Question 1122671
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ *\ +\ 56a\ +\ 49]


Has to be the square of a SUM because all the signs are positive.  The second term of the sum must be the square root of the last term, namely 7.  The missing term must have an *[tex \Large a^2] in it.  And the coefficient on the middle term is the coefficient on the first term of the binomial times the last term of the binomial times 2.  So 56 divided by 2 is 28, and 28 divided by 7 is 4.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (4a\ +\ 7)^2\ =\ \underline{16a^2}\ +\ 56a\ +\ 49]
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}

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