How do you factor ${{y}^{2}}-2y-3$ ?

Hint: In this question, we have to find the factors of the equation. A quadratic equation is given in the problem, so we will apply splitting the middle term method to get the factors for the same. We start solving this problem by finding two numbers such that the product of the two numbers is equal to the product of the coefficient of ${{y}^{2}}$ and the constant. Also, the sum of these two numbers is the coefficient of y. Thus, after finding these 2 numbers, we split the middle term as the sum of those two numbers and make the necessary calculations to get the required solution.

Complete step by step solution:
According to the question, we have to find the factors of the quadratic equation.
Thus, we will use splitting the middle term method to get the solution.
The quadratic equation given to us is ${{y}^{2}}-2y-3$ -------- (1)
As we know, the general form of quadratic equation is $a{{y}^{2}}+y+c$ -------- (2)
Thus, on comparing equation (1) and (2), we get a=1, b=-2, and c=-3
To factorize, we have to find two numbers m and n such that $m+n=b=-2$ and $mn=a.c=1.(-3)=-3$ .
We see that if $m=-3$ and $n=1$ , then we get $m+n=-3+1=-2=b$ and $mn=a.c=-3.(1)=-3$
So, we will split the middle term as the addition of -3y and y because $-3y+1y=-2y$ , thus we get
$\Rightarrow {{y}^{2}}-3y+y-3$
Now, take common y from the first two numbers and 1 from the last two numbers, we get
$\Rightarrow y\left( y-3 \right)+1\left( y-3 \right)$
So, we will take (y-3) common on both LHS and RHS of the addition sign, we get
$\Rightarrow \left( y-3 \right)\left( y+1 \right)$
Therefore, for the quadratic equation ${{y}^{2}}-2y-3$ , its factors are $\left( y-3 \right)\left( y+1 \right)$ .

Note: Do mention all the steps properly to avoid confusion and mathematical error. You can also use the Hit and Trial method, the cross multiplication method or the Discriminant formula to get the solution for the problem.