Question 627770
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First, Outside, Inside, Last


Using your example:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (2x\ -\ 3)(x\ +\ 3)]


The Firsts are the first terms in each of the binomials, namely *[tex \LARGE 2x] and *[tex \LARGE x].


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (\underline{2x}\ -\ 3)(\underline{x}\ +\ 3)]


Multiply these together and get *[tex \LARGE 2x^2]


The Outsides are the first term in the first binomial and the second term in the second binomial, namely *[tex \LARGE 2x] and *[tex \LARGE 3].


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (\underline{2x}\ -\ 3)(x\ +\ \underline{3})]


Multiply these together and get *[tex \LARGE 6x]


The Insides are the second term in the first binomial and the first term in the second binomial:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (2x\ \underline{-\ 3})(\underline{x}\ +\ 3)]


Notice that because there is a minus sign, you have to take it along for the ride.  Multiply these two together and get *[tex \LARGE -3x]


And finally, so to speak, are the Lasts, the second term in each of the binomials:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (2x\ \underline{-\ 3})(x\ +\ \underline{3})]


Multiply these together and get *[tex \LARGE -9]


Now add up all 4 results:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x^2\ +\ 6x\ -\ 3x\ -\ 9]


And finally collect like terms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x^2\ +\ 3x\ -\ 9]


Done.  By the way, the answer is NOT *[tex \LARGE 2x^3\ + 3x\ -\ 9] like you indicated.  The high order term is 2nd degree, not 3rd degree.


Now, Jesyka, I know you are trying to sound both blonde and new at the same time, perhaps in a bid to generate sympathy, but it isn't working and it really isn't necessary.  Any college that would admit a student that cannot spell words like student, math, and simplify is, in my view, hardly worth attending.  You might want to rethink your educational strategies.  Your misspellings, use of "textspeak", and general playing dumb are annoying and disrespectful.  If you truly want to be a college student, grow up and act like one.


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