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Factorize the expression $\frac{1}{16}x^2y^2-\frac{4}{49}y^2z^2$.
Given:
The given algebraic expression is $\frac{1}{16}x^2y^2-\frac{4}{49}y^2z^2$.
To do:
We have to factorize the expression $\frac{1}{16}x^2y^2-\frac{4}{49}y^2z^2$.
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.
$\frac{1}{16}x^2y^2-\frac{4}{49}y^2z^2$ can be written as,
$\frac{1}{16}x^2y^2-\frac{4}{49}y^2z^2=(\frac{1}{4}xy)^2-(\frac{2}{7}yz)^2$ [Since $\frac{1}{16}=(\frac{1}{4})^2, \frac{4}{49}=(\frac{2}{7})^2$]
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,
$\frac{1}{16}x^2y^2-\frac{4}{49}y^2z^2=(\frac{1}{4}xy)^2-(\frac{2}{7}yz)^2$
$\frac{1}{16}x^2y^2-\frac{4}{49}y^2z^2=(\frac{1}{4}xy+\frac{2}{7}yz)(\frac{1}{4}xy-\frac{2}{7}yz)$
$\frac{1}{16}x^2y^2-\frac{4}{49}y^2z^2=y(\frac{1}{4}x+\frac{2}{7}z)y(\frac{1}{4}x-\frac{2}{7}z)$
$\frac{1}{16}x^2y^2-\frac{4}{49}y^2z^2=y^2(\frac{1}{4}x+\frac{2}{7}z)(\frac{1}{4}x-\frac{2}{7}z)$
Hence, the given expression can be factorized as $y^2(\frac{1}{4}x+\frac{2}{7}z)(\frac{1}{4}x-\frac{2}{7}z)$.
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