Decide what number must be added to both sides of the equation to make the left side a perfect square trinomial, then factor th \( x^{2}-6 x=3 \) Add a number to both sides of the equation to make the left side a perfect square trinomial. \( x^{2}-6 x+\square=3+\square \) (Simplify your answers.)

To make the left side of the equation \( x^{2} - 6x \) a perfect square trinomial, we need to find the number that completes the square. The general form for completing the square for a quadratic expression \( ax^2 + bx \) is to take half of the coefficient of \( x \) (which is \( b \)), square it, and add it to both sides. In this case, the coefficient of \( x \) is \(-6\). 1. Take half of \(-6\): \[ \frac{-6}{2} = -3 \] 2. Square it: \[ (-3)^2 = 9 \] Now, we add \( 9 \) to both sides of the equation: \[ x^{2} - 6x + 9 = 3 + 9 \] This simplifies to: \[ x^{2} - 6x + 9 = 12 \] Now, the left side is a perfect square trinomial, which can be factored as: \[ (x - 3)^2 = 12 \] So, the number that must be added to both sides is \( 9 \). Thus, the completed equation is: \[ x^{2} - 6x + 9 = 3 + 9 \] or \[ x^{2} - 6x + 9 = 12 \]