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Factorize the expression $\frac{50}{x^2}-\frac{2x^2}{81}$.
Given:
The given algebraic expression is $\frac{50}{x^2}-\frac{2x^2}{81}$.
To do:
We have to factorize the expression $\frac{50}{x^2}-\frac{2x^2}{81}$.
Solution:
Factorizing algebraic expressions:
Factorizing an algebraic expression implies writing the expression as a product of two or more factors. Factorization is the reverse of distribution.
An algebraic expression is factored completely when it is written as a product of prime factors.
$\frac{50}{x^2}-\frac{2x^2}{81}$ can be written as,
$\frac{50}{x^2}-\frac{2x^2}{81}=2(\frac{25}{x^2}-\frac{x^2}{81})$ (Taking $2$ common)
$\frac{50}{x^2}-\frac{2x^2}{81}=2[(\frac{5}{x})^2-(\frac{x}{9})^2]$ [Since $\frac{25}{x^2}=(\frac{5}{x})^2, \frac{x^2}{81}=(\frac{x}{9})^2$]
Here, we can observe that the given expression is a difference of two squares. So, by using the formula $a^2-b^2=(a+b)(a-b)$, we can factorize the given expression.
Therefore,
$\frac{50}{x^2}-\frac{2x^2}{81}=2[(\frac{5}{x})^2-(\frac{x}{9})^2]$
$\frac{50}{x^2}-\frac{2x^2}{81}=2(\frac{5}{x}+\frac{x}{9})(\frac{5}{x}-\frac{x}{9})$
Hence, the given expression can be factorized as $2(\frac{5}{x}+\frac{x}{9})(\frac{5}{x}-\frac{x}{9})$.
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