How do you write ${{x}^{2}}-196$ in the factored form?

Hint: The given polynomial ${{x}^{2}}-196$ can be written as the difference of two squares. For this, we need to put $196={{14}^{2}}$ in the given polynomial so that it can be written as ${{x}^{2}}-{{14}^{2}}$. Then we need to use the algebraic identity given by ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ and apply it on the polynomial ${{x}^{2}}-{{14}^{2}}$ by putting $a=x$ and $b=14$ in the identity. After this, we will finally obtain the required factored form of the given polynomial.

Complete step by step solution:
Let us consider the quadratic polynomial given in the above question as
$\Rightarrow p\left( x \right)={{x}^{2}}-196$
According to the question, we need to factorize the above polynomial. For this, we can write the above polynomial in the form of the difference of two squares. For this, we can put $196={{14}^{2}}$ in the above polynomial, since we know that $196$ is a square of $14$. On putting this in the above polynomial we get
$\Rightarrow p\left( x \right)={{x}^{2}}-{{14}^{2}}$
Now, we know the algebraic identity given by ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$. On putting $a=x$ and $b=14$ in this identity we can write the above polynomial as
$\Rightarrow p\left( x \right)=\left( x+14 \right)\left( x-14 \right)$

Hence, we have finally factored the polynomial given in the question as $\left( x+14 \right)\left( x-14 \right)$.

Note: For solving these types of questions, we need to remember all of the important algebraic identities. Since we are given a quadratic polynomial in the question, we can also obtain its roots in order to factorize it. For this, we need to put the given quadratic polynomial to zero so that we will obtain the quadratic equation ${{x}^{2}}-196=0$ on solving which we will obtain the roots. Then finally using the factor theorem we can write the factors of the given polynomial.