How do you simplify #(z^2-z-6)/(z-6) *( z^2-6z)/(z^2+2z-15)#?

1 Answer
Jul 13, 2015

Answer:

#(z^2-z-6)/(z-6)*(z^2-6z)/(z^2+2z-15)=(z^2+2z)/(z+5)#

Explanation:

#(z^2-z-6)/(z-6)*(z^2-6z)/(z^2+2z-15)#
#color(white)("XXXX")#Combining the numerator factors and the denominator factors:
#=((z^2-z-6)(z^2-6z))/((z-6)(z^2+2z-15))#
#color(white)("XXXX")#Factoring
#= ((z-3)(z+2)(z)(z-6))/((z-6)(z+5)(z-3))#
#color(white)("XXXX")#Cancelling out matching terms in the numerator and denominator
#= (cancel((z-3))(z+2)(z)cancel((z-6)))/(cancel((z-6))(z+5)cancel((z-3)))#
#color(white)("XXXX")#Rewrite simplified:
#=(z^2+2z)/(z+5)#