# 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.

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

First, factor out an $x$ .

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

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

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

$x=\frac{-b\text{\hspace{0.17em}}±\text{\hspace{0.17em}}\sqrt{{b}^{2}\text{\hspace{0.17em}}-\text{\hspace{0.17em}}4ac}}{2a}$

$\begin{array}{l}x=\frac{-10\text{\hspace{0.17em}}±\text{\hspace{0.17em}}\sqrt{{10}^{2}\text{\hspace{0.17em}}-\text{\hspace{0.17em}}4\left(1\right)\left(169\right)}}{2\left(1\right)}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{-10\text{\hspace{0.17em}}±\text{\hspace{0.17em}}\sqrt{100\text{\hspace{0.17em}}-\text{\hspace{0.17em}}676}}{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{-10\text{\hspace{0.17em}}±\text{\hspace{0.17em}}\sqrt{-576}}{2}\end{array}$

Write the square root using imaginary numbers.

$\begin{array}{l}x=\frac{-10\text{\hspace{0.17em}}±\text{\hspace{0.17em}}24i}{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-5±12i\end{array}$

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

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

equals $0$ are .

So, the original polynomial can be factored as

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

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:

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

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

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

Example 2:

Factor completely, using complex numbers.

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

First, factor out $y$ .

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

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

$\begin{array}{l}9{x}^{2}+64=9{x}^{2}-\left(-64\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\left(3x\right)}^{2}-{\left(8i\right)}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left(3x+8i\right)\left(3x-8i\right)\end{array}$

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