Question

Find the polynomial function with roots 11 and 2i.
Steps please,

Answer 1

oh my gosh seafood needs help? RARE....

Answer 2

lol^

Answer 3

YO?

Answer 4

chay-C?

Answer 5

I'm just trying to comprehend what you're asking, give me a sec.

Answer 6

Given..

Answer 7

(x-11)(x-2i)

Answer 8

im hungry...

Answer 9

Batman, eat yourself.

Answer 10

Heyy..since wen do bats eat them selves..i got a chocolate bar instead

Answer 11

What you're asking is:
When you solve the polynomial, the answers roots are 11 and 2i.
This means that the polynomials answers must be 11^2 and (2i)^2
Therefore, the polynomials answers must be 121 and -4
(x-121)(x+4)=0
x = +121
x = -4

This can't take the root of the negative, so you get 2i, you can take the root of 121. So you get 11.

Expand that equation to get the polynomial.
(x-121)(x+4) = x^2-117-484

Does this seem logical? I've never seen a question like this and this is more than likely wrong.

Answer 12

Alright Chocoman, try eating yourself.

Answer 13

like i said..i cant eat my self..i cant wait till dinner..the whole house smells sooo good..i think ima eat my house

Answer 14

Oops - that should be x^2-117x-484. Sorry.

Answer 15

Does this seem logical and right?

Answer 16

umm... ummm.... nope. :(

Answer 17

why didn't u directly multiplied (x-11)(x-2i)

Answer 18

where is the root -2i?

Answer 19

x = 11 ------------> x -11 = 0

x = 2i -------------> x - 2i = 0

Answer 20

complex roots must occur in conjugate pairs; if 2i is a root, then -2i must also be a root; the poly for those two would be x^2+4=0

Answer 21

eww...this is wat ima deal with in wat is this calculus? eww...

Answer 22

You've jumped a step - The question says.
The polynomials ROOTS are 11 and 2i
If you square both of them you get the direct answer
x = 121 -----------------> x-121
x = -4 -----------------> x +4

Answer 23

This isn't calculus.

Answer 24

why u will square them?

Answer 25

wat is this..i wanna prepare...

Answer 26

this is basic algebra

Answer 27

x^2+4=(x-2i)(x+2i)

Answer 28

yes, right @mandolio.. then/

Answer 29

guys, i get till where mandolio ended.. then wht should i dp?

Answer 30

do*

Answer 31

If the poly is third degree then take your pick:
(x+11)(x-2i)(x+2i)=0
(x+11)(x^2+4)=0
x^3+11x^2+4x+44=0

Answer 32

(x-11)(x-2i) this is the answer

Answer 33

Alright, thanks!! :)

Answer 34

@ Mandolino

I think the roots of ur eq. are -11, 2i, -2i

Answer 35

but the roots re. are 11, 2i

Answer 36

check this out-- 2i by itself can't be a root!!!!!
http://en.wikipedia.org/wiki/Complex_conjugate_root_theorem

Answer 37

First, chaise, you're way off base with squaring the roots first.

Second, a polynomial with REAL coefficients must have complex conjugate solutions (if at least one complex solution exists). The problem doesn't say the polynomial must have real coefficients.

(x-11)(x-2i) = x^2 - 2ix - 11x + 22i = x^2 - (11+2i)x +22i
That is a polynomial. Its roots are 11 and 2i. QED