Question

form a polynomial function f(x) with real coefficients having the given degree and zeros. Answers will vary depending on the choice of the leading coefficient. Use a graphing utility to graph the function and verify the result.

Degree 6; zeros: i, 2 + i, 4-i

Short answer

The polynomial can be given as$f\left(x\right)=x^6-12x^5+55x^4-120x^3+139x^2-108x+85$

Step-by-step solution

Given information

We are given the degree of polynomial to be 6 the zeros are 2+i , i and 4-i

Find the equation

as i , 2+i and 4-i are the roots of the equation hence by conjugate properties -i, 2-i and 4+i are also the roots of the polynomial

Hence

$$\begin{gathered} f\left(x\right)=(x-i)(x+i)(x-(2+i\left)\right)(x-(2-i\left)\right)(x-(4-i\left)\right) \\ (x-(4+i\left)\right) \\ f\left(x\right)=(x^2+1)({(x-2)}^2+1)({(x-4)}^2+1) \\ f\left(x\right)=(x^2+1)(x-4x+5)(x^2-8x+17) \\ f\left(x\right)=x^6-12x^5+55x^4-120x^3+139x^2-108x+85 \end{gathered}$$

Conclusion

The polynomial can be given as$f\left(x\right)=x^6-12x^5+55x^4-120x^3+139x^2-108x+85$