## College Algebra (6th Edition)

$9x^{2}+6xy+y^{2}+6x+2y+1$
$[(3x+y)+1]^{2}$ The square of a binomial can be found using the formula $(a+b)^{2} = a^{2}+2ab+b^{2}$ $[(3x+y)+1]^{2}=(3x+y)^{2}+2(3x+y)(1)+1^{2}$ $=(3x+y)^{2}+2(3x+y)+1$ $=(3x)^{2}+2(3x)(y)+y^{2}+6x+2y+1$ $= 9x^{2}+6xy+y^{2}+6x+2y+1$