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# Dividing Polynomials

Module by: First Last. E-mail the author

Summary: In this section, you will:

• Use long division to divide polynomials.
• Use synthetic division to divide polynomials.

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The exterior of the Lincoln Memorial in Washington, D.C., is a large rectangular solid with length 61.5 meters (m), width 40 m, and height 30 m.1 We can easily find the volume using elementary geometry.

V = lwh = 61.54030 = 73,800 V = lwh = 61.54030 = 73,800

So the volume is 73,800 cubic meters ( m³ ). ( m³ ). Suppose we knew the volume, length, and width. We could divide to find the height.

h = Vlw = 73,80061.540 = 30 h = Vlw = 73,80061.540 = 30

As we can confirm from the dimensions above, the height is 30 m. We can use similar methods to find any of the missing dimensions. We can also use the same method if any, or all, of the measurements contain variable expressions. For example, suppose the volume of a rectangular solid is given by the polynomial 3 x 4 3 x 3 33 x 2 +54x. 3 x 4 3 x 3 33 x 2 +54x. The length of the solid is given by 3x; 3x; the width is given by x2. x2. To find the height of the solid, we can use polynomial division, which is the focus of this section.

## Using Long Division to Divide Polynomials

We are familiar with the long division algorithm for ordinary arithmetic. We begin by dividing into the digits of the dividend that have the greatest place value. We divide, multiply, subtract, include the digit in the next place value position, and repeat. For example, let’s divide 178 by 3 using long division.

Another way to look at the solution is as a sum of parts. This should look familiar, since it is the same method used to check division in elementary arithmetic.

dividend = (divisorquotient) + remainder 178 = (359)+1 = 177+1 = 178 dividend = (divisorquotient) + remainder 178 = (359)+1 = 177+1 = 178

We call this the Division Algorithm and will discuss it more formally after looking at an example.

Division of polynomials that contain more than one term has similarities to long division of whole numbers. We can write a polynomial dividend as the product of the divisor and the quotient added to the remainder. The terms of the polynomial division correspond to the digits (and place values) of the whole number division. This method allows us to divide two polynomials. For example, if we were to divide 2 x 3 3 x 2 +4x+5 2 x 3 3 x 2 +4x+5 by x+2 x+2 using the long division algorithm, it would look like this:

We have found

2 x 3 3 x 2 +4x+5 x+2 =2 x 2 7x+18 31 x+2 2 x 3 3 x 2 +4x+5 x+2 =2 x 2 7x+18 31 x+2

or

2 x 3 3 x 2 +4x+5 x+2 =(x+2)(2 x 2 7x+18)31 2 x 3 3 x 2 +4x+5 x+2 =(x+2)(2 x 2 7x+18)31
(5)

We can identify the dividend, the divisor, the quotient, and the remainder.

Writing the result in this manner illustrates the Division Algorithm.

### A General Note: The Division Algorithm:

The Division Algorithm states that, given a polynomial dividend f(x) f(x) and a non-zero polynomial divisor d(x) d(x) where the degree of d(x) d(x) is less than or equal to the degree of f(x) f(x), there exist unique polynomials q(x) q(x) and r(x) r(x) such that

f(x)=d(x)q(x)+r(x) f(x)=d(x)q(x)+r(x)
(6)

q(x) q(x) is the quotient and r(x) r(x) is the remainder. The remainder is either equal to zero or has degree strictly less than d(x). d(x).

If r(x)=0, r(x)=0, then d(x) d(x) divides evenly into f(x). f(x). This means that, in this case, both d(x) d(x) and q(x) q(x) are factors of f(x). f(x).

### How To:

Given a polynomial and a binomial, use long division to divide the polynomial by the binomial.

1. Set up the division problem.
2. Determine the first term of the quotient by dividing the leading term of the dividend by the leading term of the divisor.
3. Multiply the answer by the divisor and write it below the like terms of the dividend.
4. Subtract the bottom binomial from the top binomial.
5. Bring down the next term of the dividend.
6. Repeat steps 2–5 until reaching the last term of the dividend.
7. If the remainder is non-zero, express as a fraction using the divisor as the denominator.

