Question

Find the polynomial function with roots 11 and 2i.

Answer 1

if 2i is one root, then -2i is also a root

Answer 2

cuz comples roots come in pairs always

Answer 3

so the roots are 11, 2i, and -2i

the polynomial function wud be... ?

Answer 4

(x-11)(x+2)(x-2) ?

Answer 5

done forget the i :)

Answer 6

(x-11i)(x+2i)(x-2i)

Answer 7

i is oly for 2i

Answer 8

11 is a real simple root

Answer 9

it wud be :-

(x-11)(x+2i)(x-2i)

Answer 10

sooo
(x-11)(x+2i)(x-2i)
(x²-11x+24i)(x-2i)

Answer 11

we can do that, but let me show u how to expand easily :)

Answer 12

(x-11)(x+2i)(x-2i)

first expand the last two factors

Answer 13

\((x-11)(x^2+4)\)

Answer 14

cuz, \((a+ib)(a-ib) = a^2 + b^2\)

Answer 15

\((x-11)(x^2+4) \)

now you can simplify easily ?

Answer 16

x^3-11x²+4x-44

Answer 17

very good work :)

wolfram says the same :

http://www.wolframalpha.com/input/?i=%28x-11%29%28x%2B2i%29%28x-2i%29

Answer 18

you can test ur answers in that site

Answer 19

Thankyouu (:>