Question 986887
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It is exactly the same thing as finding roots.


Use synthetic division.  I'll do plus and minus 1 for you:
<pre>
  1  |  1   4    1   -6
            1    5    6
     ------------------
        1   5    6    0
The remainder is zero, so *[tex \Large x - 1] is a factor, so 1 is a root.

 -1  |  1   4    1   -6
           -1   -3    2
     ------------------
        1   3   -2    4
The remainder is not zero, so *[tex \Large x + 1] is not a factor, so -1 is not a root.
</pre>
Note that in the division that proved *[tex \Large x - 1] is a factor, the quotient is *[tex \Large x^2\ +\ 5x\ +\ 6], a factorable quadratic polynomial for which you should be able to determine the zeros.


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

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