Question 946600
{{{(3x^2+10x-8)/(3x^2+x-2)}}} ....factor completely; write {{{10x}}} in numerator as {{{12x-2x}}}, and {{{x}}} in denominator as {{{3x-2x}}}


{{{(3x^2+12x-2x-8)/(3x^2+3x-2x-2)}}} ...group


{{{((3x^2+12x)-(2x+8))/((3x^2+3x)-(2x+2))}}}...factor out {{{3x}}}


{{{(3x(x+4)-2(x+4))/(3x(x+1)-2(x+1))}}}


{{{((3x-2) (x+4))/((3x-2) (x+1))}}}

since denominator cannot be equal to zero, we will have {{{(3x-2) (x+1)=0}}} if  {{{x=2/3}}} or {{{x=-1}}}

since we can cancel  {{{(3x-2)}}}) like

{{{(cross((3x-2)) (x+4))/(cross((3x-2)) (x+1))}}}


{{{(x+4)/(x+1)}}}  then {{{x}}} cannot be equal to {{{-1}}}



since numerator is {{{(x+4)}}} and it will be zero if {{{x=-4}}}, so

solution is:

{{{x = -4}}}