There exists polynomial which is not factorable with respect to real numbers, but still its factorable with respect to the complex number. They are called complex polynomial. So in other words, in a complex polynomial, the variable present is able to take a complex value.
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, $D = b^ {2}-4ac$
                                       = $4^{2}-4*1*5$
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
$x^{2}+4x+5= (x+2-i)(x+2+i)$