## Intermediate Algebra (6th Edition)

$-9a^2-b^2+6ab+16$
RECALL: (i) $(x+y)(x-y) = x^2-y^2$ (ii) $(x-y)^2=x^2-2xy+y^2$ Use rule (i) above where $x=4$ and $y=3a-b$ to have: $\\=4^2-(3a-b)^2$ Use rule (ii) above where $x=3a$ and $y=b$ to have: $\\=16-[(3a)^2-2(3a)(b)+b^2] \\=16-(9a^2-6ab+b^2) \\=16-9a^2-(-6ab)-b^2 \\=16-9a^2+6ab-b^2 \\=-9a^2-b^2+6ab+16$