How do you factor $1-100{{x}^{2}}$?

Hint: We will use the concept of the factorization of the quadratic equation to solve the above polynomial. At first, we will make the coefficient of the ${{x}^{2}}$ as 1 by taking -100 as common and then we will find the root of the equation ${{x}^{2}}-\dfrac{1}{100}=0$. Let us say that root is a, b then the factor of ${{x}^{2}}-\dfrac{1}{100}$ is $\left( x-a \right)\left( x-b \right)$.

Complete step-by-step solution:
We know from the question that we have to factorize $1-100{{x}^{2}}$. So, we will use the concept of the factorization of the quadratic polynomial to solve the above polynomial.
At first, we will make the coefficient of the ${{x}^{2}}$as 1 of the polynomials $1-100{{x}^{2}}$ by taking -100 common from it.
Then we will get:
$\Rightarrow 1-100{{x}^{2}}=-100\left( {{x}^{2}}-\dfrac{1}{100} \right)$
Now, we will find the root of the quadratic polynomial ${{x}^{2}}-\dfrac{1}{100}$ by equating it with zero.
$\Rightarrow {{x}^{2}}-\dfrac{1}{100}=0$
Since, we know that the square root of 100 is 10, so we can write $100={{10}^{2}}$.
$\Rightarrow {{x}^{2}}-\dfrac{1}{{{10}^{2}}}=0$
$\Rightarrow {{x}^{2}}-{{\left( \dfrac{1}{10} \right)}^{2}}=0$
Since, we know that ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ .
$\Rightarrow \left( x-\dfrac{1}{10} \right)\left( x+\dfrac{1}{10} \right)=0$
$\therefore x=\dfrac{1}{10},-\dfrac{1}{10}$
So, root of the equation ${{x}^{2}}-\dfrac{1}{100}=0$ is $\dfrac{1}{10},-\dfrac{1}{10}$.
So, we can express ${{x}^{2}}-\dfrac{1}{100}$ as $\left( x-\dfrac{1}{10} \right)\left( x+\dfrac{1}{10} \right)$ .
$\Rightarrow 1-100{{x}^{2}}=-100\left( {{x}^{2}}-\dfrac{1}{100} \right)=-100\left( x-\dfrac{1}{10} \right)\left( x+\dfrac{1}{10} \right)$
Hence, the factorization of the polynomial $1-100{{x}^{2}}$ is:
 \[=-100\left( x-\dfrac{1}{10} \right)\left( x+\dfrac{1}{10} \right)\]
This is our required solution.

Note: Students are required to note that to find the root of the quadratic equation $a{{x}^{2}}+bx+c=0$ we will use middle term splitting method or the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to find the root. And , And, suppose the root is $\alpha ,\beta $ then the factor of $a{{x}^{2}}+bx+c=\left( x-\alpha \right)\left( x-\beta \right)$.