Question 187560
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  (x–4)(2x^2+4)+(x+4)(x^3–1)– (x^2–x–1)]


Use FOIL to multiply your binomials:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (2x^3 - 8x^2  + 4x -16)+(x^4+4x^3-x-4)-(x^2–x–1)]


Now you can remove the parenthesese because you no longer have to segregate things:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x^3 - 8x^2  + 4x -16 +x^4 +4x^3 -x -4 -x^2 –x –1]


"Like Terms" are terms that have the same exponent on the variable.  Start with the highest exponent.  You have 1 *[tex \Large x^4] term:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^4]


And 2 terms containing *[tex \Large x^3], namely *[tex \Large 2x^3] and *[tex \Large 4x^3] which added together make *[tex \Large 6x^3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^4 + 6x^3]


And 2 terms containing *[tex \Large x^2], namely *[tex \Large -8x^2] and *[tex \Large -x^2] which added together make *[tex \Large -9x^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^4 + 6x^3 - 9x^2]


And 3 terms containing *[tex \Large x^1 = x], namely *[tex \Large 4x], *[tex \Large -x], and *[tex \Large -x] which added together make *[tex \Large 2x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^4 + 6x^3 - 9x^2 + 2x]


And finally you have 3 terms that contain no variable at all (called the constant terms), namely -16, -4, and - 1 which added together make -21.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^4 + 6x^3 - 9x^2 + 2x - 21]


Technically speaking, the constant terms actually do have the variable in them.  It is just that the variable is *[tex \Large x^0] and anything raised to the zero power is 1, so we just leave it out.


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