How do you simplify #\frac { 2z + 11} { 4z ^ { 3} + 24z ^ { 2} + 11z }#?

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Answer 1 · 2017-12-10T10:35:36

#1/(z(2z+1))#

Begin by factoring the denominator:

#4z^3+24z^2+11z=z(4z^2+24z+11)#

The quadratic term can factor (use the numerator as a clue for how to factor, there's probably a #2z+11# factor):

#4z^2+24z+11=(2z+11)(2z+1)#

So the denominator can be rewritten as:

#z(2z+11)(2z+1)#

The original can be rewritten as:

#(2z+11)/(z(2z+11)(2z+1))#

Cancel the common #2z+11#:

#1/(z(2z+1))#

Answer 2 · 2017-12-10T10:40:11

The answer is #=1/(z(2z+1))#

The denominator is

#4z^3+24z^2+11z=z(4z^2+24z+11)#

#=z(2z(2z+11)+(2z+11))#

#=z(2z+11)(2z+1)#

Therefore,

The fraction is

#(2z+11)/(z^3+24z^2+11z)=cancel(2z+11)/((z)cancel(2z+11)(2z+1))#

#=1/(z(2z+1))#