Question 1156301
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Since the leading coefficient is 1, the Remainder theorem provides this list of possible zeros 
(all of them are divisors of the constant term 45, in this case)

   +/-1, +/-3, +/-5, +/-9, +/-15, +/-45.



Next, the plot below


    {{{graph( 330, 400, -5, 5, -50, 50,
          x^4 - 2x^3 - 10x^2 + 6x + 45
)}}}


    Plot y = {{{x^4 - 2x^3 - 10x^2 + 6x + 45}}}


shows the root x= 3 of the multiplicity at least 2.


So, I divide  {{{x^4 - 2x^3 - 10x^2 + 6x + 45}}}  by  {{{(x-3)^2}}},  and I get the quotient  {{{x^2 + 4x + 5}}}.


This quotient is a quadratic polynomial with negative discriminant, so it has no real roots.


Therefore, factoring over real numbers is


    {{{x^4 - 2x^3 - 10x^2 + 6x + 45}}} = {{{(x-3)^2*(x^2 + 4x + 5)}}}.


The quadratic polynomial  x^2 + 4x + 5  has no rational roots.

It has no real roots, too, since its discriminant d = (-4)^2 - 4*1*5 = 16 - 20 = -4 is negative.


It has two complex roots  {{{(-4 +- sqrt(-4))/2}}} = {{{-2 +- i}}}.


<U>ANSWER</U>.  The roots of the given polynomial are &nbsp;x= 3 &nbsp;of the multiplicity &nbsp;2 &nbsp;and 

         two complex roots &nbsp;{{{-2 + i}}} &nbsp;and  &nbsp;{{{-2 - i}}} &nbsp;of the multiplicity 1 each.
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Solved.