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