# College Algebra : Polynomial Inequalities

## Example Questions

### Example Question #1 : Polynomial Inequalities

Solve

Explanation:

First we substract  on each side

Now we can factor the left hand side

From this, we can see that the roots are at , and .

To make this inequality true, we will have two answers.

### Example Question #2 : Polynomial Inequalities

When is the inequality  true?

this inequality cannot be satisfied.

Explanation:

We can rewrite this as . In order to solve the inequality we need to factor the polynomial. Taking a couple educated guesses we find that this polynomial factors as . So now we have our inequality written as

.

When is this true? When both linear factors are positive, and also when they are both negative.

If we match them up then we do get that the inequality is satisfied when  and when .

### Example Question #3 : Polynomial Inequalities

Give the solution set of the inequality

The set of all real numbers

Explanation:

The boundaries of the intervals in the solution set of the polynomial inequality

are the zeroes of the equation

By the Zero Product Property, either

,

in which case

,

or

,

in which case

and 4 are the boundaries of the intervals, which are , and . To determine which intervals belong to the solution set, choose one value from each, substitute for  in the original inequality, and determine whether it is true.

: Choose

False - do not include

: Choose

True - include

: Choose

True - include

Also, the boundary values themselves should not be included, because the inequality symbol  does not allow for equality.

The solution set is therefore

### Example Question #4 : Polynomial Inequalities

Give the solution set of the inequality

Explanation:

The boundaries of the intervals in the solution set of the polynomial inequality

are the zeroes of the equation

By the Zero Product Property, either

,

in which case

,

or

,

in which case

and 5 are the boundaries of the intervals, which are , and . To determine which intervals belong to the solution set, choose one value from each, substitute for  in the original inequality, and determine whether it is true.

: Choose

False - do not include

: Choose

False - do not include

: Choose

True - include

Also, the boundary values themselves should be included, because the inequality symbol  allows for equality.

The solution set is therefore .

### Example Question #5 : Polynomial Inequalities

Give the solution set of the inequality

Explanation:

To solve the polynomial inequality

it is necessary to get it into standard form by writing all terms at the same side. Subtract  from both sides:

The values of  that serve as boundaries of the solutions set are the zeroes of the polynomial, so solve for  in the equation:

Factor the polynomial by grouping and taking out the GCFs:

The polynomial cannot be factored further, as , as the sum of squares, is prime. By the Zero Factor Property, one of the binomials is equal to 0, so

in which case

or

,

which has no real solution.

Therefore, the only boundary of the intervals is 7, which divides the real numbers into two intervals,  and . To determine which intervals belong to the solution set, choose one value from each, substitute for  in the original inequality, and determine whether it is true.

: Choose

False - do not include

: Choose

True - include

Also, since the inequality symbol  does not allow for equality, 7 is not included in the solution. The solution set is .

### Example Question #6 : Polynomial Inequalities

Give the solution set of the inequality

Explanation:

To solve the polynomial inequality

it is necessary to get it into standard form by writing all terms at the same side. Subtract  from both sides:

The values of  that serve as boundaries of the solutions set are the zeroes of the polynomial, so solve for  in the equation:

Factor the polynomial by grouping and taking out the GCFs:

is the difference of squares and can be factored according to a pattern:

By the Zero Factor Property, one of the binomials is equal to 0, so

in which case

,

in which case

,

or

in which case

.

, and 2 are the boundaries of the intervals, which are , and . To determine which intervals belong to the solution set, choose one value from each, substitute for  in the original inequality, and determine whether it is true.

: Choose

True - include

: Choose

False - do not include

: Choose

True - include

: Choose

False - do not include

Also, the boundary values themselves should not be included, because the inequality symbol  does not allow for equality.

The solution set is

### Example Question #7 : Polynomial Inequalities

Which is correct graph of the following function?

None of the other graphs.