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

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nenadmatematikaBest ResponseYou've already chosen the best response.1
on of the polynomials is (x2)(x^2+9)
 2 years ago

ineedbiohelpandquickBest ResponseYou've already chosen the best response.0
Can you explain how you got that answer please?:) thanks!
 2 years ago

nenadmatematikaBest ResponseYou've already chosen the best response.1
if 2 is the zero of the polynomial then that polynomilal can be divided with (x2) without rest....if 3i is the zero, then it's dividable with (x3i)....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....(x2)(x3i)(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
 2 years ago

ineedbiohelpandquickBest ResponseYou've already chosen the best response.0
You are amazing! Thanks :)
 2 years ago

nenadmatematikaBest ResponseYou've already chosen the best response.1
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
 2 years ago

ineedbiohelpandquickBest ResponseYou've already chosen the best response.0
Is that the final answer or do you have to multiply it out to get those zeros?
 2 years ago

ineedbiohelpandquickBest ResponseYou've already chosen the best response.0
whoops ignore what I just said. :)
 2 years ago

nenadmatematikaBest ResponseYou've already chosen the best response.1
no no, it's OK...you have zeros given, and your task is to find polynomial which zeros are given for you....
 2 years ago

ineedbiohelpandquickBest ResponseYou've already chosen the best response.0
so like the (x2) (x^2 +9) is ur final answer? sorry that is were i am confused haha
 2 years ago

nenadmatematikaBest ResponseYou've already chosen the best response.1
you can leave it like that, but most professors would like to multiply...do you know how to do it"?
 2 years ago

ineedbiohelpandquickBest ResponseYou've already chosen the best response.0
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... (3x) and (x+2+2i) and (x2+2i) ??
 2 years ago

nenadmatematikaBest ResponseYou've already chosen the best response.1
well, OK lets start like this...do you know to multiply for example (x2)(x+3)? it's some basic stuff you must know before you continue working with polynomials....multiply this so I can see...:D
 2 years ago

ineedbiohelpandquickBest ResponseYou've already chosen the best response.0
yes isn't it x^2+1x6
 2 years ago

nenadmatematikaBest ResponseYou've already chosen the best response.1
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 (xa)...OK....now try to continue my next sentence: If a is the zero of the polynomial, then your polynomial is divisible with...?
 2 years ago

nenadmatematikaBest ResponseYou've already chosen the best response.1
great :D chocolate for you :D
 2 years ago

ineedbiohelpandquickBest ResponseYou've already chosen the best response.0
haha! yeah!
 2 years ago

ineedbiohelpandquickBest ResponseYou've already chosen the best response.0
so the final answer for the first one is: x^32x^2+9x18?
 2 years ago

nenadmatematikaBest ResponseYou've already chosen the best response.1
now, If I tell you this: Polynomial has zeros a and a, so I can write that polynomial as (xa)(x+a) and if I multiply those two I'll get x^2a^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
 2 years ago

ineedbiohelpandquickBest ResponseYou've already chosen the best response.0
polynomial is (x1) (x+6) and final answer is x^2+5x6
 2 years ago

ineedbiohelpandquickBest ResponseYou've already chosen the best response.0
what is the polynomial of 2+2i though is it just (x2+2i) and (x+2+2i)
 2 years ago

nenadmatematikaBest ResponseYou've already chosen the best response.1
awesome!!! :D now the next step: If polynomial has zero i then it's divisible with (xi)...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
 2 years ago

ineedbiohelpandquickBest ResponseYou've already chosen the best response.0
so i just multiply (factor) the (x2+2i) and (x+2+2i) and (x3) together haha
 2 years ago

nenadmatematikaBest ResponseYou've already chosen the best response.1
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'' 22i so its divisible also by (x(22i)) 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
 2 years ago

ineedbiohelpandquickBest ResponseYou've already chosen the best response.0
(x3) (8) i think i did something wrong
 2 years ago

ineedbiohelpandquickBest ResponseYou've already chosen the best response.0
i will show u what I did
 2 years ago

ineedbiohelpandquickBest ResponseYou've already chosen the best response.0
(x3) (2+2i) (22i) (x3) (44i+4i4i^2) (x3) (4+4) (x3) (8)
 2 years ago

nenadmatematikaBest ResponseYou've already chosen the best response.1
no no...you have to multiply (x(22i))*((x(2+2i))...try again, and be patient..:D
 2 years ago

ineedbiohelpandquickBest ResponseYou've already chosen the best response.0
ohh! so do you distribute the x to the 22i and 2+2i
 2 years ago

nenadmatematikaBest ResponseYou've already chosen the best response.1
I didn't get that...just multiply and show me the procedure....
 2 years ago

nenadmatematikaBest ResponseYou've already chosen the best response.1
I'm back in two minutes :D
 2 years ago

ineedbiohelpandquickBest ResponseYou've already chosen the best response.0
(x3) (x2+2i) (x22i) is that right so far? :(
 2 years ago

nenadmatematikaBest ResponseYou've already chosen the best response.1
yes, go,go,go...:D
 2 years ago

ineedbiohelpandquickBest ResponseYou've already chosen the best response.0
what do you do though when you multiply the x by 2i is it just 2ix?
 2 years ago

nenadmatematikaBest ResponseYou've already chosen the best response.1
of course, continue...;D
 2 years ago

ineedbiohelpandquickBest ResponseYou've already chosen the best response.0
okay lets see... (x3) (x^22x2xi2x+44i^2+2ix4i4i4i^2) x^24x4i^2+4+44i+4 x^24x+124i
 2 years ago

nenadmatematikaBest ResponseYou've already chosen the best response.1
number ''i'' must be gone after mulitplication...try again, look out for signs, and of course use i^2=1
 2 years ago

ineedbiohelpandquickBest ResponseYou've already chosen the best response.0
is the final answer x^24x+4
 2 years ago

ineedbiohelpandquickBest ResponseYou've already chosen the best response.0
thanks for all ur help:) btw
 2 years ago

nenadmatematikaBest ResponseYou've already chosen the best response.1
no, try again.....the result you got now is (x2)^2. That's polynomial with double zero x=2 so there's no 2+2i etc....you must try again....
 2 years ago

ineedbiohelpandquickBest ResponseYou've already chosen the best response.0
well now i got something crazy... x^410x+12 :((
 2 years ago

nenadmatematikaBest ResponseYou've already chosen the best response.1
OK I'll do it...follow me carefuly: (x2+2i)(x22i)=x^22x2xi2x+4+4i+2xi4i+4=x^24x+8 do you get it? you must be careful with doing this
 2 years ago

ineedbiohelpandquickBest ResponseYou've already chosen the best response.0
okay i get that part
 2 years ago

nenadmatematikaBest ResponseYou've already chosen the best response.1
hahah...well that part is all!!! :D I hope I helped you :D
 2 years ago

ineedbiohelpandquickBest ResponseYou've already chosen the best response.0
Oh i was just combining it all wrong! thank you!!! :) yes you did!!
 2 years ago

nenadmatematikaBest ResponseYou've already chosen the best response.1
you're welcome...when you have any troubles with math be free to contact me, I'll be glad to help you :D
 2 years ago

ineedbiohelpandquickBest ResponseYou've already chosen the best response.0
awesome for sure! I am pretty weak at math so I will for sure be contacting you!
 2 years ago
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