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How can you find of real roots for $P(x) = x^4 - 4x^3 + 4x^2 - 10$?

Using the Descartes' rule of signs:

The polynomial $P(x) = x^4 - 4x^3 + 4x^2 - 10$ has three sign changes between the first, second, third and fourth terms (the sequence of pairs of successive signs is +, −, +, -), therefore it has exactly three positive roots.

The polynomial $P(-x) = x^4 + 4x^3 + 4x^2 - 10$ has one sign change between the third and fourth terms (the sequence of pairs of successive signs is +++, -), therefore it has exactly one negative root.

Finally, summing up, the polynomial $P(x) = x^4 + 4x^3 + 4x^2 - 10$ has four real roots.

However, the actual answer is two real roots. How can you find this answer and what is wrong with my implementation of the Descartes' rule of signs?

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The number of positive roots is only known to be either the number of sign changes, or less than that by an even number. Same for negative roots after changing sign on $x.$

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You have only counted the maximum possible number of those signed roots. To get all possibilities, you must decrease the number of possible roots by two. There are three sign changes in the original polynomial, so the number of positive roots is at most $3$, with the possibility of there being only $1$ positive root. There is only one sign change in $P(-x)$ (which you should indicate), so there is at most $1$ negative root. Thus it is possible to have $2$ real roots.

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  • $\begingroup$ One can say exactly one negative root. $\endgroup$ – coffeemath Oct 27 '18 at 16:25
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    $\begingroup$ "...so there is at most 1 negative root": and therefore there is exactly one negative root, because 1 decreased by two is not a possible number of roots. (I realise that you know this; I'm just pointing it out for the OP's benefit.) $\endgroup$ – TonyK Oct 27 '18 at 16:26
  • $\begingroup$ I have indicated $P(-x)$ as per your request. Also, how does this rule apply when the root is zero and hence is neither positive nor negative. $\endgroup$ – user532394 Oct 27 '18 at 16:30
  • $\begingroup$ It's easy to see when $0$ is a root: just add it to the total. $\endgroup$ – Sean Roberson Oct 27 '18 at 16:33
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Others have addressed the rule of signs.

The original quartic is easily seen to be $x^2(x-2)^2-10=[x(x-2)]^2-10$. This is very easy to sketch in this form as the turning points are clearly $x=0,1,2$ (subtracting $10$ makes no difference to the shape). Between $x=0$ and $x=2$ the function clearly remains negative with the value at the local maximum at $x=1$ being $-9$.

Since there must be a turning point between any pair of roots, there can be at most two real roots.

Note that the quartic readily factors as $[x(x-2)-\sqrt {10}][x(x-2)+\sqrt {10}]$ using the difference of two squares, and all the roots can be computed by solving the quadratics.

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Let $\>x=t+1\>$ then $$ \\P(x)=P(t+1)=(t+1)^4-4(t+1)^3+4(t+1)^2-10= \\=(t^4+4t^3+6t^2+4t+1)-(4t^3+12t^2+12t+4)+(4t^2+8t+4)-10= \\=t^4-2t^2-9 \\P(t+1)=0=>t^4-2t^2-9=0=> $$ if $\>x\>$ is real then $\>t\>$ is real => $$ \\t^2=\dfrac{2+\sqrt{40}}{2}=1+\sqrt{10} $$ Answer $\>x=1+\sqrt{1+\sqrt{10}}\>$ or $\>x=1-\sqrt{1+\sqrt{10}}\>$

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