To complete the square for the quadratic polynomial \(v^2 - 2v\), we will follow the steps below:
- Take the coefficient of \(v\), which is \(-2\).
- Divide it by 2: \(-2 / 2 = -1\).
- Square that result: \((-1)^2 = 1\).
Now, we add and also subtract this squared value (1) inside the expression to complete the square:
\[ v^2 - 2v + 1 - 1 \]
This gives us:
\[ (v - 1)^2 - 1 \]
Thus, the number that makes the polynomial \(v^2 - 2v\) a perfect-square quadratic is \(1\).
So, the completed expression is:
\[ v^2 - 2v + 1 = (v - 1)^2 \]
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