# Polynomial Inequality

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

## Graph

### Odd Exponent Function

These are the graphs of
y = x1, y = x3, y = x5.

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 = x2, y = x4, y = x6.

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

### Solution

x4 - x2 = x2(x2 - 1)

Common Monomial Factor

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

Factor the Difference of Two Squares: a2 - b2

Arrange the factors
so that
the zeros of the factors are in ascending order.

The zero of x2 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 x2:
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)x2(x - 1)
on the x-axis.

The highest degree term of y = (x + 1)x2(x - 1)
is x4.
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 x2.

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)x2(x - 1)
on the x-axis.

See (x + 1)x2(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

## Example 2

### Solution

Factor the right side
by using the synthetic division.

Factor Theorem

f(x) = x3 + x2 - 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
x2 + 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 x3.
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

## Example 3

### Solution

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

Factor (x3 + x2 - 5x + 3)
by using the synthetic division.

f(x) = x3 + x2 - 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
x2 + 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(x3 + x2 - 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 x4.
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 = x1

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

## Example 4

### Solution

x4 - x = x(x3 - 1)

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

Factor the Difference of Two Cubes: a3 - b3

See (x2 + x + 1).

To find the sign of (x2 + x + 1),
find the discriminant D.

The discriminant D is
D = -3.

D = -3

D is minus.

So y = x2 + x + 1
has no zeros.

So y = x2 + x + 1
is above the x-axis (y = 0).

So y = x2 + x + 1 > 0.

This means
(x2 + x + 1) is plus.

Then see x(x - 1)(x2 + x + 1) < 0.

(x2 + x + 1) is plus: not 0.
So divide both sides by (x2 + x + 1).

Then x(x - 1) < 0.

(x2 + x + 1) is plus.
So dividing both sides by (x2 + 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.

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