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
tell me first what you would do..
y=(x-4)(x+2i)(x-2i)
What I ment was I know How as if the roots were just basic numbers but when the i's are added Im confused
Lets just say this I need help with this equation
a polynomial has roots 4, -2i and 2i therefore x = 4 x = -2i x = 2i agree?
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
that's because it's not (x-4)(x-2i)(x+2i) ;) you need to do one extra step
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?
does i equal -1 always?
nope. i^2 = -1
hmmm
\[i = \sqrt{-1}\]
oh...not again :C
lol
\[(a+b)(a-b)=a^2-b^2\] \((x+2i)(x-2i)=x^2-(2i)^2=x^2-4i^2=x^2+4\)
I see what your saying , seee I didnt know about that extra step.
so i guess you're saying that those canceled out huh..
you really can just multiply out \[(x-4)(x-2i)(x+2i)\]
first multiply \((x+2i)(x-2i)=x^2+4\) then \((x-4)(x^2+4)=x^3-4x^2+4x-16\)
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
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 \((x+2i)(x-2i)=x^2+4\)
Thank you BOTH ;D
hmm i did not know this side.. i always used that method i stated..it was less confusing haha
I think I got both ways.
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