Question 472203: Use the given zero to find the remaining zeros of the function.
 ; zero: 3i
A. -2, -3i, 4 - i, 4 + i
B. 2, -3i, -4 - i, -4 + i
C. -2, -3i, -4 - i, -4 + i
D. 2, -3i, 4 - i, 4 + i
Answer by richard1234(7193)   (Show Source):
You can put this solution on YOUR website! We know that since 3i is a root, then its complex conjugate -3i is also a root.
From here, there are two ways to solve it:
Solution 1: Divide the polynomial by (x-3i)(x+3i) (which is equal to x^2 + 3). Here, you will obtain a cubic polynomial, in which you will have to bash out another solution since the cubic formula is long and complicated. Once you find another root r, divide by x-r to obtain a quadratic, in which the two other solutions are trivial. Given the fact that the polynomial is fifth-degree, and that the coefficients are ugly, the solution is rather long and tedious.
Solution 2: This way is somewhat "cheating" (not really)...taking advantage that the question is multiple choice, we can say that by Vieta's formulas, the sum of the roots is 10. Since 3i and its conjugate -3i add up to 0, all we need to do is check whether the other three roots add up to 10. Choice D is the only one that does so, so D must be correct.
However, if this question were not multiple choice, you would have to go with Solution 1, even though it is tiresome. Solution 2 would work rather nicely on a test such as the SAT, as it will save lots of time.
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