How do you factor \[{{s}^{2}}-16\]?

Hint: The expressions which have one square term being subtracted from another square term is called difference of squares. The algebraic form of this is \[{{a}^{2}}-{{b}^{2}}\]. The factored form of these types of expression is \[\left( a+b \right)\left( a-b \right)\]. The factored form of a polynomial expression is used to find the roots of the polynomial.

Complete step by step solution:
The given expression is \[{{s}^{2}}-16\]. It has two terms; the first term is \[{{s}^{2}}\] and the second term is 16.
As, we know that 16 is square of 4, the first term is also square of the variable s. Using this simplification in the given expression, it can be written as \[{{s}^{2}}-16={{s}^{2}}-{{4}^{2}}\].
As we can see that this expression is evaluating the difference of two square terms, it is different from square form. We know that the difference of square expression \[{{a}^{2}}-{{b}^{2}}\] is factorized as \[\left( a+b \right)\left( a-b \right)\]. Here, we have a, and b are s and 4 respectively. Substituting the values in the expansion, we get
\[\Rightarrow {{s}^{2}}-{{4}^{2}}=\left( s+4 \right)\left( s-4 \right)\]
Hence the factored form of the given expression is \[\left( s+4 \right)\left( s-4 \right)\].

Note: To solve these types of problems one should know the difference of square form, and its factored form. There are many other special expression forms like this such as, difference of cubes, addition of cubes. Algebraic form of difference of cubes and its expansion is \[{{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)\]. Similarly, for the addition of cubes, it is \[{{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)\].
It should be noted that, only if the polynomial has real roots then we can express it in factored form. Otherwise, we can not.