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Most of the questions on this site ask: Given a polynomial, how to find the number of real roots.

My question is: given a 6th degree polynomial $P_b(x)$: (b lies between 1 and 4) $$1+b+b^2-bx-2b^2x-2b^3x-b^4x+b^2x^2+3b^3x^2+3b^4x^2+2b^5x^2-b^3x^3-3b^4x^3-5b^5x^3-b^6x^3+b^4x^4+4b^5x^4+3b^6x^4-b^5x^5-3b^6x^5+b^6x^6=0$$

How can I find a condition on $b$ so that $P_b(x)$ has at least three three real roots and $|P_b^{'}(x_i)|<1$ where $x_i$ is a root?

Substituting $y=bx$ $$(1+b+b^2)-(1+2b+2b^2+b^3)y+(1+3b+3b^2+2b^3)y^2-(1+3b+5b^2+b^3)y^3+(1+4b+3b^2)y^4-(1+3b)y^5+y^6=0$$

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    $\begingroup$ first u should collect the lke powers of x :D $\endgroup$ – Shobhit Sep 2 '14 at 17:41
  • $\begingroup$ yeah they are written together $\endgroup$ – Bosnia Sep 2 '14 at 17:42
  • $\begingroup$ The substitution $y=bx$ will simplify things somewhat. $\endgroup$ – Mark Bennet Sep 2 '14 at 17:44
  • $\begingroup$ sorry it was $3b^3x^2$ $\endgroup$ – Bosnia Sep 2 '14 at 17:46
  • $\begingroup$ Shobhit means: instead of e.g. $-b^3x^3 -3b^4x^3 -5b^5x^3 - b^6x^3$, you should write $-(b^3+3b^4+5b^5+b^6)x^3$. And I agree. $\endgroup$ – TonyK Sep 2 '14 at 18:12

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