Answer
Fill the blanks with
... 4 ...
... 1 ...
... 16 ...
Work Step by Step
$x^{2}+4x+4=(x+2)^{2}$
$3(x^{2}+4x)=3(x^{2}+4x+4-4)=3(x+2)^{2}-12$
(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 ...
