Math Help Please

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

  1. 1
asked by Jacinta
  1. 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)

    posted by DD
  2. 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!

    posted by Jacinta
  3. 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.

    posted by MathMate
  4. The function would be
    f(x)=x²(x-3)(x-1+2i)(x-1-2i)
    = x5-5x4+11x³-15x²

    posted by MathMate
  5. 7+what equal 125

    posted by kimora

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