# 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*, *x*^{3}, *x*^{5}.

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* = *x*^{2}, *x*^{4}, *x*^{6}.

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 *x*^{4} - *x*^{2} < 0

To find the zeros,

factor the left side.*x*^{4} - *x*^{2}

= *x*^{2}(*x*^{2} - 1)

Common monomial factor

*x*^{2} - 1^{2} = (*x* + 1)(*x* - 1)

Factor the differnce of squares (*a*^{2} - *b*^{2})

So *x*^{2}(*x* + 1)(*x* - 1) < 0.

Graph *y* = *x*^{2}(*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 *x*^{2}.

The exponent of *x*^{2} 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* = *x*^{2}(*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* = *x*^{2}(*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 *x*^{3} + 3*x*^{2} - 16*x* + 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 *x*^{4} + *x*^{3} - 5*x*^{2} + 3*x* ≤ 0

To find the zeros,

factor the left side.*x*^{4} + *x*^{3} - 5*x*^{2} + 3*x*

= *x*(*x*^{3} + *x*^{2} - 5*x* + 3)

Common monomial factor

Use the synthetic division

to find the zeros of [*x*^{3} + *x*^{2} - 5*x* + 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 *x*^{4} - *x* < 0

To find the zeros,

factor the left side.*x*^{4} - *x*

= *x*(*x*^{3} - 1)

Common monomial factor

*x*^{3} - 1^{3} = (*x* - 1)(*x*^{2} + *x* + 1)

Factor the differnce of two cubes (*a*^{3} - *b*^{3})

So *x*(*x* - 1)(*x*^{2} + *x* + 1) < 0.

(*x*^{2} + *x* + 1) cannot be factored.

Then, to find out its sign,

find the discriminant *D*.*D* = 1^{2} - 4⋅1⋅1

The discriminant

1^{2} = 1

-4⋅1⋅1 = -4

1 - 4 = -3

*D* = -3 < 0

So *y* = *x*^{2} + *x* + 1

has no real roots.

So *y* = *x*^{2} + *x* + 1

is above the *x*-axis.

So *y* = *x*^{2} + *x* + 1 > 0.

Quadratic function - number of zeros

Get back to *x*(*x* - 1)(*x*^{2} + *x* + 1) < 0.

(*x*^{2} + *x* + 1) > 0

So you can divide both sides by (*x*^{2} + *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 *x*^{2} 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.

Solving quadratic inequalities

Find the range that satisfies*y* = *x*(*x* - 1) < 0.*y* is [less than] then 0.

So color the [lower region] of the graph,

[excluding the zeros].

Then 0 < *x* < 1.