### Example 1

#### Problem 1

##### Using Long Division to Divide a Second-Degree Polynomial

Divide 5 x 2 +3x2 5 x 2 +3x2 by x+1. x+1.

##### Solution

The quotient is 5x2. 5x2. The remainder is 0. We write the result as

5 x 2 +3x2 x+1 =5x2 5 x 2 +3x2 x+1 =5x2

or

5 x 2 +3x2=( x+1 )( 5x2 ) 5 x 2 +3x2=( x+1 )( 5x2 )
##### Analysis

This division problem had a remainder of 0. This tells us that the dividend is divided evenly by the divisor, and that the divisor is a factor of the dividend.

### Example 2

#### Problem 1

##### Using Long Division to Divide a Third-Degree Polynomial

Divide 6 x 3 +11 x 2 31x+15 6 x 3 +11 x 2 31x+15 by 3x2. 3x2.

##### Solution

There is a remainder of 1. We can express the result as:

6 x 3 +11 x 2 31x+15 3x2 =2 x 2 +5x7+ 1 3x2 6 x 3 +11 x 2 31x+15 3x2 =2 x 2 +5x7+ 1 3x2
##### Analysis

We can check our work by using the Division Algorithm to rewrite the solution. Then multiply.

(3x2)(2 x 2 +5x7)+1=6 x 3 +11 x 2 31x+15 (3x2)(2 x 2 +5x7)+1=6 x 3 +11 x 2 31x+15

Notice, as we write our result,

• the dividend is 6 x 3 +11 x 2 31x+15 6 x 3 +11 x 2 31x+15
• the divisor is 3x2 3x2
• the quotient is 2 x 2 +5x7 2 x 2 +5x7
• the remainder is 1 1

### Try It:

#### Exercise 1

Divide 16 x 3 12 x 2 +20x3 16 x 3 12 x 2 +20x3 by 4x+5. 4x+5.

##### Solution

4 x 2 8x+15 78 4x+5 4 x 2 8x+15 78 4x+5

## Using Synthetic Division to Divide Polynomials

As we’ve seen, long division of polynomials can involve many steps and be quite cumbersome. Synthetic division is a shorthand method of dividing polynomials for the special case of dividing by a linear factor whose leading coefficient is 1.

To illustrate the process, recall the example at the beginning of the section.

Divide 2 x 3 3 x 2 +4x+5 2 x 3 3 x 2 +4x+5 by x+2 x+2 using the long division algorithm.

The final form of the process looked like this:

There is a lot of repetition in the table. If we don’t write the variables but, instead, line up their coefficients in columns under the division sign and also eliminate the partial products, we already have a simpler version of the entire problem.

Synthetic division carries this simplification even a few more steps. Collapse the table by moving each of the rows up to fill any vacant spots. Also, instead of dividing by 2, as we would in division of whole numbers, then multiplying and subtracting the middle product, we change the sign of the “divisor” to –2, multiply and add. The process starts by bringing down the leading coefficient.

We then multiply it by the “divisor” and add, repeating this process column by column, until there are no entries left. The bottom row represents the coefficients of the quotient; the last entry of the bottom row is the remainder. In this case, the quotient is 2 x 2 7x+18 2 x 2 7x+18and the remainder is –31. –31. The process will be made more clear in Example 3.

### A General Note: Synthetic Division:

Synthetic division is a shortcut that can be used when the divisor is a binomial in the form xk xk where k k is a real number. In synthetic division, only the coefficients are used in the division process.

### How To:

Given two polynomials, use synthetic division to divide.

1. Write k k for the divisor.
2. Write the coefficients of the dividend.
3. Bring the lead coefficient down.
4. Multiply the lead coefficient by k. k. Write the product in the next column.
5. Add the terms of the second column.
6. Multiply the result by k. k. Write the product in the next column.
7. Repeat steps 5 and 6 for the remaining columns.
8. Use the bottom numbers to write the quotient. The number in the last column is the remainder and has degree 0, the next number from the right has degree 1, the next number from the right has degree 2, and so on.

### Example 3

#### Problem 1

##### Using Synthetic Division to Divide a Second-Degree Polynomial

Use synthetic division to divide 5 x 2 3x36 5 x 2 3x36 by x3. x3.

