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Factorize the expression $16a^4-b^4$.
Given:
The given algebraic expression is $16a^4-b^4$.
To do:
We have to factorize the expression $16a^4-b^4$.
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.
$16a^4-b^4$ can be written as,
$16a^4-b^4=(4a^2)^2-(b^2)^2$ [Since $16a^4=(4a^2)^2, b^4=(b^2)^2$]
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,
$16a^4-b^4=(4a^2)^2-(b^2)^2$
$16a^4-b^4=(4a^2+b^2)(4a^2-b^2)$
Now,
$4a^2-b^2$ can be written as,
$4a^2-b^2=(2a)^2-b^2$ [Since $4a^2=(2a)^2$]
Using the formula $a^2-b^2=(a+b)(a-b)$, we can factorize $(2a)^2-b^2$.
$(2a)^2-b^2=(2a+b)(2a-b)$.............(I)
Therefore,
$16a^4-b^4=(4a^2+b^2)(2a+b)(2a-b)$ [Using (I)]
Hence, the given expression can be factorized as $(4a^2+b^2)(2a+b)(2a-b)$.
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