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A. Long Division and Synthetic Division

# A. Long Division and Synthetic Division

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Section 4.1 Synthetic Division; the Remainder and Factor Theorems

The process illustrated is called the division algorithm, and like the division of

whole numbers, the final result can be checked by multiplication.

dividend

divisor

quotient

remainder

check: x3 Ϫ 4x2 ϩ x ϩ 6 ϭ 1x Ϫ 12 1x2 Ϫ 3x Ϫ 22 ϩ 4

ϭ 1x3 Ϫ 3x2 Ϫ 2x Ϫ x2 ϩ 3x ϩ 22 ϩ 4

#

divisor quotient

ϭ 1x3 Ϫ 4x2 ϩ x ϩ 22 ϩ 4

combine like terms

ϭ x Ϫ 4x ϩ x ϩ 6 ✓

3

2

In general, the division algorithm for polynomials says

Division of Polynomials

Given polynomials p1x2 and d1x2

such that

0, there exist unique polynomials q(x) and r(x)

p1x2 ϭ d1x2q1x2 ϩ r1x2,

where r1x2 ϭ 0 or the degree of r1x2 is less than the degree of d1x2 .

Here, d1x2 is called the divisor, q1x2 is the quotient, and r1x2 is the remainder.

In other words, “a polynomial of greater degree can be divided by a polynomial of

equal or lesser degree to obtain a quotient and a remainder.” As with whole numbers,

if the remainder is zero, the divisor is a factor of the dividend.

Synthetic Division

As the word “synthetic” implies, synthetic division not only simulates the long division process, but also condenses it and makes it more efficient when the divisor is

linear. The process works by capitalizing on the repetition found in the division algorithm. First, the polynomials involved are written in decreasing order of degree, so the

variable part of each term is unnecessary as we can let the position of each coefficient

indicate the degree of the term. For the dividend from Example 1, 1 Ϫ4 1 6 would

represent the polynomial 1x3 Ϫ 4x2 ϩ 1x ϩ 6. Also, each stage of the algorithm involves a product of the divisor with the next multiplier, followed by a subtraction.

These can likewise be computed using the coefficients only, as the degree of each term

is still determined by its position. Here is the division from Example 1 in the synthetic

division format. Note that we must use the zero of the divisor (as in x ϭ 32 for a divisor

of 2x Ϫ 3, or in this case, “1” from x Ϫ 1 ϭ 02 and the coefficients of the dividend in

the following format:

zero of the divisor

1

coefficients of the dividend

Ϫ4

1

1

6

partial products

The process of synthetic

division is only summarized

here. For a complete discussion,

see Appendix III.

1

coefficients of the quotient

remainder

As this template indicates, the quotient and remainder will be read from the last row.

The arrow indicates we begin by “dropping the leading coefficient into place.” We

then multiply this coefficient by the “divisor,” then place the result in the next column

and add. Note that using the zero of the divisor enables us to add in each column

directly, rather than subtracting then changing to algebraic addition as before.

1

1

Ϫ4

multiply 1 # 1

WORTHY OF NOTE

1

1

Ϫ3 ↓

1

6

#

multiply divisor coefficient,

place result in next column and add

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In a sense, we “multiply in the diagonal direction,” and “add in the vertical direction.” Repeat the process until the division is complete.

1

1

Ϫ4

1

1

Ϫ3

6

multiply 1 # 1Ϫ32 ϭ Ϫ3,

1

Ϫ3

Ϫ2↓

Ϫ4

1

6

1

Ϫ3

Ϫ2

Ϫ3

Ϫ2

place result in next column and add

1

1

1

384

4↓

multiply 1 # 1Ϫ22 ϭ Ϫ2,

place result in next column and add

The quotient is read from the last row by noting the remainder is 4, leaving the

coefficients 1 Ϫ3 Ϫ2, which translate back into the polynomial x2 Ϫ 3x Ϫ 2. The

final result is identical to that in Example 1, but the new process is more efficient, since

all stages are actually computed on a single template as shown here:

zero of the divisor

coefficients of the dividend

1

1

Ϫ4

1

6

1

Ϫ3

Ϫ2

1

Ϫ3

Ϫ2

coefficients of the quotient

4

remainder

EXAMPLE 2

Dividing Polynomials Using Synthetic Division

Solution

Using Ϫ2 as our “divisor” (from x ϩ 2 ϭ 02, we set up the synthetic division

template and begin.

