 # Solving Polynomial Inequalities How to solve polynomial inequalities problems: examples and their solutions.

## Example 1 Use the synthetic division
to factor the left side of the inequality.

(x - 1)(x - 2)(x + 6) > 0

Factor theorem

The zeros are 1, 2, and -6.

Mark the zeros on the number line: -6, 1, 2.
Draw empty circles on the zeros.

Graph y = (x - 1)(x - 2)(x + 6).

Start from the top right. (brown point)

(x - 1), (x - 2), and (x + 6) are all odd powers.
So the graph passes through the number line
at x = 2, 1, and -6.

Graphing polynomial functions

The left side is greater than 0.
So draw the region on the number line
where the graph is above the number line.

Then -6 < x < 1, x > 2.

## Example 2 Use the synthetic division
to factor the left side of the inequality.

x(x - 1)2(x + 3) ≤ 0

Factor theorem

The zeros are 0, 1, and -3.

Mark the zeros on the number line: -3, 0, 1.
Draw full circles on the zeros.

Graph y = x(x - 1)2(x + 3).

Start from the top right. (brown point)

(x - 1)2 is an even power.
So the graph bounces off the number line
at x = 1.

x and (x + 3) are odd powers.
So the graph passes through the number line
at x = 0, -3.

Graphing polynomial functions

The left side is lesser than (or equal to) 0.
So draw the region on the number line
where the graph is below the number line.

Then -3 ≤ x ≤ 0, x = 1.

## Example 3 Factor the left side of the inequality.

Factoring the difference of two cubes

(x2 + x + 1) is always (+).
So divide both sides by (x2 + x + 1).

It won't change the order of the inequality sign,

Let's see why (x2 + x + 1) is always (+).

The discriminant D is (-).
So x2 + x + 1 = 0 has no real roots.
And y = x2 + x + 1 is always above the x-axis.

So (x2 + x + 1) is always (+)
and has no real zeros.

Mark the zeros on the number line: 0, 1.
Draw empty circles on the zeros.

Graph y = x(x - 1) on the number line.

The left side is lesser than 0.
So draw the region on the number line
where the graph is below the number line.

Then 0 < x < 1.