Question 984208
first make {{{f ( x )=0}}}

 {{{0= x^3 + x^2-4x}}}...factor out {{{x}}}

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

{{{0 = highlight(x)highlight((x^2+x-4))}}}... this product will be equal to zero only if one or both of the factors is equal to zero

so, one root is {{{highlight(x[1]=0)}}}

the other two roots we will get solving {{{x^2+x-4=0}}}-use quadratic formula

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

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

{{{x = (-1 +- sqrt( 1+16 ))/2 }}} 

{{{x = (-1 +- sqrt( 17 ))/2 }}} 

exact solutions are:

{{{highlight(x[2] = (-1 + sqrt( 17 ))/2) }}}
and
{{{highlight(x[3] = (-1 - sqrt( 17 ))/2) }}}