SOLUTION: Using the given zero, find one other zero of f(x). Explain the process you used to find your solution. 1 - 6i is a zero of f(x) = x4 - 2x3 + 38x2

SOLUTION: Using the given zero, find one other zero of f(x). Explain the process you used to find your solution. 1 - 6i is a zero of f(x) = x4 - 2x3 + 38x2 - 2x + 37.

Question 1132901: Using the given zero, find one other zero of f(x). Explain the process you used to find your solution.
1 - 6i is a zero of f(x) = x4 - 2x3 + 38x2 - 2x + 37.
Answer by greenestamps(13209) (Show Source): You can put this solution on YOUR website!

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.

So another root is 1+6i; two of the roots are 1-6i and 1+6i.

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.

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

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.

So the four roots of the given polynomial are i, -i, 1-6i, and 1+6i.
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