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Algebra 2 Finding Zeros. HELP!

  1. Jan 30, 2009 #1
    For each of the following polynomials, one zero is given. List all zeros of the polynomials.

    1. P(x) = x^3 - 3x - 2, -1 is a zero
    A. -1, -2
    B. -1, 2
    C. -1 of multiplicity 2, 2
    D. -1, 1, 2

    2. P(x) = x^3 - 6x^2 + 11x - 6 , 3 is a zero
    A. 3, -2, -1
    B. 3, 2 of multiplicity 2
    C. 3, 1 of multiplicity 2
    D. 3, 2, 1

    3. P(x) = 6x^3 + 19x^2 + 2x - 3 , -3 is a zero
    A. There are no additional zeros
    B. -3, 3, 2
    C. -3, -1/3, 1/2
    D. -3, 1/3, -1/2

    4. P(x) = x^4 + 2x^3 - 7x^2 - 20x - 12, -2 is a zero of multiplicity 2
    A. -2 of multiplicity 2, -3, 1
    B. -2 of multiplicity 2, 6, 1
    C. -2 of multiplicity 2, -6, -1
    D. -2 of multiplicity 2, 3, -1

    5. P(x) = x^3 + x^2 - 4x - 24, -2+2i is a zero
    A. -2 + 2i, -2 - 2i, -3
    B. -2 + 2i, -2 - 2i
    C. -2 + 2i, 2 - 2i, 3
    D. -2 + 2i, -2 - 2i, 3


    I kinda understand how to work out the problems. But 3, 4, & 5 are giving me problems. Help?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 30, 2009 #2

    Dick

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    Homework Helper

    If you are having trouble with, for example, number 3, so us how you tried to do it. You divided the polynomial by (x+3), right?
     
  4. Jan 30, 2009 #3
    If you are trying to see if something is a zero... plug that value in. IF that value is a zero, what should you get?
     
  5. Jan 30, 2009 #4

    jgens

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    Gold Member

    Additionally, for the zeros of multiplicity 2, you know that the original function contains two factors of (x - a) where f(a) = 0.
     
  6. Jan 30, 2009 #5
    I'm pretty sure I got one and two right.

    3. I worked out the problem using synthetic division, which is what I was told to do, and the answer came out even. I got:

    6, 1, -1, and 0.

    No remainder, completely even. But that's not one of the options I have to chose from, so I must being doing something wrong.
     
  7. Jan 30, 2009 #6
    That's impossible. You have a cubic which means it will have 3 roots (not necessarily all real).

    Plus use what I told you, when you plug in your root, what should you expect to get? What happens when you plug in yours? For example 0.
     
  8. Jan 30, 2009 #7

    Dick

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    Loonygirl is giving you the results of the synthetic division, not the roots, so the polynomial has been factored into (x+3)(6x^2+x-1). Now solve (or factor) the quadratic.
     
  9. Jan 30, 2009 #8
    After reading Dick's response, I turned in the assignment. I thought I had everything right... And I got 2 out of 5 right. Can someone tell me where I went wrong?

    Here's the answers I turned in:

    1. D. -1, 1, 2

    2. A. 3, -2, -1

    3. C. -3, -1/3, 1/2


    And I got 4 and 5 right... the hardest ones! I used synthetic division on 1-3. I thought I had it right....
     
  10. Jan 30, 2009 #9

    Mark44

    Staff: Mentor

    For 1, the correct answer is B, since the zeroes are -1 and 2 (-1 is a zero of multiplicity 2).
    For 3, the correct answer is D. The zeroes are -3, 1/3, and -1/2.

    Be careful with synthetic division: it's easy to forget what you're doing and what all the numbers mean. One good thing about factoring polynomials is that after you have one factored, you can check your work by multiplying your factors. You should get what you started with.
     
  11. Feb 5, 2009 #10
    With number 5 - how do you do synthetic division with an imaginary number?
     
  12. Feb 5, 2009 #11

    Dick

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    You do it just like you do it with real numbers. You have to do complex arithmetic instead of real. But that's the hard way to do it. Since the polynomial is real you know that if -2+2i is a root then so is -2-2i (the complex conjugate). If you multiply (x-(-2+2i))(x-(-2-2i)) out it will give you a real quadratic that will divide P(x).
     
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