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Factorize the algebraic expression $9a^4-24a^2b^2+16b^4-256$.
Given:
The given expression is $9a^4-24a^2b^2+16b^4-256$.
To do:
We have to factorize the algebraic expression $9a^4-24a^2b^2+16b^4-256$.
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.
$9a^4-24a^2b^2+16b^4-256$ can be written as,
$9a^4-24a^2b^2+16b^4-256=(9a^4-24a^2b^2+16b^4)-256$
$9a^4-24a^2b^2+16b^4-256=[(3a^2)^2-2(3a^2)(4b^2)+(4b^2)^2]-256$ [Since $9a^4=(3a^2)^2, 16b^4=(4b^2)^2$ and $24a^2b^2=2(3a^2)(4b^2)$]
Here, we can observe that the given expression is of the form $m^2-2mn+n^2$. So, by using the formula $(m-n)^2=m^2-2mn+n^2$, we can factorize the given expression.
Here,
$m=3a^2$ and $n=4b^2$
Therefore,
$9a^4-24a^2b^2+16b^4-256=[(3a^2)^2-2(3a^2)(4b^2)+(4b^2)^2]-256$
$9a^4-24a^2b^2+16b^4-256=(3a^2-4b^2)^2-256$
Now,
$(3a^2-4b^2)^2-256$ can be written as,
$(3a^2-4b^2)^2-256=(3a^2-4b^2)^2-(16)^2$ [Since $256=(16)^2$]
Using the formula $a^2-b^2=(a+b)(a-b)$, we can factorize $(3a^2-4b^2)^2-(16)^2$ as,
$(3a^2-4b^2)^2-256=(3a^2-4b^2)^2-(16)^2$
$(3a^2-4b^2)^2-256=(3a^2-4b^2+16)(3a^2-4b^2-16)$
Hence, the given expression can be factorized as $(3a^2-4b^2+16)(3a^2-4b^2-16)$.
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