Question 1172472
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Since the coefficients of the polynomial are real numbers (they even are integers (!) ),
it implies that together with the imaginary root 5i, its complex conjugate -5i is also the root.


Hence, the polynimial f(x) is divisible by (x-5i)*(x+5i) = {{{x^2 + 25}}}.


When you perform long division, you will get the quotient  q(x) = {{{f(x)/(x^2+25)}}} = {{{x^3 + x^2 - 2x - 2}}}.


You can factor this quotient further using grouping/re-grouping


    {{{x^3 + x^2 - 2x - 2}}} = {{{(x^3 + x^2)}}} - {{{(2x+2)}}} = {{{x^2*(x+1)}}} - {{{2*(x+1)}}} = {{{(x^2-2)*(x+1)}}}.


Therefore, the full decomposition of the given polynomial over complex number domain is


    f(x) = {{{(x-5i)*(x-5i)*(x+1)*(x-sqrt(2))*(x+sqrt(2))}}},


and its roots are  5i, -5i, {{{sqrt(2)}}}, {{{-sqrt(2)}}}  and -1.    <U>ANSWER</U>
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Solved.