##### Solution

Begin by setting up the synthetic division. Write k k and the coefficients.

Bring down the lead coefficient. Multiply the lead coefficient by k. k.

Continue by adding the numbers in the second column. Multiply the resulting number by k. k. Write the result in the next column. Then add the numbers in the third column.

The result is 5x+12. 5x+12. The remainder is 0. So x3 x3 is a factor of the original polynomial.

##### Analysis

Just as with long division, we can check our work by multiplying the quotient by the divisor and adding the remainder.

(x3)(5x+12)+0=5 x 2 3x36 (x3)(5x+12)+0=5 x 2 3x36

### Example 4

#### Problem 1

##### Using Synthetic Division to Divide a Third-Degree Polynomial

Use synthetic division to divide 4 x 3 +10 x 2 6x20 4 x 3 +10 x 2 6x20 by x+2. x+2.

##### Solution

The binomial divisor is x+2 x+2 so k=−2. k=−2. Add each column, multiply the result by –2, and repeat until the last column is reached.

The result is 4 x 2 +2x10. 4 x 2 +2x10. The remainder is 0. Thus, x+2 x+2 is a factor of 4 x 3 +10 x 2 6x20. 4 x 3 +10 x 2 6x20.

##### Analysis

The graph of the polynomial function f(x)=4 x 3 +10 x 2 6x20 f(x)=4 x 3 +10 x 2 6x20 in Figure 2 shows a zero at x=k=−2. x=k=−2. This confirms that x+2 x+2 is a factor of 4 x 3 +10 x 2 6x20. 4 x 3 +10 x 2 6x20.

### Example 5

#### Problem 1

##### Using Synthetic Division to Divide a Fourth-Degree Polynomial

Use synthetic division to divide 9 x 4 +10 x 3 +7 x 2 6 9 x 4 +10 x 3 +7 x 2 6 by x1. x1.

##### Solution

Notice there is no x-term. We will use a zero as the coefficient for that term.

The result is 9 x 3 + x 2 +8x+8+ 2 x1 . 9 x 3 + x 2 +8x+8+ 2 x1 .

### Try It:

#### Exercise 2

Use synthetic division to divide 3 x 4 +18 x 3 3x+40 3 x 4 +18 x 3 3x+40 by x+7. x+7.

##### Solution

3 x 3 3 x 2 +21x150+ 1,090 x+7 3 x 3 3 x 2 +21x150+ 1,090 x+7

## Using Polynomial Division to Solve Application Problems

Polynomial division can be used to solve a variety of application problems involving expressions for area and volume. We looked at an application at the beginning of this section. Now we will solve that problem in the following example.

### Example 6

#### Problem 1

##### Using Polynomial Division in an Application Problem

The volume of a rectangular solid is given by the polynomial 3 x 4 3 x 3 33 x 2 +54x. 3 x 4 3 x 3 33 x 2 +54x. The length of the solid is given by 3x 3x and the width is given by x2. x2. Find the height, t, t, of the solid.

##### Solution

There are a few ways to approach this problem. We need to divide the expression for the volume of the solid by the expressions for the length and width. Let us create a sketch as in Figure 3.

We can now write an equation by substituting the known values into the formula for the volume of a rectangular solid.

V = lwh 3 x 4 3 x 3 33 x 2 +54x = 3x(x2)h V = lwh 3 x 4 3 x 3 33 x 2 +54x = 3x(x2)h

To solve for h, h, first divide both sides by 3x. 3x.

3x(x2)h 3x = 3 x 4 3 x 3 33 x 2 +54x 3x (x2)h = x 3 x 2 11x+18 3x(x2)h 3x = 3 x 4 3 x 3 33 x 2 +54x 3x (x2)h = x 3 x 2 11x+18

Now solve for h h using synthetic division.

h= x 3 x 2 11x+18 x2 h= x 3 x 2 11x+18 x2

The quotient is x 2 +x9 x 2 +x9 and the remainder is 0. The height of the solid is x 2 +x9. x 2 +x9.

