Question 819909
The possible rational roots/zeros of {{{f(x)= x^5 - 7x^4 + 10x^3 + 26x^2 - 47x - 39 }}} are:
<u>+</u>1, <u>+</u>3, <u>+</u>13, <u>+</u>39<br>
Using synthetic division to test these we will find that only one of these, 3, is actually a zero:
<pre>
3  |   1   -7   10   26   -47   -39
----        3  -12   -6    60    39
      ------------------------------
       1   -4   -2   20    13     0
</pre>Not only does the zero in the lower right indicate that 3 is a zero (and (x-3) is a factor) but the rest of the bottom row tells us the other factor. The "1 -4 -2 20 13" translates into {{{x^4-4x^3-2x^2+20x+13}}}. So
{{{f(x) = (x-3)(x^4-4x^3-2x^2+20x+13)}}}<br>
As mentioned earlier, none of the other possible rational roots/zeros work. And, if "given that x = 3+ 2i" is supposed to indicate that 3+2i is also a zero, then this is an error. 3+2i is <i>not</i> a zero of f(x)! It is was, then 3-2i would also be a zero. And if they were both zeros then {{{(x-(3+2i))}}} and {{{(x-(3-2i))}}} would be factors of f(x). And since {{{(x-(3+2i))*(x-(3-2i)) = (x^2-6x+13)}}}, {{{(x^2-6x+13)}}} would also be a factor of f(x). And if {{{(x^2-6x+13)}}} were a factor of f(x) then it would divide evenly into f(x). But it does not!! So either there is an error in what you posted or I am misinterpreting "given that x = 3+ 2i".<br>
In summary:<ul><li>x = 3 is the only zero we can find.</li><li>3+2i (and 3-2i) are <i>not</i> zeros of f(x).</li></ul>
P.S. The only way I know to find approximations for the remaining 4 zeros would be to<ol><li>Use a graphing calculator to graph f(x) (or {{{y = x^4-4x^3-2x^2+20x+13}}})</li><li>Use the trace function to find approximations for where the graph crosses the x-axis.</li></ol>