## College Algebra (10th Edition)

$x=1$ or $x=-1$ or $x=-4$
We solve by factoring with grouping: $x^{3}+4x^{2}-x-4=0$ $x^{2}(x+4)-1(x+4)=0$ $(x^{2}-1)(x+4)=0$ $(x-1)(x+1)(x+4)=0$ $x-1=0$ or $x+1=0$ or $x+4=0$ $x=1$ or $x=-1$ or $x=-4$