Question

Find all zeros (both real and complex) for the function f(x)=x^3-5x^2+4x+10

Answer 1

$f(x)=x^3-5x^2+4x+10\\ f(x)=x^3+x^2-6x^2-6x+10x+10\\ f(x)=x^2(x+1)-6x(x+1)+10(x+1)\\ f(x)=(x+1)(x^2-6x+10)\\\\ x+1=0\\ x=-1\\\\ x^2-6x+10=0\\ x^2-6x+9+1=0\\ (x-3)^2=-1\\ x-3=i \vee x-3=-i\\ x=3+i \vee x=3-i\\\\ x=\{-1,3-i,3+i\}$