## Precalculus (6th Edition)

$2(z+4)(z-2)$
$(z+4)$ is in both terms, we can factor it out: $(z+4)((3z+2)-(z+6))=(z+4)(2z-4)$ We can factor out a constant of 2 from the second term: $(z+4)(2z-4)=(z+4)\times2\times (z-2)=2(z+4)(z-2)$