## Precalculus (10th Edition)

$f(x)=x^4-6x^3+10x^2-6x+9$
The Conjugate Pairs Theorem says that if a polynomial has real coefficients, then any complex zeros occur in conjugate pairs. That is, if $a + bi$ is a zero then so is $a – bi$ and vice-versa. Therefore, the zeros of the function are: $\pm i$ and 3 with multiplicity 2 Hence, the function could be: $f(x) = (x-i)[x-(-i)](x-3)(x-3)\\ f(x) = (x-i)(x+i)(x^2-6x+9)\\ f(x) = (x^2-i^2)(x^2-6x+9)\\ f(x) = (x^2+1)(x^2-6x+9)\\ f(x) = x^2(x^2-6x+9)+1(x^2-6x+9)\\ f(x) = x^4-6x^3+9x^2+x^2-6x+9\\ f(x) = x^4-6x^3+10x^2-6x+9$