Any complex number can be written in the form a+ib, where $a$, $b$ are both real number and

$i=\sqrt{-1}$.Also $a$ is called the real part and $b$ is called the imaginary part of $a+ib$

For example: $2+3i$, $-2+i$, $5$, $\sqrt{5}+2i$

According to the fundamental theorem of algebra,

Considering the complex numbers, any polynomial can be factored into product of linear factors.

One of the most common example of a complex polynomial, is a quadratic equation $x^{2}+1$.

When we compare this with quadratic expression $ax^ {2} +bx + c $, $a=1$, $b=0$ and $c=1$

So the discriminant, $D = b^ {2}-4ac$

= $0^{2}-4*1*1$

= $-4$

This is a negative number.

So the condition for a quadratic polynomial to be complex is that its discriminant

must be a negative number.

Now, factoring it

$x^{2}+1 = x^{2} – (\sqrt{-1})^{2}=x^{2}-i^{2}$.

This is in the form of difference of square

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

The fundamental theorem of algebra can also be stated in terms of roots as

Considering the complex numbers, any polynomial of degree $n$ will have $n$ roots.

In terms of the root, there exist a property for all polynomial: If $a+ib$ is a root of the polynomial, the n its conjugate $a-ib$ ia also its root.

For example, factor the polynomial $x^{2}+4x+5$

Lets first check, what type of polynomial this is.

Let $a= 1$, $b=4$ and $c=5$

So thebdiscriminant, $D = b^ {2}-4ac$

= $4^{2}-4*1*5$

=$16-20$

=$-4$

So it’s again a negative number, and hence this is a complex polynomial.

Lets find its root by using the quadratic formula

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

= $\frac{-4\pm \sqrt{4^{2}-4*1*5}}{2*1}$

= $\frac{-4\pm \sqrt{-4}}{2}$

= $\frac{-4\pm \sqrt{-4}}{2}$

= $\frac{-4\pm 2\sqrt{-1}}{2}$

= ${-2\pm i}$

So it can be seen that there exist two roots $-2+i$ and $-2-i$

So it can be factored as

$x^{2}+4x+5= (x+2-i)(x+2+i)$