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

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

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

nenadmatematika
 3 years ago
Best ResponseYou've already chosen the best response.1if 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

ineedbiohelpandquick
 3 years ago
Best ResponseYou've already chosen the best response.0You are amazing! Thanks :)

nenadmatematika
 3 years ago
Best ResponseYou've already chosen the best response.1if 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

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

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

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

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

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

ineedbiohelpandquick
 3 years ago
Best ResponseYou've already chosen the best response.0Can 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) ??

nenadmatematika
 3 years ago
Best ResponseYou've already chosen the best response.1well, 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

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

nenadmatematika
 3 years ago
Best ResponseYou've already chosen the best response.1that'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...?

nenadmatematika
 3 years ago
Best ResponseYou've already chosen the best response.1great :D chocolate for you :D

ineedbiohelpandquick
 3 years ago
Best ResponseYou've already chosen the best response.0haha! yeah!

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

nenadmatematika
 3 years ago
Best ResponseYou've already chosen the best response.1now, 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

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

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

nenadmatematika
 3 years ago
Best ResponseYou've already chosen the best response.1awesome!!! :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

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

nenadmatematika
 3 years ago
Best ResponseYou've already chosen the best response.1be 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

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

ineedbiohelpandquick
 3 years ago
Best ResponseYou've already chosen the best response.0i will show u what I did

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

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

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

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

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

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

nenadmatematika
 3 years ago
Best ResponseYou've already chosen the best response.1yes, go,go,go...:D

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

nenadmatematika
 3 years ago
Best ResponseYou've already chosen the best response.1of course, continue...;D

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

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

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

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

nenadmatematika
 3 years ago
Best ResponseYou've already chosen the best response.1no, 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....

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

nenadmatematika
 3 years ago
Best ResponseYou've already chosen the best response.1OK 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

ineedbiohelpandquick
 3 years ago
Best ResponseYou've already chosen the best response.0okay i get that part

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

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

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

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