Given a polynomial of $n$th order, represented by $$f(x)=a_{0}x^{n}+a_{1}x^{n-1}+a_{2}x^{n-2}+\cdots+a_{n-2}x^{2}+a_{n-1}x+a_{n}=0$$
Is it possible to find the number of integral solutions/roots to any general polynomial like this?

  • 2
    $\begingroup$ Certainly. Apply the rational roots theorem. $\endgroup$
    – Kaj Hansen
    Commented Jan 6, 2015 at 16:24

1 Answer 1


You can apply the rational roots theorem as well as synthetic division, but there is one method I have found--Descartes' Rule of Signs. This rule tells one how many roots of a certain type to expect. In order to use this to find how many roots, you first look at the polynomial as it stands and look at the signs of the coefficients. You then count how many sign changes there are in the polynomial. Take $f(x) = 6x^3 \color{blue}{+} 5x^2 \color{red}{-} 4x \color{blue}{+} 3$. There are 2 sign changes in the positive root case. The changes are highlighted, from the positive $5x^2$ to the negative $4x$ and from the negative $4x$ and the positive 3. But, some of them can be generated via the quadratic formula, hence resulting in complex roots that are not very graphable. Because of this, you have to count down by twos to find the complete list of possible number of roots. So you can have either 0 or 2 positive roots.

Next up, the negative root case. We change the sign of $x$ to get: $f(-x) = -6x^3 + 5x^2 + 4x - 3$. In this case, there are also 2 sign changes-- the negative $6x^3$ to the positive $5x^2$ and the positive $4x$ to the negative 3. We count down by twos to get either 0 or 2 negative roots. Because of this you can expect 0 or 2 positive roots and 0 or 2 negative roots.

So, one of the possible ways to find the number of roots in a polynomial is to use Descartes' Rule of Signs.



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