Answer
$f(x)=x^5-5x^4+11x^3-13x^2+8x-2$
Work Step by Step
The Conjugate Pairs Theorem states that when a polynomial has real coefficients, then any complex zeros occur in conjugate pairs. This means that, when $(p +i \ q)$ is a zero of a polynomial function with a real number of the coefficients, then its conjugate $(p –i q)$, is also a zero of the function.
We are given that the zeros of the function are: $1 \pm i$ and $1$ with multiplicity $3$.
We write the factors of the function as $(x-zero)^{multiplicity}$ and multiply them to get the polynomial.
Therefore, we can write the equation of the function as:
$f(x)=[x-(1+i)][(x-(1-i)](x-1)(x-1)(x-1)\\=(x-1-i)(x-1+i)(x-1)^3\\=(x^2-2x+1+1)(x^3-3x^2+ 3x - 1)\\=(x^2-2x+2)(x^3 - 3 x^2 + 3 x - 1)\\=x^5-5x^4+11x^3-13x^2+8x-2$
