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While practicing a couple of problems in equality, I realized that it would be rather handy to have a method which can help one to prove whether a polynomial of degree n with real coefficients is positive or negative. In the case of quadratics, it is quite trivial as one may factorize the expression or rewrite it after perfect squaring. So, for instance consider the following quaritic equation:-

2*x^4-8*x^3+24*x^2-32*x+14 >0

Is there a way to prove that this function is always positive and in general is there a technique which could be used for all polynomials.

(Methods that involve Calculus will be really appreciated.)

PS:I think that finding the Range of the polynomial might to do the trick, if the lowest point(if it exists) is positive than the polynomial will always be positive.

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    $\begingroup$ Guess two possible roots, and reduce it then to a quadratic polynomial. $\endgroup$ – Hetebrij Oct 5 '15 at 10:30
  • $\begingroup$ For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$ – Jesse P Francis Oct 5 '15 at 10:32
  • $\begingroup$ Can we use calculus. $\endgroup$ – model_checker Oct 5 '15 at 10:32
  • $\begingroup$ We can use calculus, if you can solve a polynomial of the third degree. $\endgroup$ – Hetebrij Oct 5 '15 at 10:37
  • $\begingroup$ By differentiating the Polynomial.. $\endgroup$ – model_checker Oct 5 '15 at 10:39
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The derivative of the polynomial $p(x) = 2x^4 - 8x^3 + 24x^2 - 32x + 14$ is $$ p'(x) = 8x^3 - 24x^2 + 48x - 32 = 8(x^3 - 3x^2 + 6x - 4). $$ The possible integer roots of p'(x) are divisors of $-4$. We are lucky, $1$ is such a root. The polynomial $x - 1$ is a factor of $p'(x)$: $$ p'(x) = 8(x - 1)(x^2 - 2x + 4) = 8(x - 1)\bigl((x-1)^2+3\bigr) $$ Clearly, there are no other roots. The sign of $p'(x)$ is the same as that of $x-1$: $$ p'(x) \begin{cases} < 0 & \text{if $x < 1$},\\ = 0 & \text{if $x = 0$}, \\ > 0 & \text{if $x > 1$}. \end{cases} $$ Thus, the polynomial is strictly decreasing on interval $(-\infty,1]$, strictly increasing on interval $[1,+\infty)$ and $p(1) = 0$, so we conclude that $p(x) \geqslant 0$ for all real $x$ (and $p(x)>0$ for $x\neq 1$).

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    $\begingroup$ @JessePFrancis: yes, thanks. $\endgroup$ – Éric Guirbal Oct 5 '15 at 11:34
  • $\begingroup$ what means OP ? $\endgroup$ – Éric Guirbal Oct 5 '15 at 11:39
  • $\begingroup$ For meaning of "OP" see here! $\endgroup$ – Jesse P Francis Oct 5 '15 at 11:41

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