Question 327162
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4x^2\ -\ 8x\ +\ 6]


*[tex \Large 2x] goes into *[tex \Large 4x^2\ \ \ \ ] *[tex \Large 2x] times.


Hence the first term of *[tex \Large q(x)] is *[tex \Large 2x]


*[tex \Large 2x] times *[tex \Large -1] is *[tex \Large -2x] and *[tex \Large 2x] times *[tex \Large 2x] is *[tex \Large 4x^2].


*[tex \Large (4x^2\ -\ 8x)\ -\ (4x^2\ -\ 2x)\ =\ -6x]


Bring down the *[tex \Large 6]


*[tex \Large 2x] goes into *[tex \Large -6x\ \ \ ] *[tex \Large -3] times.


So the second term of *[tex \Large q(x)] is *[tex \Large -3]


*[tex \Large -3] times *[tex \Large -1] is *[tex \Large 3] and *[tex \Large -3] times *[tex \Large 2x] is *[tex \Large -6x].


*[tex \Large (-6x\ +\ 6)\ -\ (-6x\ -\ 3)\ =\ 9]


So *[tex \Large q(x)\ =\ 2x\ -\ 3] and *[tex \Large r(x)\ =\ 9]


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