## Precalculus (6th Edition) Blitzer

$x=-1,\pm2,5$
Step 1. As $5$ is a solution of the equation, we use synthetic division as shown in the figure. Thus, we have $(x-5)(x^3+x^2-4x-4)=0$ Step 2. Factor the second term as $x^3+x^2-4x-4=x^2(x+1)-4(x+1)=(x+1)(x^2-4)=(x+1)(x+2)(x-2)$ We have the equation as $(x-5)(x+1)(x+2)(x-2)=0$ Step 3. We can identify the solutions as $x=-1,\pm2,5$