Our variable x represents the number of hours to move.  Marvin's rate can be expressed as 30x+100.  Elliot's rate can be expressed as 55x.  We must set the expression for Marvin's rate less than Elliot's rate: .  Upon solving for x, we find that x is greater than 4, so Marvin's is cheaper for jobs greater than 4 hours.

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 .

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 

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 .

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 .

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 

First graph 

Since the inequality is , this means that the line is included in the solution, as indicated by a solid line.

Shade the area outside of the parabola since it is a negative parabola, but greater than or equal to.

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.

In order to solve this, we need to figure out what values of x makes the numerator and denominator equal zero. 

These values are , and .

Then test a few numbers greater or lesser than the above values to figure out what range of values make the inequality true.

The boundary points of a rational inequality are the zeroes of the numerator and the denominator. 

First, set the numerator equal to 0 and solve for :

Now set the denominator equal to 0 and solve for :

Factor the trinomial using the reverse-FOIL method. Look for two integers whose sum is  and whose product is 7; these are , and , so the equation can be rewritten as

Set each factor to zero and solve for :

Therefore, the boundary points are , which divide the real numbers into four intervals. Choose any value from each interval as a test point, setting  to that value and determining whether the inequality is true.

The four intervals are listed below, along with their arbitrary test points.

: Set 

True; include .

: Set 

False; exclude .

: Set :

True; include 

: Set :

False; exclude .

Since the inequality symbol is the "is greater than or equal to"  symbol, include the zero, 0, of the numerator - but not the other two boundaries, the zeroes of the denominator. This makes the solution set

.