Question 275827
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Equations/275827: Solve for x.
(x+1)(x+3)(x+5)(x+7)=-15
I've worked this numerous times and get lost every time. Please explain how to find x.
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<pre>
Your starting equation is

    (x+1)(x+3)(x+5)(x+7) = -15.    (1)


Although it leads to the fourth degree polynomial equation, it admits analytical solution
not of transcendent level of complexity.


Let y = x+4.  Then

    x+3 = y-1,  x+5 = y+1,  x+1 = y-3,  x+7 = y+3.


Therefore, equation (1) can be equivalently written in this form

    (y-3)*(y-1)*(y+1)*(y+3 = -15.    (2)


Regrouping the terms, we get

    (y^2-9)*(y^2-1) = -15,

     y^4 - 9y^2 - y^2 + 9 = -15,

     y^4 - 10y^2 +24 = 0,

     (y^2-6)*(y^2-4) = 0.


Therefore, four routs of equation (2) are

    {{{y[1]}}} = {{{sqrt(6)}}},  {{{y[2]}}} = {{{-sqrt(6)}}},  {{{y[3]}}} = -2,  {{{y[4]}}} = 2.


From here, the solutions to equation (1) are

    {{{x[1]}}} = {{{-4+sqrt(6)}}},  {{{x[2]}}} = {{{-4-sqrt(6)}}},  {{{x[3]}}} = -2-4 = -6,  {{{x[4]}}} = 2-4 = -2.



<U>ANSWER</U>.  The solutions to the given equation are {{{x[1]}}} = {{{-4+sqrt(6)}}},  {{{x[2]}}} = {{{-4-sqrt(6)}}},  {{{x[3]}}} = -2-4 = -6,  {{{x[4]}}} = 2-4 = -2.
</pre>

Solved.