Question 1181501
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Check that (2 + i) is a root of (z^4) + (2(z^3)) - (9(z^2)) - 10z + 50 = 0. What are the remaining 3 roots?
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            Yes, from the first glance,  it is almost impregnable fortress.

            Let apply a military ruse.



<pre>
Our polynomial is with real  ( even with integer (!) )  coefficients.

Hence, if (2+i) is a root, then (2-i) is a root, too (!)


If so, then our polynomial must be divisible by  (z-(2+i))*(z-(2-i)) = ((z-2)-i)*((z-2)-i) = (z-2)^2 + 1 = z^2 - 2z +5.


Lets make long division to check if it is TRUE:


    {{{(z^4 + 2z^3 -9z^2 - 10z + 50)/(z^2 - 2z + 5)}}} = {{{z^2 + 6z + 10)}}}     with NO REMAINDER (!)


Thus we checked that the given polynomial is a multiple of the polynomial z^2 - 2z + 5.


It means that (2+i) REALLY is a root of the given polynomial (without direct calculations (!))


    +------------------------------------------------------------------------------------------------+
    |    So, we know now that                                                                        |
    |                                                                                                |
    |        (2+i) really is the root of the given polynomial;                                       |
    |        (2-i) is the other root;                                                                |
    |        one factor to the given polynomial is  z^2 - 2z + 5  with the roots  (2+i) and (2-i);   |
    |        the other factor is the polynomial  z^2 +6z + 10.                                       |
    +------------------------------------------------------------------------------------------------+


At this point, the last step to do is to find the roots of the quadratic polynomial  z^2 + 6z + 10.


Apply the quadratic formula to get


        {{{z[1,2]}}} = {{{(-6 +- sqrt(6^2 - 4*10))/2}}} = {{{(-6 +- sqrt(-4))/2}}} = {{{(-6 +- 2i)/2}}} = -3 +- i.


The problem is just solved in full.


We checked and proved that 2+i is the root.

From it, we concluded that 2-i is the root, too.

And finally, we found two remaining roots  -3+i  and  -3-i.
</pre>

Solved, answered, carefully explained and totally completed.



<H3>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;T R I U M P H &nbsp;&nbsp;&nbsp;&nbsp;(!)</H3>


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In his post, &nbsp;tutor @greenestamps writes &nbsp;(cited)


<pre>
    In the hopes of making the problem easier, we can, as the other tutor did, assume ("hope") that the polynomial has real coefficients.
</pre>


I want to &nbsp;HIGHLIGHT &nbsp;that it is not an assumption: &nbsp;the given polynomial &nbsp;REALLY &nbsp;has integer &nbsp;(hence, &nbsp;real) &nbsp;coefficients.


It is not an assumption: &nbsp;it is a &nbsp;FACT.