To complete the square for the expression \( x^2 - \frac{1}{6}x \), we follow these steps:
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Identify the coefficient of the \( x \) term, which is \( -\frac{1}{6} \).
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Take half of this coefficient and square it: \[ \text{Half of } -\frac{1}{6} \text{ is } -\frac{1}{12}. \] Squaring this value: \[ \left(-\frac{1}{12}\right)^2 = \frac{1}{144}. \]
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Add and subtract this square inside the expression: \[ x^2 - \frac{1}{6}x + \frac{1}{144} - \frac{1}{144}. \]
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Rewrite the expression by grouping: \[ \left(x^2 - \frac{1}{6}x + \frac{1}{144}\right) - \frac{1}{144}. \]
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The trinomial \( x^2 - \frac{1}{6}x + \frac{1}{144} \) is a perfect square and can be factored: \[ \left(x - \frac{1}{12}\right)^2. \]
Putting it all together, we have: \[ x^2 - \frac{1}{6}x = \left(x - \frac{1}{12}\right)^2 - \frac{1}{144}. \]
So, the completed square expression is: \[ \left(x - \frac{1}{12}\right)^2 - \frac{1}{144}. \]
However, if you only want the perfect square trinomial factored and are not concerned with subtracting \( \frac{1}{144} \), then the factored form of the perfect square trinomial is: \[ \left(x - \frac{1}{12}\right)^2. \]