Question 1045631
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How to factor 8x^4-14x^2+3
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1.  Introduce new variable  u = {{{x^2}}}.
    Then the polynomial takes the form  {{{8u^2-14u+3}}}.

2.  Find the roots of the polynomial  {{{8u^2-14u+3}}}.

    For it, solve the quadratic equation  {{{8u^2-14u+3}}} = {{{0}}}.

    Use the quadratic formula. You will get 

    {{{u[1,2]}}} = {{{(14 +- sqrt(14^2 - 4*8*3))/(2*8)}}} = {{{(14 +- 10)/16}}}.

    So,  {{{u[1]}}} = {{{3/2}}}  and  {{{u[2]}}} = {{{1/4}}}.


3.  Hence, the quadratic polynomial  {{{8u^2-14u+3}}}  is

    {{{8u^2-14u+3}}} = {{{8*(u-3/2)*(u-1/4)}}} = (2u-3)*(4u-1).

    Correspondingly,  the  original polynomial factorization is 

    {{{8x^4-14x^2+3}}} = {{{(2x^2-3)*(4x^2-1)}}}.


4.  Next, you can factor further each of the two factors above to get

    {{{8x^4-14x^2+3}}} = {{{(sqrt(2)x-sqrt(3))*(sqrt(2)x+sqrt(3))*(2x-1)*(2x+1)}}}.


It is what you were asking for.
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