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ineedbiohelpandquick

  • 4 years ago

write (solve for) a polynomial with the following zeros a) 2 and -3i

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  1. nenadmatematika
    • 4 years ago
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    on of the polynomials is (x-2)(x^2+9)

  2. ineedbiohelpandquick
    • 4 years ago
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    Can you explain how you got that answer please?:) thanks!

  3. nenadmatematika
    • 4 years ago
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    if 2 is the zero of the polynomial then that polynomilal can be divided with (x-2) without rest....if 3i is the zero, then it's dividable with (x-3i)....you must know that 3i is the complex zero, and there is always one friend following him...it's -3i so this polynomial is dividable with (x-(-3i))=x+3i.....now you can multiply all the parenthesis and you get your polynomial....(x-2)(x-3i)(x+3i)....those last two give the difference of the squares...so you can write it as I did above....of course, you can multiply them also...:D

  4. ineedbiohelpandquick
    • 4 years ago
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    You are amazing! Thanks :)

  5. nenadmatematika
    • 4 years ago
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    if you'd like to see if you got this right, I can give you one example, and you can solve it here....check this one out....|dw:1330286335128:dw|

  6. ineedbiohelpandquick
    • 4 years ago
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    Is that the final answer or do you have to multiply it out to get those zeros?

  7. ineedbiohelpandquick
    • 4 years ago
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    whoops ignore what I just said. :)

  8. nenadmatematika
    • 4 years ago
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    no no, it's OK...you have zeros given, and your task is to find polynomial which zeros are given for you....

  9. ineedbiohelpandquick
    • 4 years ago
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    so like the (x-2) (x^2 +9) is ur final answer? sorry that is were i am confused haha

  10. nenadmatematika
    • 4 years ago
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    you can leave it like that, but most professors would like to multiply...do you know how to do it"?

  11. ineedbiohelpandquick
    • 4 years ago
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    Can you help explain that to me? if you have time:) .... ALSO my next question is 3 and 2+2i so the answer would just be... (3-x) and (x+2+2i) and (x-2+2i) ??

  12. nenadmatematika
    • 4 years ago
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    well, OK lets start like this...do you know to multiply for example (x-2)(x+3)? it's some basic stuff you must know before you continue working with polynomials....multiply this so I can see...:D

  13. ineedbiohelpandquick
    • 4 years ago
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    yes isn't it x^2+1x-6

  14. nenadmatematika
    • 4 years ago
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    that's right....now we continue with this....for example if I tell you that one zero of the polynomial is some number, lets call it ''a'' that means that your polynomial is divisible with (x-a)...OK....now try to continue my next sentence: If -a is the zero of the polynomial, then your polynomial is divisible with...?

  15. ineedbiohelpandquick
    • 4 years ago
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    (x+a)?

  16. nenadmatematika
    • 4 years ago
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    great :D chocolate for you :D

  17. ineedbiohelpandquick
    • 4 years ago
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    haha! yeah!

  18. ineedbiohelpandquick
    • 4 years ago
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    so the final answer for the first one is: x^3-2x^2+9x-18?

  19. nenadmatematika
    • 4 years ago
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    now, If I tell you this: Polynomial has zeros a and -a, so I can write that polynomial as (x-a)(x+a) and if I multiply those two I'll get x^2-a^2....now you anwer: Polynomial has zeros 1 and -6 then you can write you polynomial as ...??? and if you multiply those two you'll get ...????? :D try

  20. ineedbiohelpandquick
    • 4 years ago
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    polynomial is (x-1) (x+6) and final answer is x^2+5x-6

  21. ineedbiohelpandquick
    • 4 years ago
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    what is the polynomial of 2+2i though is it just (x-2+2i) and (x+2+2i)

  22. nenadmatematika
    • 4 years ago
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    awesome!!! :D now the next step: If polynomial has zero i then it's divisible with (x-i)...now I told you earlier that i is ''complex zero'', and it never appears alone, it appears together with -i....so write that polynomial....and when you multiply use very important formula i^2=-1...try :D

  23. ineedbiohelpandquick
    • 4 years ago
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    so i just multiply (factor) the (x-2+2i) and (x+2+2i) and (x-3) together haha

  24. nenadmatematika
    • 4 years ago
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    be careful: if the given zero is 2+2i that means that your polynomial is divisible by x-(2+2i)...and it will appear with the ''friend'' 2-2i so its divisible also by (x-(2-2i)) so when you multiply those two, one thing is very important...every ''i'' in your polynomial must be cancelled....try to do it, so I can be sure you're not making any mistakes...:D

  25. ineedbiohelpandquick
    • 4 years ago
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    (x-3) (8) i think i did something wrong

  26. ineedbiohelpandquick
    • 4 years ago
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    i will show u what I did

  27. ineedbiohelpandquick
    • 4 years ago
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    (x-3) (2+2i) (2-2i) (x-3) (4-4i+4i-4i^2) (x-3) (4+4) (x-3) (8)

  28. nenadmatematika
    • 4 years ago
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    no no...you have to multiply (x-(2-2i))*((x-(2+2i))...try again, and be patient..:D

  29. ineedbiohelpandquick
    • 4 years ago
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    ohh! so do you distribute the x to the 2-2i and 2+2i

  30. nenadmatematika
    • 4 years ago
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    I didn't get that...just multiply and show me the procedure....

  31. nenadmatematika
    • 4 years ago
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    I'm back in two minutes :D

  32. ineedbiohelpandquick
    • 4 years ago
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    (x-3) (x-2+2i) (x-2-2i) is that right so far? :(

  33. nenadmatematika
    • 4 years ago
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    yes, go,go,go...:D

  34. ineedbiohelpandquick
    • 4 years ago
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    what do you do though when you multiply the x by -2i is it just -2ix?

  35. nenadmatematika
    • 4 years ago
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    of course, continue...;D

  36. ineedbiohelpandquick
    • 4 years ago
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    okay lets see... (x-3) (x^2-2x-2xi-2x+4-4i^2+2ix-4i-4i-4i^2) x^2-4x-4i^2+4+4-4i+4 x^2-4x+12-4i

  37. nenadmatematika
    • 4 years ago
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    number ''i'' must be gone after mulitplication...try again, look out for signs, and of course use i^2=-1

  38. ineedbiohelpandquick
    • 4 years ago
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    is the final answer x^2-4x+4

  39. ineedbiohelpandquick
    • 4 years ago
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    thanks for all ur help:) btw

  40. nenadmatematika
    • 4 years ago
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    no, try again.....the result you got now is (x-2)^2. That's polynomial with double zero x=2 so there's no 2+2i etc....you must try again....

  41. ineedbiohelpandquick
    • 4 years ago
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    well now i got something crazy... x^4-10x+12 :((

  42. nenadmatematika
    • 4 years ago
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    OK I'll do it...follow me carefuly: (x-2+2i)(x-2-2i)=x^2-2x-2xi-2x+4+4i+2xi-4i+4=x^2-4x+8 do you get it? you must be careful with doing this

  43. ineedbiohelpandquick
    • 4 years ago
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    okay i get that part

  44. nenadmatematika
    • 4 years ago
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    hahah...well that part is all!!! :D I hope I helped you :D

  45. ineedbiohelpandquick
    • 4 years ago
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    Oh i was just combining it all wrong! thank you!!! :) yes you did!!

  46. nenadmatematika
    • 4 years ago
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    you're welcome...when you have any troubles with math be free to contact me, I'll be glad to help you :D

  47. ineedbiohelpandquick
    • 4 years ago
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    awesome for sure! I am pretty weak at math so I will for sure be contacting you!

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