Fill in the blanks to make a "perfect square." For example, in fill in 9 and 3 since \[ x^{2}+6 x+\ldots=(x+\square)^{2} \] \( x^{2}+14 x+\square=(x+\square)^{2} \). \( x^{2}-8 x+\square=(x-\square)^{2} \). \( x^{2}-20 x+\square=(x-\square)^{2} \). Note: You can earn partial credit on this problem.

To create a "perfect square," we need to find the value that completes the square for each expression. 1. For \( x^{2} + 14x + \square = (x + \square)^{2} \): We take half of 14, which is 7, and then square it. Thus, filling in the blank, we get: \[ \square = 49 \] So, \( x^{2} + 14x + 49 = (x + 7)^{2} \). 2. For \( x^{2} - 8x + \square = (x - \square)^{2} \): Taking half of -8, which is -4, and squaring it gives: \[ \square = 16 \] Therefore, \( x^{2} - 8x + 16 = (x - 4)^{2} \). 3. For \( x^{2} - 20x + \square = (x - \square)^{2} \): Half of -20 is -10. Squaring that results in: \[ \square = 100 \] Hence, \( x^{2} - 20x + 100 = (x - 10)^{2} \). In summary, the completed squares are: 1. \( \square = 49 \) 2. \( \square = 16 \) 3. \( \square = 100 \)