College pre calc help, i hate takehome tests

Gmart

Gmart

Active member
question is, with the given zeros find the fifth degree equation,

-2i, √3, 1 (negative 2 eye, square root of three, and one)

+k for serious answers
 
the zeroes are the roots of the equation, or the numbers you find when the equation is factored

(x^2+2)(x^2-3)(x-1)

i don't know if the teacher wants factored form, or standard form. If they want standard form, just multiply that out so it looks like:

ax^5+bx^4+cx^3+dx^2+ex+f=0

 
thanks for trying, still don't get it tho, i've been struggling with this course since the beginning of the semester +k tho
 
by zeros i assume you mean roots so... (r-1)(r - sqrt3) then you cant just have a -2i....it should be like 0 plus/minus 2i (real and imaginary part)

but even with those 4 roots its would be a 4th degree polynomial, unless zero is a root where r (in this case) could be factored out of the equation
 
ok i get it to an extent, so i get the x^2 - 3, that gives u √3, and x-1 gives you 1, but, how do you get the negative 2i, is √-2 = to 2i?
 
sorry, i'd rather ask a question and learn then hand in something with no answer. and what? lrn2wolfram alpha, learn to work from a lpha? no entiendo
 
This is a really weird question. If -2i is a zero, then +2i also has to be a zero. Also, in order to make it a fifth order polynomial, you're going to have to have a double root. How are you supposed to choose between 1, and root3 as your double root?
 
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