Compute the quotient of 1x3 ϩ 3x2 Ϫ 4x Ϫ 122 and 1x ϩ 22, then check your

use Ϫ2 as a “divisor”

Ϫ2

1

1

The result shows

Check

3

Ϫ2

1

Ϫ4

Ϫ2

Ϫ6

Ϫ12

12

0

multiply by divisor, place result

x3 ϩ 3x2 Ϫ 4x Ϫ 12

ϭ x2 ϩ x Ϫ 6, with no remainder.

xϩ2

x3 ϩ 3x2 Ϫ 4x Ϫ 12 ϭ 1x ϩ 22 1x2 ϩ x Ϫ 62

ϭ 1x3 ϩ x2 Ϫ 6x ϩ 2x2 ϩ 2x Ϫ 122

ϭ x3 ϩ 3x2 Ϫ 4x Ϫ 12 ✓

Now try Exercises 13 through 20 ᮣ

Note that in synthetic division, the degree of q(x) will always be one less than p(x),

since the process requires a linear divisor (degree 1).

Since the division process is so dependent on the place value (degree) of each

term, polynomials such as 2x3 ϩ 3x ϩ 7, which has no term of degree 2, must be written using a zero placeholder: 2x3 ϩ 0x2 ϩ 3x ϩ 7. This ensures that like place values

“line up” as we carry out the division.

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EXAMPLE 3

Dividing Polynomials Using a Zero Placeholder

Compute the quotient

Solution

3

use 3 as a “divisor”

2x3 ϩ 3x ϩ 7

xϪ3

2

2

0

6

6

3

18

21

2x3 ϩ 3x ϩ 7 ϭ 1x Ϫ 32 12x2 ϩ 6x ϩ 212 ϩ 70

ϭ 12x3 ϩ 6x2 ϩ 21x Ϫ 6x2 Ϫ 18x Ϫ 632 ϩ 70

ϭ 2x3 ϩ 3x ϩ 7 ✓

Now try Exercises 21 through 30 ᮣ

Many corporations now pay their

employees monthly to save on

payroll costs. If your monthly salary

was \$2037/mo, but you received a

check for only \$237, would you

complain? Just as placeholder

zeroes ensure the correct value of

each digit, they also ensure the

correct valuation of each term in

the division process.

As noted earlier, for synthetic division the divisor must be a linear polynomial and

2x3 Ϫ 3x2 Ϫ 8x ϩ 12

,

the zero of this divisor is used. This means for the quotient

2x Ϫ 3

3

we have 2x Ϫ 3 ϭ 0, and x ϭ

would be used for synthetic division [see

2

Example 6(c)]. Finally, if the divisor is nonlinear, long division must be used.

Division with a Nonlinear Divisor

Compute the quotient:

Solution

note place holder 0 for “x2” term

2x3 ϩ 3x ϩ 7 ϭ 12x2 ϩ 6x ϩ 2121x Ϫ 32 ϩ 70

WORTHY OF NOTE

EXAMPLE 4

7

63

70

70

2x3 ϩ 3x ϩ 7

ϭ 2x2 ϩ 6x ϩ 21 ϩ

. Multiplying by

xϪ3

xϪ3

The result shows

x Ϫ 3 gives

Check

385

2x4 ϩ x3 Ϫ 7x2 ϩ 3

.

x2 Ϫ 2

Write the dividend as 2x4 ϩ x3 Ϫ 7x2 ϩ 0x ϩ 3, and the divisor as x2 ϩ 0x Ϫ 2.

The quotient of leading terms gives

2x4 from dividend

ϭ 2x2 as our first multiplier.

2 from divisor

x

2x2 ϩ x Ϫ 3

x ϩ 0x Ϫ 2 ͤ 2x ϩ x Ϫ 7x2 ϩ 0x ϩ 3

2 2

Multiply 2x 1x ϩ 0x Ϫ 22

Ϫ12x4 ϩ 0x3 Ϫ 4x2 2

x3 Ϫ 3x2 ϩ 0x

2

Multiply x1x ϩ 0x Ϫ 22

Ϫ1x3 ϩ 0x2 Ϫ 2x2

2

4

3

Ϫ3x ϩ 2x ϩ 3

Ϫ1Ϫ3x2 ϩ 0x ϩ 62

2x Ϫ 3

2

Multiply Ϫ31x 2 ϩ 0x Ϫ 22

bring down next term

bring down next term

remainder is 2x Ϫ 3

Since the degree of 2x Ϫ 3 (degree 1) is less than the degree of the divisor (degree 2),

the process is complete.

