## Precalculus (6th Edition)

$(4p-5)^{2}$
Test whether this a perfect square, $(A-B)^{2}=A^{2}-2AB+B^{2}$ First term: $A^{2}=(4p)^{2}\Rightarrow A=4p$ Third term: $B^{2}=5^{2}\Rightarrow B=5$ Test: does $-2AB$ equal the middle term? $-2AB=-2(4p)(5)=-40p\qquad$ ... yes, it does. This is a perfect square, $16p^{2}-40p+25=A^{2}-2AB+B^{2}=(A-B)^{2}$ $(A-B)^{2}=(4p-5)^{2}$