## College Algebra (6th Edition)

$x^{2}+4x+4=(x+2)^{2}$ $3(x^{2}+4x)=3(x^{2}+4x+4-4)=3(x+2)^{2}-12$ (adding 4 to the first parentheses, we must subtract 3(4) to leave the value unchanged) $y^{2}-2y+1=(y-1)^{2}$ $4(y^{2}-2y)=4(y^{2}-2y+1-1)=4(y-1)^{2}-4$ (adding 1 to the first parentheses, we must subtract 4(1) to leave the value unchanged) So, the complete work looks like: $3(x+2)^{2}-12+4(y-1)^{2}-4=32\quad/+12+4$ $3(x+2)^{2}+4(y-1)^{2}=48$ Fill the blanks with ... 4 ... ... 1 ... ... 16 ...