### Try It:

#### Exercise 3

The area of a rectangle is given by 3 x 3 +14 x 2 23x+6. 3 x 3 +14 x 2 23x+6. The width of the rectangle is given by x+6. x+6. Find an expression for the length of the rectangle.

##### Solution

3 x 2 4x+1 3 x 2 4x+1

### Media:

Access these online resources for additional instruction and practice with polynomial division.

## Key Equations

 Division Algorithm f(x)=d(x)q(x)+r(x) where q(x)≠0 f(x)=d(x)q(x)+r(x) where q(x)≠0

## Key Concepts

• Polynomial long division can be used to divide a polynomial by any polynomial with equal or lower degree. See Example 1 and Example 2.
• The Division Algorithm tells us that a polynomial dividend can be written as the product of the divisor and the quotient added to the remainder.
• Synthetic division is a shortcut that can be used to divide a polynomial by a binomial in the form xk. xk. See Example 3, Example 4, and Example 5.
• Polynomial division can be used to solve application problems, including area and volume. See Example 6.

## Section Exercises

### Verbal

#### Exercise 4

If division of a polynomial by a binomial results in a remainder of zero, what can be conclude?

##### Solution

The binomial is a factor of the polynomial.

#### Exercise 5

If a polynomial of degree n n is divided by a binomial of degree 1, what is the degree of the quotient?

### Algebraic

For the following exercises, use long division to divide. Specify the quotient and the remainder.

#### Exercise 6

( x 2 +5x1 )÷( x1 ) ( x 2 +5x1 )÷( x1 )

##### Solution

x+6+ 5 x-1 ,quotient:x+6,remainder:5 x+6+ 5 x-1 ,quotient:x+6,remainder:5

#### Exercise 7

( 2 x 2 9x5 )÷( x5 ) ( 2 x 2 9x5 )÷( x5 )

#### Exercise 8

( 3 x 2 +23x+14 )÷( x+7 ) ( 3 x 2 +23x+14 )÷( x+7 )

##### Solution

3x+2,quotient: 3x+2,remainder: 0 3x+2,quotient: 3x+2,remainder: 0

#### Exercise 9

( 4 x 2 10x+6 )÷( 4x+2 ) ( 4 x 2 10x+6 )÷( 4x+2 )

#### Exercise 10

( 6 x 2 25x25 )÷( 6x+5 ) ( 6 x 2 25x25 )÷( 6x+5 )

##### Solution

x5,quotient:x5,remainder:0 x5,quotient:x5,remainder:0

#### Exercise 11

( x 2 1 )÷( x+1 ) ( x 2 1 )÷( x+1 )

#### Exercise 12

( 2 x 2 3x+2 )÷( x+2 ) ( 2 x 2 3x+2 )÷( x+2 )

##### Solution

2x7+ 16 x+2 ,quotient:2x7,remainder:16 2x7+ 16 x+2 ,quotient:2x7,remainder:16

#### Exercise 13

( x 3 126 )÷( x5 ) ( x 3 126 )÷( x5 )

#### Exercise 14

( 3 x 2 5x+4 )÷( 3x+1 ) ( 3 x 2 5x+4 )÷( 3x+1 )

##### Solution

x2+ 6 3x+1 ,quotient:x2,remainder:6 x2+ 6 3x+1 ,quotient:x2,remainder:6

#### Exercise 15

( x 3 3 x 2 +5x6 )÷( x2 ) ( x 3 3 x 2 +5x6 )÷( x2 )

#### Exercise 16

( 2 x 3 +3 x 2 4x+15 )÷( x+3 ) ( 2 x 3 +3 x 2 4x+15 )÷( x+3 )

##### Solution

2 x 2 3x+5,quotient:2 x 2 3x+5,remainder:0 2 x 2 3x+5,quotient:2 x 2 3x+5,remainder:0

For the following exercises, use synthetic division to find the quotient.

