SOLUTION: find a polynomial function of lowest degree with integer coefficients that has the given zeros 0,i,-i. (x-0) (x-i) (x+i)= p(x)=(x-0)(x-i)(xti) Im stuck after this part please help

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: find a polynomial function of lowest degree with integer coefficients that has the given zeros 0,i,-i. (x-0) (x-i) (x+i)= p(x)=(x-0)(x-i)(xti) Im stuck after this part please help      Log On


   

Question 432788: find a polynomial function of lowest degree with integer coefficients that has the given zeros 0,i,-i.
(x-0) (x-i) (x+i)= p(x)=(x-0)(x-i)(xti) Im stuck after this part please help?
Found 2 solutions by tinbar, Gogonati:
Answer by tinbar(133) About Me  (Show Source):
You can put this solution on YOUR website!
you have it right so far. now just expand and simplify (x)*(x-i)*(x+i), which gives x(x^2+1) = x^3+x. so let your p(x) = a*x^3+b*x, where a,b are Integers and your done. This will be the LOWEST degree polynomial to satisfy the conditions

Answer by Gogonati(855) About Me  (Show Source):
You can put this solution on YOUR website!
Since the zeros of this polynomial are:0,i,-i we write:
x%28x-i%29%28x%2Bi%29=x%28x%5E2-i%5E2%29=x%28x%5E2%2B1%29=x%5E3%2Bx, (remember that i^2=-1).
Done.