Question

complete the square. fill in the number that makes the polynomial a perfect - square quadratic.
$f^{2}-6f + \square$
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Explanation:

Step1: Recall the perfect - square formula

The perfect - square quadratic formula is \((a - b)^2=a^{2}-2ab + b^{2}\). For the given polynomial \(f^{2}-6f+\square\), we can compare it with \(a^{2}-2ab + b^{2}\). Here, \(a = f\) and \(-2ab=-6f\). Since \(a = f\), we substitute \(a\) into \(-2ab=-6f\), getting \(-2\times f\times b=-6f\).

Step2: Solve for b

We can cancel out \(f\) from both sides of the equation \(-2\times f\times b=-6f\) (assuming \(f
eq0\); if \(f = 0\), the equation still holds in a general sense for the formula). So we have \(-2b=-6\). Dividing both sides by \(-2\), we get \(b = 3\).

Step3: Find the value of \(b^{2}\)

From the perfect - square formula \(a^{2}-2ab + b^{2}\), we know that the constant term we need is \(b^{2}\). Since \(b = 3\), then \(b^{2}=3^{2}=9\). So the polynomial \(f^{2}-6f + 9=(f - 3)^{2}\), which is a perfect - square quadratic.

Answer:

9