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anonymous

  • 5 years ago

Find a polynomail equation with real coefficients that has the given zeros. 1-5i and 1+5i

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  1. amistre64
    • 5 years ago
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    1-5i + n = 0 1+5i +m = 0 n= -1+5i m= -1-5i (1-5i -1+5i) (1+5i -1-5i)

  2. amistre64
    • 5 years ago
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    1 +5i -1 -5i -5i -25i^2 +5i +25i^2 -1 -5i +1 +5i 5i +25i^2 -5i -25i^2

  3. amistre64
    • 5 years ago
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    guess it aint gonna work out like i thought :) gotta head to marketing class now; Ciao :)

  4. anonymous
    • 5 years ago
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    ? Ok.

  5. anonymous
    • 5 years ago
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    \[ f(x) = (x-(1+5i))(x-(1-5i)) \] \[ = x^2 - x(1-5i) -x(1+5i) + (1+5i)(1-5i)\] \[ = x^2 -x +5xi - x - 5xi + 1 -5i + 5i - 25i^2\] \[ = x^2 - 2x + 1 + 25 \] \[ = x^2 - 2x + 26 \] \[ \rightarrow f(1-5i) = 0, f(1+5i) = 0\]

  6. anonymous
    • 5 years ago
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    When you are doing this problem: I understand the first equation is just subtracting from x then the secong equation you are distributing... why do you add the additonal (1+5i)(1-5i)??

  7. anonymous
    • 5 years ago
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    It's all the same equation. I didn't add the (1+5i)(1-5i). That term is the result of multiplying the last term in each factor.. \[ (a - b)(a - c) = a^2 -ac -ab +bc\] Where b in this case is (1+5i) and c is (1-5i)

  8. anonymous
    • 5 years ago
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    Ok..I dont understand. But thank you so much for you help. I have the answer just not sure how I got it... thanks again.

  9. anonymous
    • 5 years ago
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    Well wait. Which part don't you understand?

  10. anonymous
    • 5 years ago
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    It doesn't do any good to have the answer and not understand how you got it. Otherwise you won't be able to use this tool on more complicated tasks.

  11. anonymous
    • 5 years ago
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    Do you agree that \[ (x-a)(x-b) = x^2 - ax -bx +ab \]

  12. anonymous
    • 5 years ago
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    Yes I agree. That would be just simply distrubtng....

  13. anonymous
    • 5 years ago
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    Right. Do you also agree that if x=a or x=b then the product (x-a)(x-b) would be 0 because one of the factors would be 0?

  14. anonymous
    • 5 years ago
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    Did I lose you?

  15. anonymous
    • 5 years ago
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    Kinda.. because I dont know where the zero comes from??

  16. anonymous
    • 5 years ago
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    \[ If\ x = a, then\ (x-a) = (a-a) = 0 \rightarrow (x-a)(x-b) = 0*(a-b) = 0\]

  17. anonymous
    • 5 years ago
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    Thanks for your help.. I dont want to waste anymore of your time...

  18. anonymous
    • 5 years ago
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    ... You're not wasting my time. Did you understand my last sentence?

  19. anonymous
    • 5 years ago
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    \[ \rightarrow \] just means 'therefore'

  20. anonymous
    • 5 years ago
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    I wish I did...

  21. anonymous
    • 5 years ago
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    Let x = a. Then (x-a) = (a-a). (a-a) = ?

  22. anonymous
    • 5 years ago
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    0?

  23. anonymous
    • 5 years ago
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    Yes. a - a = 0. No matter what 'a' is.

  24. anonymous
    • 5 years ago
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    Ok.. so now that I know that.. what signifigance does that have in my problem?

  25. anonymous
    • 5 years ago
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    So what we're saying is that when x = a, (x-a) = 0. If (x-a)=0 then (x-a)(x-b) would be 0*(x-b) which is?

  26. anonymous
    • 5 years ago
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    The point of the problem is to find a polynomial that has 2 roots. That means that there are two different values for x at which the polynomial will be 0. We construct such a polynomial by writing an expression which has two factors (x-a)(x-b) Where a and b are the two roots (0's). When x is equal to either a or b, the whole expression will be equal to 0.

  27. anonymous
    • 5 years ago
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    Does that make sense?

  28. anonymous
    • 5 years ago
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    Thanks for your help. But I really need to go. Im not getting it. But appreciate your help so much...

  29. anonymous
    • 5 years ago
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    Give it some thought. And try to follow through what I've written. If you have questions, feel free to come back and ask.

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