Question 926612
1. 

{{{x^3-6x^2+4x=0}}} factor out {{{x}}}

{{{x(x^2-6x+4)=0}}} => one solution is {{{x=0}}}

{{{x^2-6x+4=0}}} use quadratic formula to find other two solutions:

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

{{{x = (-(-6) +- sqrt( (-6)^2-4*1*4 ))/(2*1) }}} 

{{{x = (6 +- sqrt(36-16))/2 }}} 

{{{x = (6 +- sqrt(20))/2 }}} 

{{{x = (6 +- sqrt(4*5))/2 }}} 

{{{x = (cross(6)3 +- cross(2)sqrt(5))/cross(2) }}} 

{{{x = (3 +- sqrt(5)) }}} 

solutions:

{{{highlight(x = 3 + sqrt(5)) }}} 

{{{highlight(x = 3 - sqrt(5)) }}} 


{{{ graph( 600, 600, -10, 10, -10, 10, x^3-6x^2+4x) }}}




2. 

{{{x^4-5x^2+4=0}}}....factor completely

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

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

{{{x^2(x^2-1)-4(x^2-1)=0}}}

{{{(x^2-4)(x^2-1)=0}}}

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

solutions:

if {{{(x-2)=0}}}=>{{{highlight(x=2)}}}

if {{{(x+2)=0}}}=>{{{highlight(x=-2)}}}

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

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


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


3. 

{{{3x^3+12x^2-x-4=0}}}

{{{3x^2(x+4)-(x+4)=0}}}

{{{(3x^2-1)(x+4)=0}}}


solutions:

if {{{(3x^2-1)=0}}}=>{{{3x^2=1}}}=>{{{x^2=1/3}}}=>{{{x=sqrt(1/3)}}}=>{{{highlight(x=1/sqrt(3))}}} or {{{x=highlight(-1/sqrt(3))}}}

if {{{(x+4)=0}}}=>{{{highlight(x=-4)}}}

{{{ graph( 600, 600, -10, 10, -10, 10,(3x^2-1)(x+4)) }}}



4. 
{{{(3x+5)(x^2+5x-6)=0 }}}

{{{(3x+5)(x^2-x+6x-6)=0 }}}

{{{(3x+5)((x^2-x)+(6x-6))=0 }}}

{{{(3x+5)(x(x-1)+6(x-1))=0 }}}

{{{(3x+5)(x+6)(x-1)=0 }}}


solutions:

if {{{(3x+5)=0 }}}=>{{{3x=-5}}}=>{{{highlight(x=-5/3)}}}

if {{{(x+6)=0 }}}=> {{{highlight(x=-6)}}}

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


{{{ graph( 600, 600, -10, 10, -10, 10,(3x+5)(x+6)(x-1)) }}}