SOLUTION: Find a fourth-degree polynomial equation with integer coefficients that has the given numbers as roots. The given numbers are 3+ i and -2i

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Question 136630This question is from textbook Prentice Hall Algebra 2
: Find a fourth-degree polynomial equation with integer coefficients that has the given numbers as roots.
The given numbers are 3+ i and -2i
This question is from textbook Prentice Hall Algebra 2

Found 2 solutions by Fombitz, solver91311:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Complex roots always come in conjugate pairs.
If 3+i is a root, so is 3-i.
If -2i is a root, so is 2i.
f%28x%29=%28x-2i%29%28x%2B2i%29%28x-%283%2Bi%29%29%28x-%283-i%29%29
f%28x%29=%28x%5E2%2B4%29%28x%5E2%2B%283%2Bi%29%283-i%29%29
Work out the final steps to get the solution.
Post another question if you get stuck.

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
If a polynomial equation has a complex root of the form a%2Bbi, then the conjugate of the complex number, a-bi is also a root. Therefore, your four roots are 3%2Bi, 3-i, 0-2i, and 0%2B2i

If any number a is a root of a polynomial equation, then x-a is a factor of the polynomial. Therefore, the factors of your polynomial are x-%283%2Bi%29, x-%283-i%29, x-%28-2i%29, and x-2i. Just multiply the 4 factors together and you will have your required polynomial. The problem asks for a polynomial equation so remember to set the 4th degree polynomial result equal to 0 at the end.

Hint: Be very careful with your signs when multiplying. Remember that i%5E2=-1, so something like %28-2i%29%282i%29=-4%28-1%29=4.