Question 1000692
{{{f(x)=x^3+4x^2-5}}}; if {{{x[1]=1}}} means {{{x-1}}} is a factor of {{{f(x)=x^3+4x^2-5}}}

so, factor {{{f(x)=x^3+4x^2-5}}}

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

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

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


{{{f(x)=(x-1)(x^2+5x+5)}}}=>{{{x-1}}} is a factor and {{{x=1}}} is a zero

now use quadratic formula to find other two zeros:

if {{{0=x^2+5x+5}}}, others zeros are:

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

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

{{{x = (-5 +- sqrt( 25-20 ))/2 }}} 

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

solutions:

{{{x = (-5 + sqrt( 5 ))/2 }}}=>{{{x =-1.4 }}}
or
{{{x = (-5 - sqrt( 5 ))/2 }}}=>{{{x = -3.6}}}


{{{ graph( 600, 600, -10, 10, -10, 10, x^3+4x^2-5) }}}