Question 206051


{{{((z^2-4z-5)/(z^2-3z-4))/((z^2+9z+14)/(z^2-2z-8))}}} Start with the given expression.



{{{((z^2-4z-5)/(z^2-3z-4))((z^2-2z-8)/(z^2+9z+14))}}} Multiply the first fraction {{{(z^2-4z-5)/(z^2-3z-4)}}} by the reciprocal of the second fraction {{{(z^2+9z+14)/(z^2-2z-8)}}}.



{{{(((z+1)*(z-5))/(z^2-3z-4))((z^2-2z-8)/(z^2+9z+14))}}} Factor {{{z^2-4z-5}}} to get {{{(z+1)*(z-5)}}}.



{{{(((z+1)*(z-5))/((z+1)*(z-4)))((z^2-2z-8)/(z^2+9z+14))}}} Factor {{{z^2-3z-4}}} to get {{{(z+1)*(z-4)}}}.



{{{(((z+1)*(z-5))/((z+1)*(z-4)))(((z+2)*(z-4))/(z^2+9z+14))}}} Factor {{{z^2-2z-8}}} to get {{{(z+2)*(z-4)}}}.



{{{(((z+1)*(z-5))/((z+1)*(z-4)))(((z+2)*(z-4))/((z+7)*(z+2)))}}} Factor {{{z^2+9z+14}}} to get {{{(z+7)*(z+2)}}}.



{{{((z+1)*(z-5)(z+2)*(z-4))/((z+1)*(z-4)(z+7)*(z+2))}}} Combine the fractions. 



{{{(highlight((z+1))(z-5)highlight((z+2))highlight((z-4)))/(highlight((z+1))highlight((z-4))(z+7)highlight((z+2)))}}} Highlight the common terms. 



{{{(cross((z+1))(z-5)cross((z+2))cross((z-4)))/(cross((z+1))cross((z-4))(z+7)cross((z+2)))}}} Cancel out the common terms. 



{{{(z-5)/(z+7)}}} Simplify. 



So {{{((z^2-4z-5)/(z^2-3z-4))/((z^2+9z+14)/(z^2-2z-8))}}} simplifies to {{{(z-5)/(z+7)}}}.



In other words, {{{((z^2-4z-5)/(z^2-3z-4))/((z^2+9z+14)/(z^2-2z-8))=(z-5)/(z+7)}}}