Posted by **Jacinta** on Wednesday, July 22, 2009 at 11:20pm.

Suppose that a polynomial function of degree 5 with rational coefficients has 0 (with multiplicity 2), 3, and 1 –2i as zeros. Find the remaining zero.

A. –2

B. –1 – 2i

C. 0

D. 1 + 2i

- Math Help Please -
**DD**, Wednesday, July 22, 2009 at 11:28pm
So the funtion is:

x^2(x - 7)(x + 3 - 5i)(x - 3 + 5i)

For it to have rational coefficients, you must get rid of the -3 + 5i term.

Therefore the remaining zero is:

3 - 5i

so your answer would be (C)

- Math Help Please -
**Jacinta**, Wednesday, July 22, 2009 at 11:37pm
How can the answer be 0 ? I don't get it, I figured to be -1-2i Choice B. What did i do wrong? Thanks!

- Math Help Please -
**MathMate**, Wednesday, July 22, 2009 at 11:49pm
There is a little rule of thumb that you can count on for finding zeroes of polynomials with complex roots.

Complex roots always come in pairs. Each root of the pair is the complex conjugate of the other.

For example, if you have a root as 4+3i, the other root must be 4-3i. If another complex root is -2-i, then its conjugate is -2+i, etc.

To get the correct answer to the given problem, you only need to choose the complex conjugate of the given complex root.

- Math Help Please -
**MathMate**, Wednesday, July 22, 2009 at 11:56pm
The function would be

f(x)=x²(x-3)(x-1+2i)(x-1-2i)

= x^{5}-5x^{4}+11x³-15x²

- Math Help Please -
**kimora**, Thursday, April 23, 2015 at 4:59pm
7+what equal 125

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