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Factorize the expression $4(xy+1)^2-9(x-1)^2$.
Given:
The given algebraic expression is $4(xy+1)^2-9(x-1)^2$.
To do:
We have to factorize the expression $4(xy+1)^2-9(x-1)^2$.
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.
$4(xy+1)^2-9(x-1)^2$ can be written as,
$4(xy+1)^2-9(x-1)^2=[2(xy+1)]^2-[3(x-1)]^2$ [Since $4=2^2, 9=3^2$]
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,
$4(xy+1)^2-9(x-1)^2=[2(xy+1)]^2-[3(x-1)]^2$
$4(xy+1)^2-9(x-1)^2=[2(xy+1)+3(x-1)][2(xy+1)-3(x-1)]$
$4(xy+1)^2-9(x-1)^2=[2xy+2+3x-3][2xy+2-3x+3]$
$4(xy+1)^2-9(x-1)^2=(2xy+3x-1)(2xy-3x+5)$
Hence, the given expression can be factorized as $(2xy+3x-1)(2xy-3x+5)$.
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