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Factorize the expression $49(a-b)^2-25(a+b)^2$.
Given:
The given algebraic expression is $49(a-b)^2-25(a+b)^2$.
To do:
We have to factorize the expression $49(a-b)^2-25(a+b)^2$.
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.
$49(a-b)^2-25(a+b)^2$ can be written as,
$49(a-b)^2-25(a+b)^2=[7(a-b)]^2-[5(a+b)]^2$ [Since $49=(7)^2, 25=5^2$]
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,
$49(a-b)^2-25(a+b)^2=[7(a-b)]^2-[5(a+b)]^2$
$49(a-b)^2-25(a+b)^2=[7(a-b)+5(a+b)][7(a-b)-5(a+b)]$
$49(a-b)^2-25(a+b)^2=(7a-7b+5a+5b)(7a-7b-5a-5b)$
$49(a-b)^2-25(a+b)^2=(12a-2b)(2a-12b)$
$49(a-b)^2-25(a+b)^2=2(6a-b)2(a-6b)$
$49(a-b)^2-25(a+b)^2=4(6a-b)(a-6b)$
Hence, the given expression can be factorized as $4(6a-b)(a-6b)$.
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