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Factorize the algebraic expression $p^2q^2-6pqr+9r^2$.
Given:
The given expression is $p^2q^2-6pqr+9r^2$.
To do:
We have to factorize the algebraic expression $p^2q^2-6pqr+9r^2$.
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.
$p^2q^2-6pqr+9r^2$ can be written as,
$p^2q^2-6pqr+9r^2=(pq)^2-2(pq)(3r)+(3r)^2$ [Since $p^2q^2=(pq)^2, 9r^2=(3r)^2$ and $6pqr=2(9pq)(3r)$]
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,
$m=pq$ and $n=3r$
Therefore,
$p^2q^2-6pqr+9r^2=(pq)^2-2(pq)(3r)+(3r)^2$
$p^2q^2-6pqr+9r^2=(pq-3r)^2$
$p^2q^2-6pqr+9r^2=(pq-3r)(pq-3r)$
Hence, the given expression can be factorized as $(pq-3r)(pq-3r)$.
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