How do you find a polynomial function of lowest degree with rational coefficients that has the given number of some of it's zeros. -5i, 3?

1 Answer
Jul 27, 2015

Answer:

#f(x) = (x-5i)(x+5i)(x-3) = x^3-3x^2+25x-75#

Explanation:

If the coefficients are real (let alone rational), then any complex zeros will occur in conjugate pairs.

So the roots of #f(x) = 0# are at least #+-5i# and #3#.

Hence

#f(x) = (x-5i)(x+5i)(x-3)#

#= (x^2+25)(x-3)= x^3-3x^2+25x-75#

Any polynomial in #x# with these zeros will be a multiple of #f(x)#