# Solving Polynomial Inequalities

How to solve the polynomial inequalities by using the exponents of the factors: basic graphs, examples, and their solutions.

## Graphs of the Odd Exponent Functions

These are the graphs of y = x, x3, x5.

The exponents of the factors are all [odd].

So the graphs of the functions
[pass through] the x-axis
at the zero (x = 0).

## Graphs of the Even Exponent Functions

These are the graphs of y = x2, x4, x6.

The exponents of the factors are all [even].

So the graphs of the functions
[bounce off] the x-axis
at the zero (x = 0).

## Example 1: Solve x4 - x2 < 0

To find the zeros,
factor the left side.

x4 - x2
= x2(x2 - 1)

Common monomial factor

x2 - 12 = (x + 1)(x - 1)

Factor the differnce of squares (a2 - b2)

So x2(x + 1)(x - 1) < 0.

Graph y = x2(x + 1)(x - 1)
on the x-axis.

The zeros are 0, -1, and 1.
Draw the zeros on the x-axis.

x = 1 is the zero of the factor (x - 1).

The exponent of (x - 1) is 1.
It's [odd].

So, starting from the upper right side of the x-axis,
draw the graph
that [passes through] the x-axis
at x = 1.

x = 0 is the zero of the factor x2.

The exponent of x2 is 2.
It's [even].

So draw the graph
that [bounces off] the x-axis
at x = 0.

x = -1 is the zero of the factor (x + 1).

The exponent of (x + 1) is 1.
It's [odd].

So draw the graph
that [passes through] the x-axis
at x = -1.

So this is the graph of
y = x2(x + 1)(x - 1).

The graph passes through the x-axis
at x = 1 and -1.

And the graph bounces off the x-axis
at x = 0.

Find the range that satisfies
y = x2(x + 1)(x - 1) < 0.

y is [less than] then 0.

So color the [lower region] of the graph,
[excluding the zeros].

Then -1 < x < 0, 0 < x < 1.

You can also write the answer as
-1 < x < 1, x ≠ 0.

## Example 2: Solve x3 + 3x2 - 16x + 12 ≥ 0

To find the zeros,
use synthetic division to factor the left side.

Factor theorem

Write the coefficients of the terms:
1 3 -16 12.

Write the L shape form like this.

Pick a number
that seems to make the remainder 0.

1 seems to be the zero.

So write 1
next to the form.

: 1
↗: 1⋅1 = 1
: 3 + 1 = 4
↗: 4⋅1 = 4
: -16 + 4 = -12
↗: -12⋅1 = -12
: 12 - 12 = 0

The remainder is 0.

So 1 is the zero of the left side.

Do synthetic division again.

Pick a number
that seems to make the remainder 0.

2 seems to be the zero.

So write 2
next to the form.

: 1
↗: 1⋅2 = 2
: 4 + 2 = 6
↗: 6⋅2 = 12
: -12 + 12 = 0

The remainder is 0.

So 2 is the zero of the left side.

Do synthetic division again.

Pick a number
that seems to make the remainder 0.

-6 seems to be the zero.

So write -6
next to the form.

: 1
↗: 1⋅(-6) = -6
: 6 - 6 = 0

The remainder is 0.

So -6 is the zero of the left side.

1, 2, and -6 are the zeros.
And the quotient is 1.

So (x - 1)(x - 2)(x - (-6))⋅1 ≥ 0.

Graph y = (x - 1)(x - 2)(x - (-6))
on the x-axis.

The zeros are 1, 2, and -6.
Draw the zeros on the x-axis.

x = 2 is the zero of the factor (x - 2).

The exponent of (x - 2) is 1.
It's [odd].

So, starting from the upper right side of the x-axis,
draw the graph
that [passes through] the x-axis
at x = 2.

x = 1 is the zero of the factor (x - 1).

The exponent of (x - 1) is 1.
It's [odd].

So draw the graph
that [passes through] the x-axis
at x = 1.

x = -6 is the zero of the factor (x - (-6)).

The exponent of (x - (-6)) is 1.
It's [odd].

So draw the graph that
[passes through] the x-axis
at x = -6.

So this is the graph of
y = (x - 1)(x - 2)(x - (-6)).

