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Polynomials with Complex Roots

The Fundamental Theorem of Algebra assures us that any polynomial with real number coefficients can be factored completely over the field of complex numbers .

In the case of quadratic polynomials , the roots are complex when the discriminant is negative.

Example 1:

Factor completely, using complex numbers.

$x^{3} + 10 x^{2} + 169 x$

First, factor out an $x$ .

$x^{3} + 10 x^{2} + 169 x = x (x^{2} + 10 x + 169)$

Now use the quadratic formula for the expression in parentheses, to find the values of $x$ for which $x^{2} + 10 x + 169 = 0$ .

Here $a = 1, b = 10$ and $c = 169$ .

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

$\begin{array}{l} x = \frac{- 10 \pm \sqrt{10^{2} - 4 (1) (169)}}{2 (1)} \\ = \frac{- 10 \pm \sqrt{100 - 676}}{2} \\ = \frac{- 10 \pm \sqrt{- 576}}{2} \end{array}$

Write the square root using imaginary numbers.

$\begin{array}{l} x = \frac{- 10 \pm 24 i}{2} \\ = - 5 \pm 12 i \end{array}$

We now know that the values of $x$ for which the expression

$x^{2} + 10 x + 169$

equals $0$ are $x = - 5 + 12 i and x = - 5 - 12 i$ .

So, the original polynomial can be factored as

$x^{3} + 10 x^{2} + 169 x = x (x - [- 5 + 12 i]) (x - [- 5 - 12 i])$

You can verify this using FOIL .

Sometimes, you can factor a polynomial using complex numbers without using the quadratic formula. For instance, the difference of squares rule:

$x^{2} - a^{2} = (x + a) (x - a)$

This can also be used with complex numbers when $a^{2}$ is negative, as follows:

$x^{2} + 25 = (x + 5 i) (x - 5 i)$

Example 2:

Factor completely, using complex numbers.

$9 x^{2} y + 64 y$

First, factor out $y$ .

$9 x^{2} y + 64 y = y (9 x^{2} + 64)$

Now, use the difference of square rule to factor $9 x^{2} + 64$ .

$\begin{array}{l} 9 x^{2} + 64 = 9 x^{2} - (- 64) \\ = {(3 x)}^{2} - {(8 i)}^{2} \\ = (3 x + 8 i) (3 x - 8 i) \end{array}$

Therefore, $9 x^{2} y + 64 y = y (3 x + 8 i) (3 x - 8 i)$ .