Question 987565
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Per the rational roots theorem, if this function has any rational zeros they must belong to the set *[tex \Large \left\{-1,\,-\frac{1}{2},\,\frac{1}{2},\,1\right\}]
<pre>
1  |  2   -5   4   -1
           2  -3    1
---------------------
      2   -3   1    0
</pre>
Hence, 1 is a zero, and *[tex \Large (x\ -\ 1)] and *[tex \Large (2x^2\ -\ 3x\ +\ 1)] are factors.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  2x^2\ -\ 3x\ +\ 1\ =\ 0]


which factors to


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  (x\ -\ 1)(2x\ -\ 1)\ =\ 0]


That gives us zeros for the original function at 1, 1, and *[tex \Large \frac{1}{2}]
<pre>
                        -&#8734;  |  1/2  |   1   |  &#8734;
-------------------------------------------------
x - 1                       -       -       +
x - 1                       -       -       +
2x - 1                      -       +       +
-------------------------------------------------
(2x-1)(x-1)^2               -       +       +
</pre>
*[illustration cubic.jpg].

John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \