Question

Write the polynomial of the smallest degree with roots 8, 2i, and -2i

Answer 1

(x-8) (x-2i)

Right?

Answer 2

mhmm

Answer 3

is a poly defined with complex coeffs? or just real coeefs?

Answer 4

(x-8) (x-2i) (x+2i)

Answer 5

expand that

Answer 6

x^3-8*x^2+4*x-32

Answer 7

hmmm, the definition of a poly seems to only deal with the variables and not the coeffs; there are complex polynomials - or, polys with complex coeffs

Answer 8

you are thinking too much @amistre64

Answer 9

i agree that my conundrum has no real bearing in the question, but it does seem to be an oddity for me :)

Answer 10

Her q is Write the polynomial of the smallest degree with roots 8, 2i, and -2i

there are 3 roots so just fill in the blanks of this form
(x-a) (x-b) (x-c)

what a b and c is 8, 2i, and -2i

That simple then just expand

Answer 11

the graph even looks OK

https://www.google.com/search?q=x%5E3-8*x%5E2%2B4*x-32&rlz=1C1CHFX_enUS470US470&oq=x%5E3-8*x%5E2%2B4*x-32&aqs=chrome.0.57.1170&sugexp=chrome,mod=1&sourceid=chrome&ie=UTF-8

Answer 12

ah, i missed the last root on the end

i forgo the swapping signs and just subtract x from each rot, then expand

8, 2i, -2i

8-x, 2i-x, -2i-x

f(x) = (8-x)(2i-x)(-2i-x)

Answer 13

lol well roots of imaginaries always come in pairs right? :D

Answer 14

(x-8)(x^2+4) Then expand. (X^2+4) is the imaginary roots. Imaginary roots always take this form (x^2 + b^2) where bi is a root

Answer 15

http://www.wolframalpha.com/input/?i=%28x-8%29%28x-2i%29

i think pairs of complex only exist with real polys; this construct only has one complex root

Answer 16

you may also use sum of roots,product in pairs and product of roots alternately..

Answer 17

alternatively*

Answer 18

8, 2i, -2i

8-x, 2i-x, -2i-x

f(x) = (8-x)(2i-x)(-2i-x)

Answer 19

?

Answer 20

that looks wrong or just harder mines is right
(x-8) (x-2i) (x+2i)
Expand
x^3-8*x^2+4*x-32