(x+1)(x-√3)(x+√3)(3x-11)
no way to tell the 5th root.
Suppose that a polynomial function of degree 5 with rational coefficients has the given numbers as zeros. Find the other zero( s):
-1, radical 3, 11/3
4 answers
One of the rational roots would have to be a double root, so it could be
y = (x+1)^2(x-√3)(x+√3)(3x-11) = (x+1)^2(x^2+9)(3x-11)
or
y = (x+1)(x-√3)(x+√3)(3x-11)^2 = (x+1)(x^2+9)(3x-11)^2
y = (x+1)^2(x-√3)(x+√3)(3x-11) = (x+1)^2(x^2+9)(3x-11)
or
y = (x+1)(x-√3)(x+√3)(3x-11)^2 = (x+1)(x^2+9)(3x-11)^2
did I miss an i somewhere?
(x-√3)(x+√3) = x^2-3
(x-√3)(x+√3) = x^2-3
Suppose that a polynomial function of degree 5 with rational coefficients has -2, 5, and 3 - I as zeros. Find the other zeros.
5 answers