#### Exercise 17

( 3 x 3 2 x 2 +x4 )÷( x+3 ) ( 3 x 3 2 x 2 +x4 )÷( x+3 )

#### Exercise 18

( 2 x 3 6 x 2 7x+6 )÷(x4) ( 2 x 3 6 x 2 7x+6 )÷(x4)

##### Solution

2 x 2 +2x+1+ 10 x4 2 x 2 +2x+1+ 10 x4

#### Exercise 19

( 6 x 3 10 x 2 7x15 )÷(x+1) ( 6 x 3 10 x 2 7x15 )÷(x+1)

#### Exercise 20

( 4 x 3 12 x 2 5x1 )÷(2x+1) ( 4 x 3 12 x 2 5x1 )÷(2x+1)

##### Solution

2 x 2 7x+1 2 2x+1 2 x 2 7x+1 2 2x+1

#### Exercise 21

( 9 x 3 9 x 2 +18x+5 )÷(3x1) ( 9 x 3 9 x 2 +18x+5 )÷(3x1)

#### Exercise 22

( 3 x 3 2 x 2 +x4 )÷( x+3 ) ( 3 x 3 2 x 2 +x4 )÷( x+3 )

##### Solution

3 x 2 11x+34 106 x+3 3 x 2 11x+34 106 x+3

#### Exercise 23

( 6 x 3 + x 2 4 )÷( 2x3 ) ( 6 x 3 + x 2 4 )÷( 2x3 )

#### Exercise 24

( 2 x 3 +7 x 2 13x3 )÷( 2x3 ) ( 2 x 3 +7 x 2 13x3 )÷( 2x3 )

##### Solution

x 2 +5x+1 x 2 +5x+1

#### Exercise 25

( 3 x 3 5 x 2 +2x+3 )÷(x+2) ( 3 x 3 5 x 2 +2x+3 )÷(x+2)

#### Exercise 26

( 4 x 3 5 x 2 +13 )÷(x+4) ( 4 x 3 5 x 2 +13 )÷(x+4)

##### Solution

4 x 2 21x+84 323 x+4 4 x 2 21x+84 323 x+4

#### Exercise 27

( x 3 3x+2 )÷( x+2 ) ( x 3 3x+2 )÷( x+2 )

#### Exercise 28

( x 3 21 x 2 +147x343 )÷( x7 ) ( x 3 21 x 2 +147x343 )÷( x7 )

##### Solution

x 2 14x+49 x 2 14x+49

#### Exercise 29

( x 3 15 x 2 +75x125 )÷( x5 ) ( x 3 15 x 2 +75x125 )÷( x5 )

#### Exercise 30

( 9 x 3 x+2 )÷( 3x1 ) ( 9 x 3 x+2 )÷( 3x1 )

##### Solution

3 x 2 +x+ 2 3x1 3 x 2 +x+ 2 3x1

#### Exercise 31

( 6 x 3 x 2 +5x+2 )÷( 3x+1 ) ( 6 x 3 x 2 +5x+2 )÷( 3x+1 )

#### Exercise 32

( x 4 + x 3 3 x 2 2x+1 )÷( x+1 ) ( x 4 + x 3 3 x 2 2x+1 )÷( x+1 )

##### Solution

x 3 3x+1 x 3 3x+1

#### Exercise 33

( x 4 3 x 2 +1 )÷( x1 ) ( x 4 3 x 2 +1 )÷( x1 )

#### Exercise 34

( x 4 +2 x 3 3 x 2 +2x+6 )÷( x+3 ) ( x 4 +2 x 3 3 x 2 +2x+6 )÷( x+3 )

##### Solution

x 3 x 2 +2 x 3 x 2 +2

#### Exercise 35

( x 4 10 x 3 +37 x 2 60x+36 )÷( x2 ) ( x 4 10 x 3 +37 x 2 60x+36 )÷( x2 )

#### Exercise 36

( x 4 8 x 3 +24 x 2 32x+16 )÷( x2 ) ( x 4 8 x 3 +24 x 2 32x+16 )÷( x2 )

##### Solution

x 3 6 x 2 +12x8 x 3 6 x 2 +12x8

#### Exercise 37

( x 4 +5 x 3 3 x 2 13x+10 )÷( x+5 ) ( x 4 +5 x 3 3 x 2 13x+10 )÷( x+5 )

