# Precalculus : Polynomial Functions

## Example Questions

### Example Question #8 : Rational Equations And Partial Fractions

Solve the inequality:

1 < z < 31

z < 1

No solution

z < 1 or z > 31

z < 1 or z > 31

Explanation:

First, find the LCD, which is 6(z-1). Then, multiply the entire equation by this:

This simplifies to:

Now, make a number line with points at 1 and 31, and vertical lines in between zones:

Let's test three points, one in each zone:

f(0):

f(10):

f(35):

Therefore, the inequality is true for values of z such that z < 1 or z > 31.

### Example Question #9 : Rational Equations And Partial Fractions

Decompose  into partial fractions.

Explanation:

Decomposing a fraction into partial fractions shows what fractions were added or subtracted to result in the expression we see. First, factor the denominator.

Next, we'll express the factored form as the sum of two fractions. We don't know these fractions' numerators, so we'll call them X and Y.

Next, eliminate all of the denominators by multiplying both sides of the fraction by the LCD.

Looking back to our original fraction, note that the values p = 1 and b = -1 cannot be plugged in, as they will make the function undefined. However, if we take each of these values and plug them in, we can solve for X and Y.

First, let p=1.

Second, let p=-1

Now, substitute these values in for X and Y.

Therefore, .

### Example Question #10 : Rational Equations And Partial Fractions

Decompose  into partial fractions.

Explanation:

Decomposing a fraction into partial fractions shows what fractions were added or subtracted to result in the expression we see. First, factor the denominator.

Next, we'll express the factored form as the sum of two fractions. We don't know these fractions' numerators, so we'll call them X and Y.

Next, eliminate all of the denominators by multiplying both sides of the fraction by the LCD.

Looking back to our original fraction, note that the values b = 0 and b = -1 cannot be plugged in, as they will make the function undefined. However, if we take each of these values and plug them in, we can solve for X and Y.

First, let b=0.

Second, let b=-1

Now, substitute these values in for X and Y.

Therefore, .

### Example Question #1 : Radical Functions

Solve for  and use the solution to show where the radical functions intersect:

Explanation:

To solve, first square both sides of the equation to reverse the square-rooting of the binomials, then simplify:

Now solve for :

The x-coordinate for the intersection point is .

Choose one of the two radical functions that compose the equation, and set the function equal to y.  The more simple a function is, the easier it is to use:

Now substitute  into the function.

The y-coordinate of the intersection point is .

The intersection point of the two radical functions is .

Now graph the two radical functions:

,

### Example Question #2 : Radical Functions

Solve for

Explanation:

In order to solve this equation, we need to isolate the radical. In order to do so, we subtract 3 from both sides which leaves us with:

To get rid of the radical, we square both sides:

the radical is then canceled out leaving us with

We solve for  by dividing by 4:

### Example Question #3 : Radical Functions

Solve for

Explanation:

In order to get rid of the radical, we square both sides:

Since the radical cancels out, we're left with

Subtracting both sides by 1 gives us

We then divide both sides by 6 to get

### Example Question #4 : Radical Functions

Which of the following is a solution to the following equation?

Explanation:

We begin by sqaring both sides of the equation.  On the left side, the square root simply disappears, while on the right side we square the term.

We then set the left side equal to 0 by subtracting everything on that side.

We then factor

Therefore,

With problems of this type, it is always wise to double check for any extraneous roots (answers that don't actually work for some reason).  However, in this case both answers work.  Since is the only option among our choices, we should go with it.

### Example Question #5 : Radical Functions

Explanation:

When dealing with a radical equation, do the inverse operation to isolate the variable. In this case, the inverse operation of a square root is to square the expression. Thus we square both sides to continue. This yields the following.

### Example Question #6 : Radical Functions

Solve:

Explanation:

1) To remove the radicals, raise both sides of the equation to the second power:

2) To remove the radical, raise both side of the equation to the second power:

3) Now simplify, write as a quadratic equation, and solve:

4) Checking for extraneous solutions.

Plugging in

Plugging in

Since the square root of negative 5.4 gives us an imaginary solution we conclude that the only real solution is x=3.

### Example Question #7 : Radical Functions

Solve the rational equation: