Question 765779
{{{7x^3+3x^2-29x+11=0 }}}...factor, write {{{3x^2}}} as {{{14x^2-11x^2}}} and {{{-29x}}} as {{{-7x-22x}}}

{{{7x^3+14x^2-7x-11x^2-22x+11=0 }}}......group

{{{(7x^3-11x^2)+(14x^2-22x)-(7x-11)=0 }}}

{{{x^2(7x-11)+2x(7x-11)-(7x-11)=0 }}}

{{{(7x-11)(x^2+2x-1) = 0}}}


solutions:

if {{{7x-11= 0}}} => {{{7x=11}}}=> {{{x=11/7}}}

to find other two roots, use quadratic formula and solve {{{x^2+2x-1 = 0}}} for {{{x}}}

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

{{{x = (-2 +- sqrt( 2^2-4*1*(-1) ))/(2*1) }}}

{{{x = (-2 +- sqrt( 4+4))/2 }}}

{{{x = (-2 +- sqrt( 8))/2 }}}

{{{x = (-2 +- 2.83)/2 }}}

solutions:

{{{x = (-2 +2.83)/2 }}}

{{{x = 0.83/2 }}}

{{{x = 0.415 }}}

and

{{{x = (-2 -2.83)/2 }}}

{{{x =-4.83/2 }}}

{{{x = -2.415 }}}


so, your solutions are: {{{x=11/7}}},{{{x = 0.415 }}}, and {{{x = -2.415 }}}


{{{ graph( 600, 600, -5, 5, -10, 35,7x^3+3x^2-29x+11) }}}