Decide what number must be added to both sides of the equation to make the left side a perfect square trinomial, then factor the left side. $$ x^{2}-6 x=3 $$ 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.)

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Add $$ 9 $$ to both sides to make the left side a perfect square trinomial. The left side then factors to $$ (x - 3)^{2} $$. 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 formula 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 $$-3$$: $$ (-3)^2 = 9 $$ 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 $$ x^{2} - 6x + 9 $$ can be factored as: $$ (x - 3)^{2} $$ So, the completed equation is: $$ (x - 3)^{2} = 12 $$ In summary, the number that must be added to both sides is $$ 9 $$, and the left side factors to $$ (x - 3)^{2} $$.

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