Algebra and Trigonometry 10th Edition

$f(x)=-x^3+2x^2-x+2$
If $i$ is a zero of $f$ then the complex conjugate $-i$ is also a zero. We have three zeros ($2,~i,~-i$). We can find a three-degree polynomial. $f(x)=a(x-2)[(x-i)[x-(-i)]$ $f(x)=a(x-2)(x^2-i^2)$ $f(x)=a(x-2)(x^2+1)$ $f(x)=a(x^3+x-2x^2-2)$ $f(x)=a(x^3-2x^2+x-2)$ $a=-1$ $f(x)=-1(x^3-2x^2+x-2)$ $f(x)=-x^3+2x^2-x+2$