Question 999717
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Find a cubic polynomial in standard form with real coefficients, having the zeros 2 and 6i. Let the leading coefficient be 1
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(x-2)*(x-6i)*(x-(-6i)) = {{{(x-2)*(x^2 + 36)}}}.


1.  If a polynomial with real coefficients has a complex root,  it has the conjugate complex number as a root,  too.


2.  If a polynomial  f(x)  of a degree  n  with leading coefficient  1  has  n  roots  {{{x[1]}}}, {{{x[2]}}}, . . . , {{{x[n]}}}  then the polynomial is the product 


      f(x) = {{{(x-x[1])*(x-x[2])* ellipsis * (x-x[n])}}}.