#### Exercise 38

( x 4 12 x 3 +54 x 2 108x+81 )÷( x3 ) ( x 4 12 x 3 +54 x 2 108x+81 )÷( x3 )

##### Solution

x 3 9 x 2 +27x27 x 3 9 x 2 +27x27

#### Exercise 39

( 4 x 4 2 x 3 4x+2 )÷( 2x1 ) ( 4 x 4 2 x 3 4x+2 )÷( 2x1 )

#### Exercise 40

( 4 x 4 +2 x 3 4 x 2 +2x+2 )÷( 2x+1 ) ( 4 x 4 +2 x 3 4 x 2 +2x+2 )÷( 2x+1 )

##### Solution

2 x 3 2x+2 2 x 3 2x+2

For the following exercises, use synthetic division to determine whether the first expression is a factor of the second. If it is, indicate the factorization.

#### Exercise 41

x2,4 x 3 3 x 2 8x+4 x2,4 x 3 3 x 2 8x+4

#### Exercise 42

x2,3 x 4 6 x 3 5x+10 x2,3 x 4 6 x 3 5x+10

##### Solution

Yes ( x2 )(3 x 3 5) ( x2 )(3 x 3 5)

#### Exercise 43

x+3,4 x 3 +5 x 2 +8 x+3,4 x 3 +5 x 2 +8

#### Exercise 44

x2,4 x 4 15 x 2 4 x2,4 x 4 15 x 2 4

##### Solution

Yes ( x2 )(4 x 3 +8 x 2 +x+2) ( x2 )(4 x 3 +8 x 2 +x+2)

#### Exercise 45

x 1 2 ,2 x 4 x 3 +2x1 x 1 2 ,2 x 4 x 3 +2x1

#### Exercise 46

x+ 1 3 ,3 x 4 + x 3 3x+1 x+ 1 3 ,3 x 4 + x 3 3x+1

No

### Graphical

For the following exercises, use the graph of the third-degree polynomial and one factor to write the factored form of the polynomial suggested by the graph. The leading coefficient is one.

#### Exercise 47

Factor is x 2 x+3 x 2 x+3

#### Exercise 48

Factor is ( x 2 +2x+4) ( x 2 +2x+4)

##### Solution

(x1)( x 2 +2x+4) (x1)( x 2 +2x+4)

#### Exercise 49

Factor is x 2 +2x+5 x 2 +2x+5

#### Exercise 50

Factor is x 2 +x+1 x 2 +x+1

##### Solution

(x5)( x 2 +x+1) (x5)( x 2 +x+1)

#### Exercise 51

Factor is x 2 +2x+2 x 2 +2x+2

For the following exercises, use synthetic division to find the quotient and remainder.

#### Exercise 52

4 x 3 33 x2 4 x 3 33 x2

##### Solution

Quotient:4 x 2 +8x+16,remainder:1 Quotient:4 x 2 +8x+16,remainder:1

#### Exercise 53

2 x 3 +25 x+3 2 x 3 +25 x+3

#### Exercise 54

3 x 3 +2x5 x1 3 x 3 +2x5 x1

##### Solution

Quotient:3 x 2 +3x+5,remainder:0 Quotient:3 x 2 +3x+5,remainder:0

#### Exercise 55

4 x 3 x 2 12 x+4 4 x 3 x 2 12 x+4

#### Exercise 56

x 4 22 x+2 x 4 22 x+2

##### Solution

Quotient: x 3 2 x 2 +4x8,remainder:6 Quotient: x 3 2 x 2 +4x8,remainder:6

### Technology

For the following exercises, use a calculator with CAS to answer the questions.

#### Exercise 57

Consider x k 1 x1 x k 1 x1 with k=1, 2, 3. k=1, 2, 3. What do you expect the result to be if k=4? k=4?

#### Exercise 58

Consider x k +1 x+1 x k +1 x+1 for k=1, 3, 5. k=1, 3, 5. What do you expect the result to be if k=7? k=7?

##### Solution

x 6 x 5 + x 4 x 3 + x 2 x+1 x 6 x 5 + x 4 x 3 + x 2 x+1

#### Exercise 59

Consider x 4 k 4 xk x 4 k 4 xk for k=1, 2, 3. k=1, 2, 3. What do you expect the result to be if k=4? k=4?

