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  • phi
a polynomial with real coefficients (that kind people normally work with) may have complex roots. But if they do have complex roots they come in "complex conjugate pairs" that means if one root is a+bi there is another root a-bi (the imaginary part is negated) in your case you have a 3rd root 6-i (the complex conjugate of the 6+i)
  • phi
if you have a root x= -4 for example that means x - (-4) =0 if you had another root x= 3 that means x-3 = 0 (x- -4)(x-3)=0 would be the polynomial (x+4)(x-3)=0 you have to use this idea to find your polynomial
  • phi
yes now multiply the (x-6-i)(x-6+i) I would write this as ( (x-6) -i) ((x-6) + i) (I just put parens around the real part) now use (a+b)(a-b)= a^2 - b^2 here a= (x-6) and b= i then multiply that mess by (x+4)

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thankyou :)

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