Question 1124403


a polynomial equation of degree {{{n}}} has {{{exactly}}} {{{n}}} roots

you have a polynomial equation of degree {{{3}}} has {{{exactly}}} {{{highlight(3)}}} roots

{{{5x^3-4x+1=0 }}}

here are they:

{{{5x^3-4x+1=0 }}}...factor; add {{{5x^2-5x^2}}}, and write {{{-4x}}} as {{{x-5x}}}

{{{5x^3+5x^2-5x^2−5x+x+1=0 }}}...group

{{{(5x^3+5x^2)-(5x^2+5x)+(x+1)=0 }}}


{{{5x^2(x+1)-5x(x+1)+(x+1)=0 }}}


{{{(x + 1) (5x^2 - 5x + 1) = 0}}}


one real solution is: {{{(x + 1)  = 0}}}=>{{{x=-1}}}

and, since your function is 3rd degree, there will be two more solutions  


 use quadratic formula to solve {{{(5x^2 - 5x + 1) = 0}}}:


In this case {{{a=5}}},{{{b=-5}}},{{{c=1}}}


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

{{{x = (-(-5) +- sqrt( (-5)^2-4*5*1 ))/(2*5) }}} 

{{{x = (5 +- sqrt( 25-20 ))/10 }}} 

{{{x = (5 +- sqrt( 5 ))/10 }}} 

{{{x = (5/10 +- sqrt( 5 )/10) }}} 

{{{x = (1/2 +- sqrt( 5 )/10) }}} 

solutions:

{{{x = 1/2 + sqrt( 5 )/10 }}}

{{{x = 1/2 - sqrt( 5 )/10 }}}  



{{{ graph( 600, 600, -5, 5, -5, 5, 5x^3-4x+1) }}}