Question 326078
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Use the caret mark, "^" (Shift-6) to indicate raising to a power.  I'm presuming you meant:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(8x^2\ +\ 5x\ -\ 3\right)\ +\ \left(12x^2\ +\ 7x\ -\ 14\right)]


which can be rendered using a standard keyboard by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ](8x^2 + 5x - 3) + (12x^2 + 7x - 14)


Enough housekeeping, let's do the math.


First get rid of the parentheses.  In front of the first set of parentheses there is an understood coefficient of 1.  Multiplying anything by 1 doesn't change anything, so just drop the parentheses altogether.  In this case you also have an understood coefficient of 1 in front of the second set of parentheses -- they go away too:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8x^2\ +\ 5x\ -\ 3\ +\ 12x^2\ +\ 7x\ -\ 14]


Terms are the chunks that are separated by + and - signs.  You can combine like terms.  Like terms have the same variable(s) raised to the same power(s).  The *[tex \Large 8x^2] and the *[tex \Large 12x^2] are like terms because *[tex \Large x^2] exactly matches *[tex \Large x^2].  Sort of like saying I have 8 bananas and 12 bananas so I have 20 bananas.  (but you can't add in the *[tex \Large 5x] term because that would be adding in the apples while you are counting bananas) So we have *[tex \Large 8x^2\ +\ 12x^2\ =\ 20x^2].  Likewise, add up the terms that have *[tex \Large x] in them, namely *[tex \Large 5x] and *[tex \Large 7x].  And finally add up the terms that are just constant coefficients, the *[tex \Large -3] and the *[tex \Large -14].


All together you should have:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 20x^2\ +\ 12x\ -\ 17]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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