Question

What polynomial function has the roots 11 and 2i?

Answer 1

multiply the roots to get the polynomial
keep in mind that complex roots, 2i, aren't all by their lonesome in roots usually, they have a conjugate companion

Answer 2

isn't 2i the same thing as the square root of -2? @jdoe0001

Answer 3

no, \(\bf i = \sqrt{-1}\)

Answer 4

@destinc1215 \[\sqrt{-4} = \sqrt{4i^2} = \pm 2i\]

Answer 5

so its 2 multiplied times the square root of -1? @jdoe0001

Answer 6

so you have the roots of 11, that is x = 11 => (x-11) = 0

and you have the root of 2i, that is x = 2i AND IT'S CONJUGATE, x = -2i
thus
(x-2i) = 0 AND (x+2i) =0

Answer 7

so x=2i and x=-2i and x=11? @jdoe0001

Answer 8

destinyc1215 yes it's, but as pointed out, it's not alone, it has a conjugate component which comes along and it's also one of the roots, as you can see in campbell_st 's line above

Answer 9

yes

Answer 10

then i multiply them all together and get my function?

Answer 11

those are the roots, and the factors the will be
(x-11)(x-2i)(x+2i) = 0

Answer 12

@jdoe0001

Answer 13

\(\bf (x-11)(x-2i)(x+2i) = 0\\
\text{keep in mind that }(a-b)(a+b) = (a^2-b^2)\\
(x-11)\left[(x-2i)(x+2i)\right] \implies
(x-11)\left[(x^2-(2i)^2)\right]\)

Answer 14

\(\bf (x-11)[(x^2-(4i^2))]\\
\color{blue}{i = \sqrt{-1}\implies i^2 = \sqrt{(-1)^2} \implies -1}\\
(x-11)(x^2-(4(-1)))
\)

Answer 15

expand that, and you'd get your polynomial

Answer 16

f(x) = x^3 - 11x^2 + 4x - 44 <===ANSWER