Question 513731
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It is not a quadratic equation.  Quadratic equations have polynomials of degree 2.  This equation has a polynomial (if you multiply it out) of degree 4.  It is a quartic equation.  


Be that as it may, since most of the difficult factoring work has been done, take two factors of *[tex \Large (x\ -\ 2)] out.  The difference is that the end result will have four roots instead of just two (The Fundamental Theorem of Algebra said it, I believe it, that settles it)


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x(x\ -\ 2)^3\ -\ 35(x\ -\ 2)^2\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ 2)^2\left(x(x\ -\ 2)\ -\ 35\right)\ =\ 0]


Get rid of the inner parentheses:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ 2)^2\left(x^2\ -\ 2x\ -\ 35\right)\ =\ 0]


You can see right away that two of your roots are 2 and 2.  Use the fact that -7 plus 5 is -2 and -7 times 5 is -35 to factor the quadratic term, giving you 7 and -5 as the additional two roots.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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