Chapter 3 - Polynomial and Rational Functions - 3.3 Zeros of Polynomial Functions - 3.3 Exercises - Page 337: 51

$x=-2$ with a multiplicity of $5$ $x=1$ with a multiplicity of $5$ $x=1-\sqrt 3$ each with a multiplicity of $2$

Step 1. Give $f(x)=(x^2+x-2)^5(x-1+\sqrt 3)^2=(x+2)^5(x-1)^5(x-1+\sqrt 3)^2$, we can find the first zero as $x=-2$ with a multiplicity of $5$ Step 2. We can find the second zero as $x=1$ with a multiplicity of $5$ Step 3. Let $((x-1+\sqrt 3)=0$, we have $x=1-\sqrt 3$ each with a multiplicity of $2$