At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
on of the polynomials is (x-2)(x^2+9)
Can you explain how you got that answer please?:) thanks!
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
You are amazing! Thanks :)
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|
Is that the final answer or do you have to multiply it out to get those zeros?
whoops ignore what I just said. :)
no no, it's OK...you have zeros given, and your task is to find polynomial which zeros are given for you....
so like the (x-2) (x^2 +9) is ur final answer? sorry that is were i am confused haha
you can leave it like that, but most professors would like to multiply...do you know how to do it"?
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) ??
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
yes isn't it x^2+1x-6
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...?
great :D chocolate for you :D
so the final answer for the first one is: x^3-2x^2+9x-18?
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
polynomial is (x-1) (x+6) and final answer is x^2+5x-6
what is the polynomial of 2+2i though is it just (x-2+2i) and (x+2+2i)
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
so i just multiply (factor) the (x-2+2i) and (x+2+2i) and (x-3) together haha
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
(x-3) (8) i think i did something wrong
i will show u what I did
(x-3) (2+2i) (2-2i) (x-3) (4-4i+4i-4i^2) (x-3) (4+4) (x-3) (8)
no no...you have to multiply (x-(2-2i))*((x-(2+2i))...try again, and be patient..:D
ohh! so do you distribute the x to the 2-2i and 2+2i
I didn't get that...just multiply and show me the procedure....
I'm back in two minutes :D
(x-3) (x-2+2i) (x-2-2i) is that right so far? :(
what do you do though when you multiply the x by -2i is it just -2ix?
of course, continue...;D
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
number ''i'' must be gone after mulitplication...try again, look out for signs, and of course use i^2=-1
is the final answer x^2-4x+4
thanks for all ur help:) btw
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....
well now i got something crazy... x^4-10x+12 :((
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
okay i get that part
hahah...well that part is all!!! :D I hope I helped you :D
Oh i was just combining it all wrong! thank you!!! :) yes you did!!
you're welcome...when you have any troubles with math be free to contact me, I'll be glad to help you :D
awesome for sure! I am pretty weak at math so I will for sure be contacting you!