Question

The factorization of $p^2{q}^2-6pqr+9r^2$ is

• A. $(pq-3r)(pq-3r)$
• B. $(pq-3r)(pq+3r)$
• C. $(pq+3r)(pq+3r)$
• D. $(pq-3r)(3r-pq)$

Answer 1

The correct option is A $(pq-3r)(pq-3r)$

$p^2{q}^2-6pqr+9r^2=0$

$\Rightarrow$ $p^2{q}^2-3pqr-3pqr+9r^2=0$

$\Rightarrow$ $pq(pq-3r)-3r(pq-3r)=0$

$\Rightarrow$ $(pq-3r)(pq-3r)=0$