# Inequalities With Polynomial and Rational Functions

## Polynomial Inequalities

Polynomials can be expressed as inequalities, the solutions for which can be determined from the polynomial's zeros.

### Learning Objectives

Solve for the zeros of a polynomial inequality to find its solution

### Key Takeaways

#### Key Points

• To solve a polynomial inequality, first rewrite the polynomial in its factored form to find its zeros.
• For each zero, input the value of the zero in place of
$x$
in the polynomial. Determine the sign (positive or negative) of the polynomial as it passes the zero in the rightward direction.
• Determine the intervals between these roots which satisfy the inequality.

#### Key Terms

• inequality: A statement that of two quantities, one is specifically less than or greater than another. Symbols: < or ≤ or > or ≥, as appropriate.

### 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
$\infty$
.

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.

## Rational Inequalities

Rational inequalities can be solved much like polynomial inequalities.

### Learning Objectives

Solve for the zeros and asymptotes of a rational inequality to find its solution

### Key Takeaways

#### Key Points

• First factor the numerator and denominator polynomial to reveal the zeros in each.
• Substitute
$x$
with a zero (root) to determine whether the rational function is positive or negative to the right of that point. Repeat for all zeros.
• The intervals that satisfy the inequality symbol will be the answer. Note that for any
$\geq$
or
$\leq$
, the interval will only be closed to include the zero if the zero is found in the numerator. If the zero is found in the denominator, that point is undefined, and cannot be included in the solution.

#### Key Terms

• zero: Also known as a root, a zero is an
$x$
value at which the function of
$x$
is equal to zero.
• inequality: A statement that of two quantities one is specifically less than or greater than another. Symbols:
$<$
or
$\leq$
or
$>$
or
$\geq$
, as appropriate.

As with solving polynomial inequalities, the first step to solving rational inequalities is to find the zeros.  Because a rational expression consists of the ratio of two polynomials, the zeroes for both polynomials will be needed.

The zeros in the numerator are
$x$
-values at which the rational inequality crosses from negative to positive or from positive to negative. The zeros in the denominator are
$x$
-values are at which the rational inequality is undefined, the result of dividing by zero.

### Example

Consider the rational inequality:

$\frac{x^2+2x-3}{x^2-4}>0$

This equation can be factored to give:

$\frac{(x+3)(x-1)}{(x+2)(x-2)}\geq 0$

As
$x$
crosses rightward past
$-3$
,
$(x+3)$
becomes positive. At that same point,
$(x-1)$
,
$(x+2)$
, and
$(x-2)$
are all negative. The product of a positive and three negatives is negative, so the rational expression becomes negative as it crosses
$x=-3$
in the rightward direction.

The same process can be used to determine that the rational expression is positive after passing the zero at
$x=-2$
, is negative after passing
$x=1$
, and is positive after passing
$x=2$
.

Thus we can conclude that for
$x$
values on the open interval from
$-\infty$
to
$-3$
, the rational expression is negative. From
$-3$
to
$-2$
, it is positive; from
$-2$
to
$1$
it is negative; from
$1$
to
$2$
it is positive, and from
$2$
to
$\infty$
it is negative.

Because the inequality is written as
$\geq0$
as opposed to
$>0$
, we will need to evaluate the
$x$
values at zeros to determine whether the function is defined.

In the case of
$x=-2$
and
$x=2$
, the rational function has a denominator equal to zero and becomes undefined.

In the case of
$x=-3$
and
$x=1$
, the rational function has a numerator equal to zero, which makes the function overall equal to zero, making it inclusive in the solution.

Thus, the full solution is:

[-3, -2), [1, 2)