The graph passes through the x-axis
at x = -6, 1, and 2.

Find the range that satisfies
y = (x - 1)(x - 2)(x - (-6)) ≥ 0.

y is [greater than or equal to] then 0.

So color the [upper region] of the graph,
[including the zeros].

Then -6 ≤ x ≤ 1, x ≥ 2.

## Example 3: Solve x4 + x3 - 5x2 + 3x ≤ 0

To find the zeros,
factor the left side.

x4 + x3 - 5x2 + 3x
= x(x3 + x2 - 5x + 3)

Common monomial factor

Use the synthetic division
to find the zeros of [x3 + x2 - 5x + 3].

Write the coefficients of the terms:
1 1 -5 3.

Write the L shape form like this.

Pick a number
that seems to make the remainder 0.

1 seems to be the zero.

So write 1
next to the form.

: 1
↗: 1⋅1 = 1
: 1 + 1 = 2
↗: 2⋅1 = 2
: -5 + 2 = -3
↗: -3⋅1 = -3
: 3 - 3 = 0

The remainder is 0.

So 1 is the zero of the left side.

Do synthetic division again.

Pick a number
that seems to make the remainder 0.

Another 1 seems to be the zero.

So write 1
next to the form.

: 1
↗: 1⋅1 = 1
: 2 + 1 = 3
↗: 3⋅1 = 3
: -3 + 3 = 0

The remainder is 0.

So 1 is the zero of the left side.

Do synthetic division again.

Pick a number
that seems to make the remainder 0.

-3 seems to be the zero.

So write -3
next to the form.

: 1
↗: 1⋅(-3) = -3
: 3 - 3 = 0

The remainder is 0.

So -3 is the zero of the left side.

1, 1, and -3 are the obtained zeros.
And the quotient is 1.

So x(x - 1)2(x - (-3))⋅1 ≤ 0.

(Don't forget the the factor x.)

Graph y = x(x - 1)2(x - (-3))
on the x-axis.

The zeros are 0, 1, and -3.
Draw the zeros on the x-axis.

x = 1 is the zero of the factor (x - 1)2.

The exponent of (x - 1)2 is 2.
It's [even].

So, starting from the upper right side of the x-axis,
draw the graph
that [bounces off] the x-axis
at x = 1.

x = 0 is the zero of the factor x.

The exponent of x is 1.
It's [odd].

So draw the graph
that [passes through] the x-axis
at x = 0.

x = -3 is the zero of the factor (x - (-3)).

The exponent of (x - (-3)) is 1.
It's [odd].

So draw the graph
that [passes through] the x-axis
at x = -3.

So this is the graph of
y = x(x - 1)2(x - (-3)).

The graph passes through the x-axis
at x = -3 and 0.

And the graph bounces off the x-axis
at x = 1.

Find the range that satisfies
y = x(x - 1)2(x - (-3)) ≤ 0.

y is [less than or equal to] then 0.

So color the [lower region] of the graph,
[including the zeros].

Then -3 ≤ x ≤ 0, x = 1.

## Example 4: Solve x4 - x < 0

To find the zeros,
factor the left side.

x4 - x
= x(x3 - 1)

Common monomial factor

x3 - 13 = (x - 1)(x2 + x + 1)

Factor the differnce of two cubes (a3 - b3)

So x(x - 1)(x2 + x + 1) < 0.

(x2 + x + 1) cannot be factored.

Then, to find out its sign,
find the discriminant D.

D = 12 - 4⋅1⋅1

The discriminant

12 = 1
-4⋅1⋅1 = -4

1 - 4 = -3

D = -3 < 0

So y = x2 + x + 1
has no real roots.

So y = x2 + x + 1
is above the x-axis.

So y = x2 + x + 1 > 0.

Quadratic function - number of zeros

Get back to x(x - 1)(x2 + x + 1) < 0.

(x2 + x + 1) > 0

So you can divide both sides by (x2 + x + 1).
The order of the inequality sign won't change.

Then x(x - 1) < 0.

Graph y = x(x - 1)
on the x-axis.

The coefficient of x2 is (+): 1.
And its zeros are 0 and 1.

So draw a parabola
that is opened upward
and that passes through x = 0, 1 on the x-axis.