The root of the polynomial $x^2-4x-5$ is:

The roots of a polynomial can be found by substituting the suitable values of a variable which equates the given polynomial to zero.

Let's substitute $x=4$ in the polynomial:

$p\left(4\right)={\left(4\right)}^2-4\left(4\right)-5$

$p\left(4\right)=16-16-5$

Therefore, $p\left(4\right)=-5$

Hence 4 is not the root of the given polynomial.

Let's substitute $x=2$ in the polynomial:

$p\left(2\right)={\left(2\right)}^2-4\left(2\right)-5$

$p\left(2\right)=4-8-5$

Therefore, $p\left(2\right)=-9$

Hence 2 is not the root of the given polynomial.

Let's substitute $x=5$ in the polynomial:

$p\left(5\right)={\left(5\right)}^2-4\left(5\right)-5$

$p\left(5\right)=25-20-5$

Therefore, $p\left(5\right)=0$

Therefore, 5 is the root of the given polynomial.

Let's substitute $x=3$ in the polynomial:

$p\left(3\right)={\left(3\right)}^2-4\left(3\right)-5$

$p\left(3\right)=9-12-5$

Therefore, $p\left(3\right)=-8$

Therefore, 3 is not a root of the given polynomial.

Thus, 5 is the correct answer.