## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$g(5i-1)=50-120i$
Factoring the $GCF=z^2$ and using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the given function, $g(z)=\dfrac{z^4-z^2}{z-1} ,$ simplifies to \begin{array}{l}\require{cancel} g(z)=\dfrac{z^4-z^2}{z-1} \\\\ g(z)=\dfrac{z^2(z^2-1)}{z-1} \\\\ g(z)=\dfrac{z^2(z+1)(z-1)}{z-1} \\\\ g(z)=\dfrac{z^2(z+1)(\cancel{z-1})}{\cancel{z-1}} \\\\ g(z)=z^2(z+1) .\end{array} Substituting $z$ with $5i-1$ in the function above results to \begin{array}{l}\require{cancel} g(z)=z^2(z+1) \\\\ g(5i-1)=(5i-1)^2(5i-1+1) \\\\ g(5i-1)=[(5i)^2-2(5i)(1)+(1)^2](5i) \\\\ g(5i-1)=(25i^2-10i+1)(5i) \\\\ g(5i-1)=25i^2(5i)-10i(5i)+1(5i) \\\\ g(5i-1)=125i^3-50i^2+5i .\end{array} Using $i^2=-1,$ the function above is equivalent to \begin{array}{l}\require{cancel} g(5i-1)=125i^3-50i^2+5i \\\\ g(5i-1)=125i^2\cdot i-50i^2+5i \\\\ g(5i-1)=125(-1)\cdot i-50(-1)+5i \\\\ g(5i-1)=-125i+50+5i \\\\ g(5i-1)=50-120i .\end{array}