the values of  b, c  and  d.

Solution

As complex roots come in conjugate pairs, we know that the value of the 3rd solution is   x_{\tiny{3}} = 3 \space {\text{–}} \space 2i.

So the cubic polynomial in factored form is,    (x \space {\text{–}} \space 1)(x \space {\text{–}} \space (3 + 2i))(x \space {\text{–}} \space (3 \space {\text{–}} \space 2i)).

Multiplying out and expanding this factored form will give us the original polynomial with all the correct values of the coefficients  b, c  and  d.

(x \space {\text{–}} \space (3 + 2i))(x \space {\text{–}} \space (3 \space {\text{–}} \space 2i)) \space\space = \space\space x^2 \space {\text{–}} \space 3x + 2ix \space {\text{–}} \space 3x \space {\text{–}} \space 2ix + 9 + 6i \space {\text{–}} \space 6i \space {\text{–}} \space 4i^2

= \space x^2 \space {\text{–}} \space 6x + 13
Now:
(x \space {\text{–}} \space 1)(x^2 \space {\text{–}} \space 6x + 13) \space\space = \space\space x^3 \space{\text{–}} \space 6x^2 +13x \space {\text{–}} \space x^2 + 6x \space {\text{–}} \space 13 \space\space = \space\space x^3 \space {\text{–}} \space 7x^2 + 19x \space {\text{–}} \space 13

b = {\text{-}}7 \space , \space c = 19 \space , \space d = {\text{-}}13

(2.2)

If    x_{\tiny{1}} = 3i    and    x_{\tiny{2}} = 1 \space {\text{–}} \space i    are complex roots of a polynomial    x^4 + bx^3 + cx^2 + dx + e = 0.

Find the values of  b, c, d   and  e.

Solution

Like in example (2.1), the two roots already given also tell us the value of the remaining roots.

x_{\tiny{1}} = 3i   ,  x_{\tiny{2}} = 1 \space {\text{–}} \space i  ,  x_{\tiny{3}} = {\text{-}}3i  ,  x_{\tiny{4}} = 1 + i

In factored form,    (x \space {\text{–}} \space 3i)(x + 3i)(x \space {\text{–}} \space (1 + i))(x \space {\text{–}} \space (1 \space {\text{–}} \space i)).

We can expand both pairs of roots separately, then use long multiplication to get the original quartic polynomial.

(x \space {\text{–}} \space 3i)(x + 3i)(x \space {\text{–}} \space (1 + i))(x \space {\text{–}} \space (1 \space {\text{–}} \space i)) \space\space = \space\space x^2 + 9

(x \space {\text{–}} \space (1 + i))(x \space {\text{–}} \space (1 \space {\text{–}} \space i)) \space\space = \space\space x^2 \space {\text{–}} \space 2x + 2

Now:
(x^2 + 9) \times (x^2 \space {\text{–}} \space 2x + 2)

\begin{array}{r} \space\space\space\space x^{\tt{2}} \space {\text{–}} \space\space 2x + \space 2 \space\\ \times \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space x^{\tt{2}} + \space\space 0 \space + \space 9 \space\\ \hline \space 9x^{\tiny{\tt{2}}} \space {\text{–}} 18x + 18\\ x^{\tt{4}} \space {\text{–}} \space 2x^{\tt{3}} + 2x^{\tiny{\tt{2}}} \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space \\ \end{array}

=>     \begin{array}{r} \space 9x^{\tiny{\tt{2}}} \space\space {\text{–}} \space 18x+ 18\\ {\color{blue}+} \space\space\space\space x^{\tt{4}} \space {\text{–}} \space 2x^{\tt{3}} + \space 2x^{\tiny{\tt{2}}} \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space \\ \hline x^{\tiny{4}} \space {\text{–}} \space 2x^{\tiny{3}} + 11x^{\tiny{\tt{2}}} \space {\text{–}} \space 18x + 18\\ \end{array}

So we get.    x^4 \space {\text{–}} \space 2x^3 + 11x^2 \space {\text{–}} \space 18x + 18

Which from gives us,    b = {\text{-}}2 \space\space , \space\space c = 11 \space\space , \space\space d = {\text{-}}18 \space\space , \space\space e = 18.

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