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Solve for $x$, it has four different solutions:

$$x^4 -2x^3-6x^2-2x+1=0$$

MPW
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Rahul Kumar
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    Please write your problem out, don't use images (nobody wants to chase down your problem statement), and also show what you have tried and where you got stuck. – MPW Nov 02 '18 at 13:38
  • The question says "solve for x" x^4 - 2x^3 + 6x^2 - 2x + 1 = 0 – Rahul Kumar Nov 02 '18 at 13:39
  • See https://math.stackexchange.com/questions/480102/quadratic-substitution-question-applying-substitution-p-x-frac1x-to-2x4x – lab bhattacharjee Nov 02 '18 at 13:47
  • Okay, I transcribed it for you. You should read the MathJax quick help (the math formatting language you use here). – MPW Nov 02 '18 at 13:47
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    This is a quartic equation. A biquadratic equation has the form $ax^4+bx^2+c=0$. – Bernard Nov 02 '18 at 13:50

3 Answers3

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Hint:

Grouping $-2x^3=2x^2(-x),-2x=2(-x)1$

$$(x^2-x+1)^2=(3x)^2$$

lab bhattacharjee
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Using the rational roots theorem, you find that $-1$ is a root. Furthermore, you obtain by long division that $\;x^4 -2x^3-6x^2-2x+1=(x+1)(x^3-3x2-3x+1)$ and can check that $-1$ is actually a double root: $$x^4 -2x^3-6x^2-2x+1=(x+1)^2(x^2-4x+1).$$

Bernard
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1]1just ssolve the last part, hope this helps