Solving Polynomial Inequalities

Like any other function, a polynomial may be written as an inequality, giving a large range of solutions.

The best way to solve a polynomial inequality is to find its zeros. The easiest way to find the zeros of a polynomial is to express it in factored form. At these points, the polynomial's value goes from negative to positive or positive to negative.  This knowledge can then be used to determine the solutions of the inequality.  Much of the work involved with solving inequalities is based in observation and judgement of a particular mathematical situation, and is therefore best demonstrated with an example.

Example

Consider the polynomial inequality:

$x^3+2x^2-5x-6>0$

This can be expressed as the product of three terms:

$(x-2)(x+1)(x+3)>0$

The three terms reveal zeros at $x=-3$, $x=-1$, and $x=2$. We know that the lower limit of the inequality crosses the x-axis at each of these $x$ values, but now have to determine which direction (positive or negative) it takes at each crossing.

$x+3>0$ for $x>-3$

$x+1>0$ for $x>-1$

$x-2>0$ for $x>2$

Thus, as the polynomial crosses the x-axis at $x=-3$, the term $(x+3)$ equals zero, becoming positive to the right. At the same point, $(x+1)$ and $(x-2)$ are negative. The product of a positive and two negatives is positive, so we can conclude that the polynomial becomes positive as it passes $x=-3$.

The next zero is at $x=-1$. From the explanation above, we know that the polynomial is positive as it approaches its next zero, but we can use the same reasoning for proof.  At $x=-1$, $(x+1)$ equals zero, becoming positive to the right. The term $(x+3)$ is positive, while $(x-2)$ is negative. The product of two positives and a negative is negative, so we can conclude that the polynomial becomes negative as it passes $x=-1$.

The same process can be used to show that the polynomial becomes positive again at $x=2$.

Recalling the initial inequality, we can now determine the solution of exactly where the polynomial is greater than zero. Because there is no zero to the left of $x=-3$, we can assume that the polynomial is negative for all $x$ values $-\infty$ to $-3$. The polynomial is positive from $x=-3$ to $x=-1$ before becoming negative once more. It becomes positive at $x=2$, and because there are no more zeros to the right, we can assume the polynomial remains positive as $x$ approaches $\infty$.

Graph of example

Graph of the third-degree polynomial with the equation $y=x^3+2x^2-5x-6$. This polynomial has three roots. It is positive in two segments and also negative in two. If it were a polynomial inequality with the condition that all values are greater than zero, the two negative segments would be removed.

Thus, the solution is: $(-3,-1),(2,\infty)$

For inequalities that are not expressed relative to zero, expressions can be added or subtracted from each side to take it into the desired form.