#### Exercise 60

Consider x k x+1 x k x+1 with k=1, 2, 3. k=1, 2, 3. What do you expect the result to be if k=4? k=4?

##### Solution

x 3 x 2 +x1+ 1 x+1 x 3 x 2 +x1+ 1 x+1

#### Exercise 61

Consider x k x1 x k x1 with k=1, 2, 3. k=1, 2, 3. What do you expect the result to be if k=4? k=4?

### Extensions

For the following exercises, use synthetic division to determine the quotient involving a complex number.

#### Exercise 62

x+1 xi x+1 xi

##### Solution

1+ 1+i xi 1+ 1+i xi

#### Exercise 63

x 2 +1 xi x 2 +1 xi

#### Exercise 64

x+1 x+i x+1 x+i

##### Solution

1+ 1i x+i 1+ 1i x+i

#### Exercise 65

x 2 +1 x+i x 2 +1 x+i

#### Exercise 66

x 3 +1 xi x 3 +1 xi

##### Solution

x 2 ix1+ 1i xi x 2 ix1+ 1i xi

### Real-World Applications

For the following exercises, use the given length and area of a rectangle to express the width algebraically.

#### Exercise 67

Length is x+5, x+5, area is 2 x 2 +9x5. 2 x 2 +9x5.

#### Exercise 68

Length is 2x + 5, 2x + 5, area is 4 x 3 +10 x 2 +6x+15 4 x 3 +10 x 2 +6x+15

##### Solution

2 x 2 +3 2 x 2 +3

#### Exercise 69

Length is 3x4, 3x4, area is 6 x 4 8 x 3 +9 x 2 9x4 6 x 4 8 x 3 +9 x 2 9x4

For the following exercises, use the given volume of a box and its length and width to express the height of the box algebraically.

#### Exercise 70

Volume is 12 x 3 +20 x 2 21x36, 12 x 3 +20 x 2 21x36, length is 2x+3, 2x+3, width is 3x4. 3x4.

2x+3 2x+3

#### Exercise 71

Volume is 18 x 3 21 x 2 40x+48, 18 x 3 21 x 2 40x+48, length is 3x4, 3x4, width is 3x4. 3x4.

#### Exercise 72

Volume is 10 x 3 +27 x 2 +2x24, 10 x 3 +27 x 2 +2x24, length is 5x4, 5x4, width is 2x+3. 2x+3.

x+2 x+2

#### Exercise 73

Volume is 10 x 3 +30 x 2 8x24, 10 x 3 +30 x 2 8x24, length is 2, 2, width is x+3. x+3.

For the following exercises, use the given volume and radius of a cylinder to express the height of the cylinder algebraically.

#### Exercise 74

Volume is π(25 x 3 65 x 2 29x3), π(25 x 3 65 x 2 29x3), radius is 5x+1. 5x+1.

x3 x3

#### Exercise 75

Volume is π(4 x 3 +12 x 2 15x50), π(4 x 3 +12 x 2 15x50), radius is 2x+5. 2x+5.

#### Exercise 76

Volume is π(3 x 4 +24 x 3 +46 x 2 16x32), π(3 x 4 +24 x 3 +46 x 2 16x32), radius is x+4. x+4.

3 x 2 2 3 x 2 2

## Footnotes

1. National Park Service. "Lincoln Memorial Building Statistics." http://www.nps.gov/linc/historyculture/lincoln-memorial-building-statistics.htm. Accessed 4/3/2014

## Glossary

Division Algorithm:
given a polynomial dividend f(x) f(x) and a non-zero polynomial divisor d(x) d(x) where the degree of d(x) d(x) is less than or equal to the degree of f(x) f(x), there exist unique polynomials q(x) q(x) and r(x) r(x) such that f(x)=d(x)q(x)+r(x) f(x)=d(x)q(x)+r(x) where q(x) q(x) is the quotient and r(x) r(x) is the remainder. The remainder is either equal to zero or has degree strictly less than d(x). d(x).
synthetic division:
a shortcut method that can be used to divide a polynomial by a binomial of the form xk xk

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