# Polynomial Inequality

How solve a polynomial inequality: graph, 4 examples, and their solutions.

## Graph

### Odd Exponent Function

These are the graphs of

y = x^{1}, y = x^{3}, y = x^{5}.

If the exponent is odd,

then the graph passes through the x-axis

at the origin.

### Even Exponent Function

These are the graphs of

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

If the exponent is even,

then the graph bounces off the x-axis

at the origin.

So, when drawing the polynomial function,

see the exponents of the factors.

If the exponent is odd,

then the graph passes through the x-axis.

If the exponent is even,

then the graph bounces off the x-axis.

## Example 1

### Example

### Solution

x^{4} - x^{2} = x^{2}(x^{2} - 1)

Common Monomial Factor

x^{2} - 1

= x^{2} - 1^{2}

= (x + 1)(x - 1)

Factor the Difference of Two Squares: a^{2} - b^{2}

Arrange the factors

so that

the zeros of the factors are in ascending order.

The zero of x^{2} is 0.

The zero of (x + 1) is -1.

The zero of (x - 1) is 1.

So first write (x + 1):

the zero is -1.

Write x^{2}:

the zero is 0.

And write (x - 1):

the zero is 1.

Write the zeros.

x = -1, 0, 1

Draw the x-axis.

Point the zeros x = -1, 0, 1.

Draw y = (x + 1)x^{2}(x - 1)

on the x-axis.

The highest degree term of y = (x + 1)x^{2}(x - 1)

is x^{4}.

The coefficient is (+).

So starting from the top right of the x-axis,

draw the graph

that goes toward the nearest zero:

x = 1.

See the factor (x - 1).

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

The exponent is 1.

It's odd.

Then draw the graph

that passes through the x-axis

at x = 1.

Next, see the factor x^{2}.

The exponent is 2.

It's even.

Then draw the graph

that bounces off the x-axis

at x = 0.

See the factor (x + 1).

(x + 1) = (x + 1)^{1}

The exponent is 1.

It's odd.

Then draw the graph

that passes through the x-axis

at x = -1.

So this is the graph of the polynomial

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

on the x-axis.

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

The left side is less than 0.

So color the region

where the graph is below the x-axis (y = 0).

The inequality sign does not include equal to [=].

So draw empty circles on the zeros:

x = -1, 0, 1.

The colored regions are

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

You can also write the region as

-1 < x < 1, x ≠ 0.

So

-1 < x < 0, 0 < x < 1

is the answer.

## Example 2

### Example

### Solution

Factor the right side

by using the synthetic division.

Factor Theorem

f(x) = x^{3} + x^{2} - 10x + 8

Write the coefficients of the terms

in descending order:

1 1 -10 8.

Draw an L shape form like this.

Pick a number

that seems to make the remainder 0.

1 seems to be good.

So write 1

on the left side of the form.

↓: 1 = 1

↗: 1⋅1 = 1

↓: 1 + 1 = 2

↗: 2⋅1 = 2

↓: -10 + 2 = -8

↗: -8⋅1 = -8

↓: 8 - 8 = 0

The remainder is 0.

So 1 is the right number.

1 2 -8 means

x^{2} + 2x - 8.

This seems to be factorable.

So do the synthetic division again.

Draw an L shape form like this.

Pick a number

that seems to make the remainder 0.

2 seems to be good.

So write 2

on the left side of the form.

↓: 1 = 1

↗: 1⋅2 = 2

↓: 2 + 2 = 4

↗: 4⋅2 = 8

↓: -8 + 8 = 0

The remainder is 0.

So 2 is the right number.

1 4 means

x + 4.

(x + 4) is a binomial factor.

So stop finding the factors.

The left side numbers are

1 and 2.

So write the factors

(x - 1)(x - 2).

1 4 means

x + 4.

So write (x + 4).

See the given inequality.

There's ≥ 0.

So write ≥ 0.

So the given inequality is

(x - 1)(x - 2)(x + 4) ≥ 0.

Arrange the factors

so that

the zeros of the factors are in ascending order.

The zero of (x - 1) is 1.

The zero of (x - 2) is 2.

The zero of (x + 4) is -4.

So first write (x + 4):

the zero is -4.

Write (x - 1):

the zero is 1.

And write (x - 2):

the zero is 2.

Write the zeros.

x = -4, 1, 2

Draw the x-axis.

Point the zeros x = -4, 1, 2.

Draw y = (x + 4)(x - 1)(x - 2)

on the x-axis.

The highest degree term of y = (x + 4)(x - 1)(x - 2)

is x^{3}.

The coefficient is (+).

So starting from the top right of the x-axis,

draw the graph

that goes toward the nearest zero:

x = 2.

See the factor (x - 2).

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

The exponent is 1.

It's odd.

Then draw the graph

that passes through the x-axis

at x = 2.

See the factor (x - 1).

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

The exponent is 1.

It's odd.

Then draw the graph

that passes through the x-axis

at x = 1.

See the factor (x + 4).

