Question #69994

1 Answer
Oct 26, 2017

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

Explanation:

Assuming that your question was stating that the polynomial has roots of -3 and #i# (the imaginary unit) as zeros, we can proceed as follows.

All complex number roots of the form #a+bi# of a polynomial are always accompanied by their complex conjugate, or the complex number #a-bi#. Because of this, we know that #-i# must also be a zero of #f(x)#, since #i# was a zero of #f(x)#.

We now have all three zeros of the function: -3, #i#, and #-i#. Expressing these as factors, we can then multiply out the factors to derive one such polynomial:

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

# = (x+3)(x^2+ix-ix-i^2) = (x+3)(x^2-i^2) #

# = (x+3)(x^2+1) = (x^3+x+3x^2+3) #

# = x^3+3x^2+x+3 #

We can also verify that #f(1) = 8# as requested:

#f(1) = (1)^3 + 3(1)^2 + (1) + 3 = 1 + 3 + 1 + 3 = 8#