Question 324418
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Sorry, you don't have any equations here -- there are no equal signs.  You only have expressions.  You can factor the polynomial expressions but you cannot solve anything.


I'm skipping the first one because it can't be done.  In the first one you have two different variables:  *[tex \Large x] and *[tex \Large X]


ii: *[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^5\ +\ 3x^4\ +\ 4x^2\ +\ 4x\ -\ 16]


Possible rational factors:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \pm1,\ \pm2,\ \pm4,\ \pm8,\ \pm16]


Unfortunately, none of them actually are factors.  The expression does not factor over the rational numbers, and since there is no general solution for the quintic equation, you would be forced to use numerical analysis techniques to approximate the values of the three real number zeros of this polynomial.


For the third one, I'm pretty sure you have a typo.  You wrote:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^4\ +\ 4x^3\ +\ *x^2\ +\ 16x\ +\ 16]


But I think perhaps you meant:


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


Again, the possible rational zeros are:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \pm1,\ \pm2,\ \pm4,\ \pm8,\ \pm16]


Using synthetic division, -2 works twice, leaving a quotient of 1 0 4


Hence, the factors are:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ +\ 2)(x\ +\ 2)\left(x^2\ +\ 4\right)]


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