(x + 4) = (x + 4)^{1}

The exponent is 1.

It's odd.

Then draw the graph

that passes through the x-axis

at x = -4.

So this is the graph of the polynomial

y = (x + 4)(x - 1)(x - 2)

on the x-axis.

See (x + 4)(x - 1)(x - 2) ≥ 0.

The left side is greater than or equal to 0.

So color the region

where the graph is above the x-axis (y = 0).

The inequality sign does include equal to [=].

So draw full circles on the zeros:

x = -4, 1, 2.

The colored regions are

-4 ≤ x ≤ 1, x ≥ 2.

So

-4 ≤ x ≤ 1, x ≥ 2

is the answer.

## Example 3

### Example

### Solution

x^{4} + x^{3} - 5x^{2} + 3x

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

Factor (x^{3} + x^{2} - 5x + 3)

by using the synthetic division.

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

Write the coefficients of the terms

in descending order:

1 1 -5 3.

Draw an L shape form like this.

Pick a number

that seems to make the remainder 0.

1 seems to be good.

So write 1

on the left side of the form.

↓: 1 = 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 right number.

1 2 -3 means

x^{2} + 2x - 3.

This seems to be factorable.

So do the synthetic division again.

Draw an L shape form like this.

Pick a number

that seems to make the remainder 0.

1 seems to be good.

So write 1

on the left side of the form.

↓: 1 = 1

↗: 1⋅1 = 1

↓: 2 + 1 = 3

↗: 3⋅1 = 3

↓: -3 + 3 = 0

The remainder is 0.

So 1 is the right number.

1 3 means

x + 3.

(x + 3) is a binomial factor.

So stop finding the factors.

Then, write the front x.

The left side numbers are

1 and 1.

So write the factor

(x - 1)^{2}.

1 3 means

x + 3.

So write (x + 3).

Write ≤ 0.

So x(x^{3} + x^{2} - 5x + 3) ≤ 0 is

x(x - 1)^{2}(x + 3) ≤ 0.

Arrange the factors

so that

the zeros of the factors are in ascending order.

The zero of x is 0.

The zero of (x - 1)^{2} is 1.

The zero of (x + 3) is -3.

So first write (x + 3):

the zero is -3.

Write x:

the zero is 0.

And write (x - 1)^{2}:

the zero is 1.

Write the zeros.

x = -3, 0, 1

Draw the x-axis.

Point the zeros x = -3, 0, 1.

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

on the x-axis.

The highest degree term of y = (x + 3)x(x - 1)^{2}

is x^{4}.

The coefficient is (+).

So starting from the top right of the x-axis,

draw the graph

that goes toward the nearest zero:

x = 1.

See the factor (x - 1)^{2}.

The exponent is 2.

It's even.

Then draw the graph

that bounces off the x-axis

at x = 1.

See the factor x.

x = x^{1}

The exponent is 1.

It's odd.

Then draw the graph

that passes through the x-axis

at x = 0.

See the factor (x + 3).

(x + 3) = (x + 3)^{1}

The exponent is 1.

It's odd.

Then draw the graph

that passes through the x-axis

at x = -3.

So this is the graph of the polynomial

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

on the x-axis.

See (x + 3)x(x - 1)^{2} ≤ 0.

The left side is less than or equal to 0.

So color the region

where the graph is below the x-axis (y = 0).

The inequality sign does include equal to [=].

So draw full circles on the zeros:

x = -3, 0, 1.

The colored regions are

-3 ≤ x ≤ 0, x = 1.

So

-3 ≤ x ≤ 0, x = 1

is the answer.

## Example 4

### Example

### Solution

x^{4} - x = x(x^{3} - 1)

x^{3} - 1

= x^{3} - 1^{3}

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

Factor the Difference of Two Cubes: a^{3} - b^{3}

See (x^{2} + x + 1).

To find the sign of (x^{2} + x + 1),

find the discriminant D.

The discriminant D is

D = -3.

D = -3

D is minus.

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

has no zeros.

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

is above the x-axis (y = 0).

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

This means

(x^{2} + x + 1) is plus.

Quadratic Function: Number of Zeros

Then see x(x - 1)(x^{2} + x + 1) < 0.

(x^{2} + x + 1) is plus: not 0.

So divide both sides by (x^{2} + x + 1).

Then x(x - 1) < 0.

(x^{2} + x + 1) is plus.

So dividing both sides by (x^{2} + x + 1)

does not change the order of the inequality sign.

Write the zeros.

x = 0, 1

Draw y = x(x - 1)

on the x-axis.

First point the zeros x = 0, 1.

And draw a parabola

that passes through x = 0, 1.

Quadratic Inequality

See x(x - 1) < 0.

The left side is less than to 0.

So color the region

where the graph is below the x-axis (y = 0).

The inequality sign does not include equal to [=].

So draw empty circles on the zeros:

x = 0 and x = 1.

The colored region is

0 < x < 1.

So

0 < x < 1

is the answer.