Question

Find all zeros (real and complex) for the function f(x)=x^3-5x^2+9x-5

Answer 1

$f(x)=x^3-5x^2+9x-5\\ f(x)=x^3-x^2-4x^2+4x+5x-5\\ f(x)=x^2(x-1)-4x(x-1)+5(x-1)\\ f(x)=(x-1)(x^2-4x+5)\\\\ (x-1)(x^2-4x+5)=0\\\\ x-1=0\\ x=1\\\\ x^2-4x+5=0\\ x^2-4x+4+1=0\\ (x-2)^2=-1\\ x-2=i \vee x-2=-i\\ x=2+i \vee x=2-i\\\\ x=\{1,2-i,2+i\}$