Question 1124747
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The roots of the polynomial equation 2x^3 - 8x^2 + 3x + 5 = 0 are alpha, beta and gamma.
Find the polynomial equation with roots alpha^2, beta^2, gamma^2
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<pre>
The given equation

    {{{2x^3 - 8x^2 + 3x + 5}}} = 0        (1)

is equivalent to

    {{{x^3 - 4x^2 + 1.5x + 2.5}}} = 0     (2)  (all the coefficients of (1) are divided by 2)


Equation (2) has the same roots  {{{alpha}}},  {{{beta}}}  and  {{{gamma}}}  as equation (1).  Therefore, 

    {{{x^3 - 4x^2 + 1.5x + 2.5}}} = {{{(x-alpha)*(x-beta)*(x-gamma)}}},                       (3)

and, according to Vieta's theorem

    {{{alpha + beta + gamma}}} = 4,  {{{alpha*beta+alpha*gamma+beta*gamma}}} = 1.5,  {{{alpha*beta*gamma}}} = -2.5.      (4)


Now, an equation with the roots  {{{alpha^2}}},  {{{beta^2}}}  and  {{{gamma^2}}}  is

    {{{(x-alpha^2)*(x-beta^2)*(x-gamma^2)}}} = 0.                     (5)


By the Vieta's theorem (or by applying FOIL directly), the coefficients of the left side polynomial are

    {{{-(alpha^2+beta^2+gamma^2)}}}  at  x^2;                          (6)

    {{{alpha^2*beta^2+alpha^2*gamma^2+beta^2*gamma^2}}}  at x;   and                 (7)

    {{{-alpha^2*beta^2*gamma^2}}}  as the constant term.                (8)


So, my task now is to express the coefficient (6), (7) and (8)  via  the coefficients (4) of the equation (2).


Regarding   {{{(alpha^2+beta^2+gamma^2)}}},  it is easy:

    {{{(alpha^2+beta^2+gamma^2)}}} = {{{(alpha+beta+gamma)^2-2*(alpha*beta+alpha*gamma+beta*gamma)}}} = {{{4^2 - 2*1.5}}} = 16-3 = 13.


So, the coefficient at x^2 of the polynomial (5)  is  {{{-(alpha^2+beta^2+gamma^2)}}} = -13.


Regarding  {{{-alpha^2*beta^2*gamma^2}}},  it is easy, too :

    {{{alpha^2*beta^2*gamma^2}}} = {{{(alpha*beta*gamma)^2}}} = {{{(-2.5)^2}}} = 6.25.


So, the constant term of the polynomial (5)  is  {{{-(alpha^2*beta^2*gamma^2)}}} = -6.25.


Regarding  {{{alpha^2*beta^2+alpha^2*gamma^2+beta^2*gamma^2}}}, it is slightly more long way :

    {{{alpha*beta+alpha*gamma+beta*gamma}}} = 1.5  of (4)  implies (squaring both sides)

    2.25 = {{{alpha^2*beta^2 + alpha^2*beta^2 + beta^2*gamma^2 + 2*alpha^2*beta*gamma + 2*alpha*beta^2*gamma + 2*alpha*beta*gamma^2}}} = 

         = {{{alpha^2*beta^2 + alpha^2*beta^2 + beta^2*gamma^2}}} + {{{2*(alpha*beta*gamma)*(alpha+beta+gamma)}}} = substituting the known values from (4) = 

         = {{{alpha^2*beta^2 + alpha^2*beta^2 + beta^2*gamma^2}}} + 2*(-2.5)*4,

which implies

    {{{alpha^2*beta^2 + alpha^2*beta^2 + beta^2*gamma^2}}} = 2.25 + 20 = 22.25.


Thus we know all three coefficients of the polynomial (5)

    {{{-(alpha^2+beta^2+gamma^2)}}} = -13  at  x^2;          

    {{{alpha^2*beta^2+alpha^2*gamma^2+beta^2*gamma^2}}} = 22.25 at x;   and 

    {{{-alpha^2*beta^2*gamma^2}}} = -6.25 as the constant term.   


<U>Answer</U>.  The polynomial equation under the question is  {{{x^3 -13x^2 + 22.25x - 6.25}}} = 0.
</pre>

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