Question 953637
1. 

{{{4x^5 -8x^3 + 4x = 0}}}

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

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

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

{{{4x(x - 1)(x+1)(x - 1)(x+1) = 0}}}

solutions:

if {{{4x = 0}}}=>{{{x=0}}}
if{{{(x - 1) = 0}}}=>{{{x=1}}}...double solution since we have factor {{{x-1}}} twice
{{{(x+1) = 0}}}=>{{{x=-1}}}...double solution since we have factor {{{x+1}}} twice


{{{ graph( 600, 600, -10, 10, -10, 10,4x^5 -8x^3 + 4x) }}}



2. 

{{{x^4 -12x^2 = -36}}}

{{{x^4 -12x^2 +36=0}}}.............recall the rule {{{(a-b)^2=a^2-2ab+b^2}}}; in your case {{{a=1}}} and {{{2ab=12}}}=> {{{b=6}}}
now write it as a square of difference

{{{(x^2 -6)^2=0}}}

{{{(x^2 -6)(x^2 -6)=0}}}...since factors are same, you will have two pairs of double roots

if {{{(x^2 -6)=0}}}=>{{{x^2=6}}}=>{{{x=sqrt(6)}}} or {{{x=- sqrt(6)}}} exact solution

=>{{{x=2.5}}} or {{{x=- 2.5}}} approximate

{{{ graph( 600, 600, -10, 10, -10, 50,(x^2 -6)(x^2 -6)) }}}



3. 


{{{4x^3 + 7x^2 -5x = 6}}}

{{{4x^3 + 7x^2 -5x -6=0}}}

{{{4x^3-x^2+8x^2-2x-3x-6=0}}}

{{{(4x^3+8x^2)-(x^2+2x)-(3x+6)=0}}}

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

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

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

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

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

solutions:

if {{{4x+3=0}}}=>{{{4x=-3}}}=>{{{x=-3/4}}}

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

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



{{{ graph( 600, 600, -10, 10, -10, 10,4x^3 + 7x^2 -5x -6) }}}