## Precalculus (6th Edition)

$5(2p-5q)^{2}$
Factor out the greatest common factor, $5$ $20p^{2}-100pq+125q^{2}=5(4p^{2}-20pq+25q^{2})=*$ Test whether the parentheses hold a perfect square, $(A-B)^{2}=A^{2}-2AB+B^{2}$ First term: $A^{2}=(2p)^{2}\Rightarrow A=2p$ Third term: $B^{2}=(5q)^{2}\Rightarrow B=5q$ Test:$\qquad$ does $-2AB$ equal the middle term? $-2AB=-2(2p)5q)=-20pq\qquad$ ... yes, it does. $\Rightarrow$The parentheses hold a perfect square, $(2p-5q)^{2}$ $*...= 5(2p-5q)^{2}$