In a question like this, Which of the following is a polynomial with roots 4, –2i, and 2i ? I know how to do, but I dont understand what to do with the "i's" . I need a explanation on this question

Question

lgbasallote · Answer

tell me first what you would do..

anonymous · Answer

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

anonymous · Answer

What I ment was I know How as if the roots were just basic numbers but when the i's are added Im confused

anonymous · Answer

Lets just say this I need help with this equation

lgbasallote · Answer

a polynomial has roots 4, -2i and 2i

therefore 

x = 4
x = -2i
x = 2i

agree?

anonymous · Answer

Agreed. I know to set it up in (x-4)(x-2i)(x+2i) but if you distribute it doesnt cancel out the i"s , if it does I guess Imdoing it wrong becasue Every explanation on equations with the i's It ends up cancelling out or atleast it looks like it does

lgbasallote · Answer

that's because it's not (x-4)(x-2i)(x+2i) ;)

you need to do one extra step

lgbasallote · Answer

you were right that x = 4 in factor form would be x - 4

but x = 2i and x = -2i dont automatically do that. it's not that easy...

x = 2i

first you need to make this into a real number so square it

$$x^2 = (2i)^2$$
$$x^2 = 4(-1)$$
$$x^2 = -4$$
$$x^2 + 4 = 0$$
do you get that?

anonymous · Answer

does i equal -1 always?

lgbasallote · Answer

nope. i^2 = -1

anonymous · Answer

hmmm

lgbasallote · Answer

$$i = \sqrt{-1}$$

lgbasallote · Answer

oh...not again :C

anonymous · Answer

lol

anonymous · Answer

$$(a+b)(a-b)=a^2-b^2$$
$(x+2i)(x-2i)=x^2-(2i)^2=x^2-4i^2=x^2+4$

anonymous · Answer

I see what your saying , seee I didnt know about that extra step.

lgbasallote · Answer

so i guess you're saying that those canceled out huh..

anonymous · Answer

you really can just multiply out 
$$(x-4)(x-2i)(x+2i)$$

anonymous · Answer

first multiply $(x+2i)(x-2i)=x^2+4$ then 
$(x-4)(x^2+4)=x^3-4x^2+4x-16$

lgbasallote · Answer

but i think it should be simplified to $$(x-4)(x^2 + 4)$$ no? i mean if i remember my algebra right..you cant have i in the final answer

anonymous · Answer

yes there is no $i$ in the answer because $x-(a+bi))(x-(a-bi)=x^2-2ax+(a^2+b^2)$
in this case it is easier because $a=0$ and you get $(x+2i)(x-2i)=x^2+4$

anonymous · Answer

Thank you BOTH ;D

lgbasallote · Answer

hmm i did not know this side.. i always used that method i stated..it was less confusing haha

anonymous · Answer

I think I got both ways.