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$$x^4 -2x^3-6x^2-2x+1=0$$ $\left[\text{Hint let } v = x + \frac{1}{x}\right]$

I am stumped, and have no idea how to proceed. I have tried solving it, but have had no success.
 

P.S: This question is meant to be solved, using only techniques for solving quadratic equations.

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5 Answers

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I think you meant for $$x^4 -2x^3-6x^2-2x+1=0$$

Obviously $x=0$ is not a solution. So we divide by $x^2$ and get

$$x^2 -2x-6-2x^{-1}+x^{-2}=0$$ $$(x^2+2+x^{-2}) - 2(x+x^{-1}) -8=0$$ $$v^2-2v-8=0$$ $$(v-4)(v+2)=0$$

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since $x=0$ is not a solution we can divide by $x^2$ $$x^2+\frac{1}{x^2}-2\left(x+\frac{1}{x}\right)-6=0$$ and no set $$t=x+\frac{1}{x}$$ then you will get $$t^2=x^2+\frac{1}{x^2}+2$$ and $$t^2-2=x^2+\frac{1}{x^2}$$ and you have to solve $$t^2-2t-8=0$$

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I would not bother with the substitution. The only possible rational roots are $\pm 1$ and it is easy to check that $-1$ works. A quick calculation shows that your polynomial is $$(x + 1)^2 \,(x^2 - 4 x + 1)$$ and we are done.

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

Divide all terms by $x^2$

$x^2-2x-6-\dfrac{2}{x}+\dfrac{1}{x^2}=0$

Rearrange in this way

$\left(x^2+\dfrac{1}{x^2}\right)-2\left(x+\dfrac{1}{x}\right)-6=0$

Now set $v=x+\dfrac{1}{x}$

squaring you get

$v^2=x^2+\dfrac{1}{x^2}+2\to x^2+\dfrac{1}{x^2}=v^2-2$

Plug in the equation

$v^2-2 -2v -6=0\to v^2-2v-8=0\to v_1=4;\;v_2=-2$

As we want to solve for $x$ two more steps

For $v=4\to x+\dfrac{1}{x}=4 \to x^2-4x+1=0 \to x= 2\pm\sqrt{3}$

for $v=-2\to x+\dfrac{1}{x}=-2\to x^2+2x+1\to x=-1$

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Sometimes you can add and subtract the right thing and land in Pascal's triangle. The 4th row of Pascal's triangle is $1 4 6 4 1$, so add and subtract $6x^2+12x^2+6x$ to get

$$x^4+4x^3+6x^2+4x+1 - 6x^3-12x^2-6x = 0$$

or

$$(x+1)^4 - 6x(x+1)^2=0.$$

Factor out $(x+1)^2$ to get

$$(x+1)^2((x+1)^2 - 6x) = 0$$

or $$(x+1)^2(x^2-4x+1)=0.$$

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