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Given a polynomial along with the information that it has no real roots, then how can we say then whether it is greater or less than zero?

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  • $\begingroup$ Look at the coefficient of the largest term, so if $p(x)=a_nx^n+\cdots+a_1x+a_0$, then the sign of the polynomial $p$ will be the sign of $a_n$, since for large $x$, the $x^n$ term will always "win". On the other hand, the sign of $p$ is the same as $p(x)$ for any $x$, so one might find $x=0$ a useful choice since then $p(0)=a_0$. $\endgroup$ – vrugtehagel May 13 '17 at 8:33
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Well..just put $x=0$, or any other convenient value of $x$, and check the sigh of function, if it is always positive then every value of $x$ will result in positive value of function. Similarly if function is negative, every value of $x$ will result in negative value of function.

In short if the constant is $>0$ then the function will be positive and if it is $<0$ then the function will be negative $~\forall x \in \mathbb R$

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If a polynomial has no real roots it does never change sign, so take its value for a given $x$, e.g.$x=0$ and the sign of this value is the sign of the polynomial.

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Hint #1: If $y=f(x)$ is a polynomial function with no zeroes, does the graph of $f$ intersect/touch the $x$-axis? (That is, will $f$ change its sign in its domain?)

Hint #2: If $y=f(x)$ is a polynomial function with no zeroes, and furthermore, $f(x_0) > 0$ for a particular $x_0$ in the domain of $f$, does this mean that $f(x) > 0$ for all $x$ in the domain of $f$? (Similarly for the case when $f(x_1) < 0$ for a particular $x_1$ in the domain of $f$, does that mean that $f(x) < 0$ for all $x$ in the domain of $f$?)

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