Question 939889

{{{(3b+1)/(b^2-3b-4)}}} to have a real solution, denominator cannot be equal to zero; so, find the value(s) of {{{b}}} that makes denominator equal to zero

{{{b^2-3b-4=0}}} ...factor completely

{{{b^2+b-4b-4=0}}} ...group

{{{(b^2+b)-(4b+4)=0}}} 

{{{b(b+1)-4(b+1)=0}}} 

{{{(b-4)(b+1)=0}}} 

solutions:

if {{{(b-4)=0}}} => {{{b=4}}}

if {{{(b+1)=0}}} => {{{b=-1}}}


real numbers {{{b=4}}} and {{{b=-1}}} cannot be used in place of the variable in the rational expression