Question 201042

{{{((z^2+4z+4)/(z^2+9z+14))/((z^2+2z)/(z^2+4z-21))}}} Start with the given expression.



{{{((z^2+4z+4)/(z^2+9z+14))((z^2+4z-21)/(z^2+2z))}}} Multiply the first fraction {{{(z^2+4z+4)/(z^2+9z+14)}}} by the reciprocal of the second fraction {{{(z^2+2z)/(z^2+4z-21)}}}.



{{{(((z+2)(z+2))/(z^2+9z+14))((z^2+4z-21)/(z^2+2z))}}} Factor {{{z^2+4z+4}}} to get {{{(z+2)(z+2)}}}.



{{{(((z+2)(z+2))/((z+7)*(z+2)))((z^2+4z-21)/(z^2+2z))}}} Factor {{{z^2+9z+14}}} to get {{{(z+7)*(z+2)}}}.



{{{(((z+2)(z+2))/((z+7)*(z+2)))(((z+7)*(z-3))/(z^2+2z))}}} Factor {{{z^2+4z-21}}} to get {{{(z+7)*(z-3)}}}.



{{{(((z+2)(z+2))/((z+7)*(z+2)))(((z+7)*(z-3))/(z*(z+2)))}}} Factor {{{z^2+2z}}} to get {{{z*(z+2)}}}.



{{{((z+2)(z+2)(z+7)*(z-3))/(z*(z+7)*(z+2)(z+2))}}} Combine the fractions. 



{{{(highlight((z+2))highlight((z+2))highlight((z+7))(z-3))/(z*highlight((z+7))highlight((z+2))highlight((z+2)))}}} Highlight the common terms. 



{{{(cross((z+2))cross((z+2))cross((z+7))(z-3))/(z*cross((z+7))cross((z+2))cross((z+2)))}}} Cancel out the common terms. 



{{{(z-3)/z}}} Simplify. 




So {{{((z^2+4z+4)/(z^2+9z+14))/((z^2+2z)/(z^2+4z-21))}}} simplifies to {{{(z-3)/z}}}.



In other words, {{{((z^2+4z+4)/(z^2+9z+14))/((z^2+2z)/(z^2+4z-21))=(z-3)/z}}}