Answer
$(2+i)(2-i)(4+3i)=20+15i$
Work Step by Step
$(2+i)(2-i)(4+3i)$
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.
$(2+i)(2-i)(4+3i)=(4-i^{2})(4+3i)=...$
Substitute $i^{2}$ by $-1$ and simplify the first factor:
$...=[4-(-1)](4+3i)=(4+1)(4+3i)=5(4+3i)=...$
Evaluate the remaining product:
$...=20+15i$
