Question 254570
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ x^3\ +\ 3x^2\ +\ 25x\ +\ 75]


The integer factors of the constant term are 1, 3, and 5 and this is a monic polynomial, therefore the only possible rational roots are:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \pm 1,\,\pm 2,\,\pm 5]


Use either polynomial long division or synthetic division to discover the one real (and fortunately rational) root.  The divisor that yields the zero remainder will be the rational factor.  The quotient will be a quadratic with a negative discriminant.  Simply solve it for the conjugate pair of complex roots.


Alternatively, you can group and factor your original function.  Take *[tex \LARGE x^2] out of the first two terms and take 25 out of the last two terms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ x^2\left(x\ +\ 3\right)\ +\ 25\left(x\ +\ 3\right)\ =\ \left(x^2\ +\ 25\right)\left(x\ +\ 3\right)] 


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