Question 350361
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Divide the coefficient on the *[tex \Large x] term by 2.  (that would be the "b" in *[tex \Large ax^2\ +\ bx\ +\ c])  Square the result.  Put that result in as the missing third term of your trinomial.


If you ever have a lead coefficient different than 1, then you need to divide the whole polynomial by the lead coefficient first.


In general, *[tex \Large ax^2\ +\ bx\ +\ c] is a perfect square if and only if *[tex \Large c\ =\ \frac{b^2}{4a^2}] 



John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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