## College Algebra (11th Edition)

$-30-40i$
$\bf{\text{Solution Outline:}}$ To simplify the given expression, $-5i(3-i)^2 ,$ use the special product on squaring binomials and the Distributive Property. Then use the equivalence $i^2=-1.$ Finally, combine combine the real parts and the imaginary parts. $\bf{\text{Solution Details:}}$ Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to \begin{array}{l}\require{cancel} -5i[(3)^2-2(3)(i)+(i)^2] \\\\= -5i[9-6i+i^2] .\end{array} Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to \begin{array}{l}\require{cancel} -5i(9)-5i(-6i)-5i(i^2) \\\\= -45i+30i^2-5i(i^2) .\end{array} Since $i^2=-1,$ the expression above is equivalent to \begin{array}{l}\require{cancel} -45i+30(-1)-5i(-1) \\\\= -45i-30+5i .\end{array} Combining the real parts and the imaginary parts results to \begin{array}{l}\require{cancel} -30+(-45i+5i) \\\\= -30-40i .\end{array}