Question 722815
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Actually, No and Yes.  In that order.  You erred by not distributing *[tex \LARGE p] across both terms of the binomial *[tex \LARGE 3p\ +\ 3].


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ p(3p\ +\ 3)\ \neq\ 3p^2\ +\ 3]


rather


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ p(3p\ +\ 3)\ =\ 3p^2\ +\ 3p]


Please review the <a href="http://www.purplemath.com/modules/numbprop.htm">Distributive Property</a>


But then you undid the error by making the same mistake in reverse when you divided your initial erroneous result by *[tex \LARGE p]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{3p^2\ +\ 3}{p}\ \neq\ 3p\ +\ 3]


But


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{3p^2\ +\ 3p}{p}\ =\ 3p\ +\ 3]


Capisce?


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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