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

- anonymous

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

- katieb

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- nenadmatematika

on of the polynomials is (x-2)(x^2+9)

- anonymous

Can you explain how you got that answer please?:) thanks!

- nenadmatematika

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

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- anonymous

You are amazing! Thanks :)

- nenadmatematika

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|

- anonymous

Is that the final answer or do you have to multiply it out to get those zeros?

- anonymous

whoops ignore what I just said. :)

- nenadmatematika

no no, it's OK...you have zeros given, and your task is to find polynomial which zeros are given for you....

- anonymous

so like the (x-2) (x^2 +9) is ur final answer? sorry that is were i am confused haha

- nenadmatematika

you can leave it like that, but most professors would like to multiply...do you know how to do it"?

- anonymous

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) ??

- nenadmatematika

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

- anonymous

yes isn't it x^2+1x-6

- nenadmatematika

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...?

- anonymous

(x+a)?

- nenadmatematika

great :D chocolate for you :D

- anonymous

haha! yeah!

- anonymous

so the final answer for the first one is: x^3-2x^2+9x-18?

- nenadmatematika

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

- anonymous

polynomial is (x-1) (x+6) and final answer is x^2+5x-6

- anonymous

what is the polynomial of 2+2i though is it just (x-2+2i) and (x+2+2i)

- nenadmatematika

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

- anonymous

so i just multiply (factor) the (x-2+2i) and (x+2+2i) and (x-3) together haha

- nenadmatematika

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

- anonymous

(x-3) (8) i think i did something wrong

- anonymous

i will show u what I did

- anonymous

(x-3) (2+2i) (2-2i)
(x-3) (4-4i+4i-4i^2)
(x-3) (4+4)
(x-3) (8)

- nenadmatematika

no no...you have to multiply (x-(2-2i))*((x-(2+2i))...try again, and be patient..:D

- anonymous

ohh! so do you distribute the x to the 2-2i and 2+2i

- nenadmatematika

I didn't get that...just multiply and show me the procedure....

- nenadmatematika

I'm back in two minutes :D

- anonymous

(x-3) (x-2+2i) (x-2-2i) is that right so far? :(

- nenadmatematika

yes, go,go,go...:D

- anonymous

what do you do though when you multiply the x by -2i is it just -2ix?

- nenadmatematika

of course, continue...;D

- anonymous

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

- nenadmatematika

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

- anonymous

is the final answer x^2-4x+4

- anonymous

thanks for all ur help:) btw

- nenadmatematika

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....

- anonymous

well now i got something crazy... x^4-10x+12 :((

- nenadmatematika

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

- anonymous

okay i get that part

- nenadmatematika

hahah...well that part is all!!! :D I hope I helped you :D

- anonymous

Oh i was just combining it all wrong! thank you!!! :) yes you did!!

- nenadmatematika

you're welcome...when you have any troubles with math be free to contact me, I'll be glad to help you :D

- anonymous

awesome for sure! I am pretty weak at math so I will for sure be contacting you!

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