This is assumed to be:
4x^2+12xy+9y^2+16x+24y+16=(2x+3y+4)^2 (12x should be 12xy).
The clue is the presence of perfect squares 4x^2, 9y^2, 16 which have square roots 2x, 3y, 4. All the signs are plus, so we try the sum of these factors 2x+3y+4. This is the trial method, and by multiplying the proposed square root by itself we can check if we get the original expression.
DIVISION METHOD
We write out the expression like the dividend in arithmetic long division, leaving a space on the left to write in a varying "divisor". We write the square root of the first squared variable. 4x^2 has a square root of 2x. Write 2x in the "quotient" and in the "divisor:
2x
2x | 4x^2+12xy+9y^2+16x+24y+16
Then we multiply the "quotient" and "divisor" and subtract the result from the "dividend";
2x
2x | 4x^2+12xy+9y^2+16x+24y+16
4x^2
12xy+...
We double the "divisor" to make it 4x, and divide it into 12xy, to give 3y. This is the next term in the "quotient" and also is added to the "divisor":
2x + 3y
2x | 4x^2+12xy+9y^2+16x+24y+16
4x^2
4x+3y | 12xy+9y^2+...
12xy+9y^2
16x+24y+16
We multiply 4x+3y by 3y=12xy+py^2 and subtract this from the remainder of the "dividend". To continue, we then double the last term of the "divisor" and divide it into the remaining 16x+24y+16, giving us 16x/4x=4:
2x + 3y + 4
2x | 4x^2+12xy+9y^2+16x+24y+16
4x^2
4x+3y | 12xy+9y^2+...
12xy+9y^2
4x+6y+4 | 16x+24y+16
16x+24y+16
0
As before, we add the latest term, 4, as the next term to the "divisor" and the "quotient". When we multiply the "divisor" by 4 and subtract the product from the remainder of the "dividend", we get zero, meaning that we have arrived at the square root 2x+3y+4.