Question 1132901
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The polynomial has integer coefficients, and one of the roots is complex.  That means another of the roots is the conjugate of the given complex number.<br>
So another root is 1+6i; two of the roots are 1-6i and 1+6i.<br>
The problem would have been more interesting -- and there would have been a lot more useful mathematics in it -- if it had asked us to find ALL the roots.  So I'll go ahead and show you how that can be done.<br>
Use Vieta's theorem to find the quadratic with those two roots:
(1) coefficient of the linear term is the opposite of the sum of the roots: -2
(2) constant term is the product of the roots: (1-6i)(1+6i) = 1+36 = 37<br>
So the given polynomial is divisible by the quadratic polynomial x^2-2x+37.  Long division (or any other method) then shows that the other quadratic factor is x^2+1; the roots of that quadratic are i and -i.<br>
So the four roots of the given polynomial are i, -i, 1-6i, and 1+6i.