How do you factor \[1-4{{y}^{2}}\]?

Hint: In order to find the solution to the given question that is to find the factor of given expression \[1-4{{y}^{2}}\] apply the identity named as difference of squares that is \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\] and get the factorisation or factoring is defined as the breaking or decomposition of an entity (for example a number, a matrix, or a polynomial) into a product of another entity, or factors, which when multiplied together give the original number or a matrix, etc. It is simply the resolution of an integer or polynomial into factors such that when multiplied together they will result in an original or initial integer or polynomial. In the factorisation method, we reduce any algebraic or quadratic equation into its simpler form, where the equations are represented as the product of factors instead of expanding the brackets. The factors of any equation can be an integer, a variable or an algebraic expression itself.

Complete step by step solution:
According to the question given expression in the question is as follows:
\[\Rightarrow 1-4{{y}^{2}}\]
To find the factor, first rewrite the above expression in the form difference of squares, we get:
\[\Rightarrow {{1}^{2}}-{{\left( 2y \right)}^{2}}\]
In order to find the factor of the above expression , apply the method of factorisation which states that we reduce any algebraic or quadratic equation into its simpler form, where the equations are represented as the product of factors instead of expanding the brackets. The factors of any equation can be an integer, a variable or an algebraic expression itself. Apply the identity named as difference of squares that is \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\] in the above expression where \[a=1\] and \[b=2y\], we get:
\[\Rightarrow {{1}^{2}}-{{\left( 2y \right)}^{2}}=\left( 1+2y \right)\left( 1-2y \right)\]
\[\Rightarrow 1-4{{y}^{2}}=\left( 1+2y \right)\left( 1-2y \right)\]
Therefore, factors of the given expression \[1-4{{y}^{2}}\] are \[\left( 1+2y \right)\] and \[\left( 1-2y \right)\].

Note: Students generally make mistakes while applying the same identity everywhere in order to find the factor of the expressions which is completely wrong. It’s important to understand that different expressions require different identities to find its factors from certain expressions. Like here in the given expression, we have used the difference of square identity like this there are other methods also that is long division of polynomials and splitting of the middle term.