Question 775977

{{{(x+1)(2x+1)^3 -3(2x+1)^2 (x+1)=0}}}..factor out {{{(x+1)(2x+1)^2}}}

{{{(x+1)(2x+1)^2((2x+1) -3)=0}}}

{{{(x+1)(2x+1)^2(2x+1 -3)=0}}}

{{{(x+1)(2x+1)^2(2x-2)=0}}}

{{{(x+1)(2x+1)^2*2(x-1)=0}}}

{{{2(x+1)(2x+1)(2x+1)(x-1)=0}}}..use zero product rule to find solutions

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

if {{{(2x+1)=0}}}=> {{{2x=-1}}}=>{{{x=-1/2}}}...this solution doubles

if {{{(x-1)=0}}}=> {{{x=1}}}

so, here you have three possible real solutions


{{{ graph( 600, 600, -5, 5, -10, 10,(x+1)(2x+1)^3 -3(2x+1)^2 (x+1)) }}}