How do you write a polynomial function in standard form with real coefficients whose zeros include #1#, #9i#, and #-9 i#?

1 Answer
Apr 19, 2017

Answer:

The simplest such polynomial is:

#f(x) = x^3-x^2+81x-81#

Explanation:

Given zeros #1#, #9i#, #-9i#.

Any polynomial in #x# with these zeros will have linear factors including #(x-1)#, #(x-9i)# and #(x+9i)#.

So let:

#f(x) = (x-1)(x-9i)(x+9i)#

#color(white)(f(x)) = (x-1)(x^2-(9i)^2)#

#color(white)(f(x)) = (x-1)(x^2+81)#

#color(white)(f(x)) = x^3-x^2+81x-81#

So this #f(x)# is a suitable example and any polynomial in #x# with these zeros will be a multiple (scalar or polynomial) of this #f(x)#.

#color(white)()#
Footnote

If only the zeros #1# and #9i# had been requested, we would still require #-9i# to be a zero too in order to get real coefficients.

One way to think of this is to recognise that #i# and #-i# are essentially indistinguishable from the perspective of the real numbers. So if #a+ib# is a zero of a polynomial with real coefficients then so is #a-ib#.