Question 316701
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Use the following facts:


The Fundamental Theorem of Algebra, namely that every *[tex \Large n]-th degree polynomial function has exactly *[tex \Large n] zeros, counting all multiplicities.


Complex roots ALWAYS come in conjugate pairs.  That means that if *[tex \Large 0\ +\ 5i] is a zero, then *[tex \Large 0\ -\ 5i] is also a zero of the desired polynomial function.


If *[tex \Large \alpha] is a zero of a polynomial function in *[tex \Large x], then *[tex \Large x\ -\ \alpha] is a factor of the polynomial.


A family of polynomial functions of the form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ p(x)\ =\ k\left(a_0x^n\ +\ a_1x^{n-1}\ +\ \cdots\ +\ a_ix^{n-i}\ +\ \cdots\ +\ a_{n-1}x\ +\ a_n\right)]


all have the same zeros.


So we know the following things:


1.  The desired polynomial function has exactly 3 zeros.


2.  The zeros are *[tex \Large 1], *[tex \Large 5i], and *[tex \Large -5i].


3.  The factors of the polynomial, including any possible common factor multiplier are:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ k(x\ -\ 1)(x\ -\ 5i)(x\ +\ 5i)]


Multiply the factors:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ p(x)\ =\ k(x\ -\ 1)(x^2\ +\ 25)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ p(x)\ =\ k(x^3\ -\ x^2\ +\ 25x\ -\ 25)]


Since *[tex \LARGE f(-1)\ =\ -104]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ k((-1)^3\ -\ (-1)^2\ +\ 25(-1)\ -\ 25)\ =\ k\,\cdot\,(-52)]


Hence *[tex \LARGE k\ =\ 2] and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ 2x^3\ -\ 2x^2\ +\ 50x\ -\ 50]




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