Question

-1, 6, 3 - 2i find a polynomial function with real coefficients a that has the given zeros (there are many correct answers)

Answer 1

f(x) = x⁴-11x³+37x²-29x-78

Explanation:

The zeros -1 and 6 are real zeros, so we'll use them directly.

The zero 3-2i is complex, and ALL complex zeros come in pairs. So we use the complex conjugate of 3-2i for the other zero: 3+2i.

Now, set up your factors. There are four zeros, so four factors:

(x- -1)(x-6)(x-(3-2i))(x-(3+2i)) = (x+1)(x-6)(x- 3-2i)(x- 3+2i)

Next, the instructions say "with real coefficients" so we must multiply everything out and combine like terms. I recommend starting with the two real factors: (x+1)(x-6) = x²-5x-6.

then multiply the two complex factors:

(x- 3-2i)(x- 3+2i) = x²-3x+2ix-3x+9-6i-2ix+6i-4i² [remember that i²=-1!!]

= x²-6x+13

and finally multiply the two trinomials that you get; you'll have 9 terms to combine.