## Precalculus (6th Edition)

$(3-i)(3+i)(2-6i)=20-60i$
$(3-i)(3+i)(2-6i)$ Begin by evaluating the product between the first two factors. The product of the first two factors represents the factored form of a difference of two squares. $(3-i)(3+i)(2-6i)=(9-i^{2})(2-6i)=...$ Substitute $i^{2}$ by $-1$ and simplify the first factor: $...=[9-(-1)](2-6i)=(9+1)(2-6i)=10(2-6i)$ Evaluate the remaining product: $...=20-60i$