2x4 ϩ x3 Ϫ 7x2 ϩ 3

2x Ϫ 3

ϭ 12x2 ϩ x Ϫ 32 ϩ 2

x2 Ϫ 2

x Ϫ2

Now try Exercises 31 through 34 ᮣ

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p1x2

r1x2

ϭ q1x2 ϩ

,

Note the we elected to keep the solution to Example 4 in the form

d1x2

d1x2

instead of multiplying both sides by d(x).

A. You’ve just seen how

we can divide polynomials

using long division and

synthetic division

B. The Remainder Theorem

In Example 2, we saw that 1x3 ϩ 3x2 Ϫ 4x Ϫ 122 Ϭ 1x ϩ 22 ϭ x2 ϩ x Ϫ 6, with

remainder zero. Similar to whole number division, this means x ϩ 2 must be a factor of

x3 ϩ 3x2 Ϫ 4x Ϫ 12, a fact made clear as we checked our answer: x3 ϩ 3x2 Ϫ 4x Ϫ 12 ϭ

1x ϩ 221x2 ϩ x Ϫ 62. Now consider the functions p1x2 ϭ x3 ϩ 5x2 ϩ 2x Ϫ 8,

p1x2

x3 ϩ 5x2 ϩ 2x Ϫ 8

ϭ

d1x2 ϭ x ϩ 3, and their quotient

. Using Ϫ3 as the divisor

xϩ3

d1x2

in synthetic division gives

use Ϫ3 as a “divisor”

Ϫ3

1

1

5

Ϫ3

2

2

Ϫ6

Ϫ4

Ϫ8

12

4

This shows x ϩ 3 is not a factor of p(x), since it didn’t divide evenly (the remainder is not zero). However, from the result p1x2 ϭ 1x ϩ 321x2 ϩ 2x Ϫ 42 ϩ 4, we make

a remarkable observation—if we evaluate p1Ϫ32, the quotient portion becomes zero,

showing p1Ϫ32 ϭ 4 (the remainder).

p1Ϫ32 ϭ 1Ϫ3 ϩ 32 3 1Ϫ32 2 ϩ 21Ϫ32 Ϫ 44 ϩ 4

Figure 4.1

ϭ 102 1Ϫ12 ϩ 4

ϭ4

This result can be verified by evaluating p1Ϫ32 in its original form

(also see Figure 4.1):

p1x2 ϭ x3 ϩ 5x2 ϩ 2x Ϫ 8

p1Ϫ32 ϭ 1Ϫ32 3 ϩ 51Ϫ32 2 ϩ 21Ϫ32 Ϫ 8

ϭ Ϫ27 ϩ 45 ϩ 1Ϫ62 Ϫ 8

ϭ4

The result is no coincidence, and illustrates the conclusion of the remainder theorem.

The Remainder Theorem

If a polynomial p(x) is divided by 1x Ϫ c2 using synthetic division,

the remainder is equal to p(c).

This gives us a powerful tool for evaluating polynomials. Where a direct evaluation involves powers of numbers and a long series of calculations, synthetic division

reduces the process to simple products and sums.

EXAMPLE 5

Using the Remainder Theorem to Evaluate Polynomials

Use the remainder theorem to find p1Ϫ52 for p1x2 ϭ x4 ϩ 3x3 Ϫ 8x2 ϩ 5x Ϫ 6.

Verify the result using a substitution.

Solution

use Ϫ5 as a “divisor”

Ϫ5

1

1

The result shows p1Ϫ52 ϭ 19.

3

Ϫ5

Ϫ2

Ϫ8

10

2

5

Ϫ10

Ϫ5

Ϫ6

25

19

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Verification using algebra

Verification using technology

p1Ϫ52 ϭ 1Ϫ52 ϩ 31Ϫ52 Ϫ 81Ϫ52 ϩ 51Ϫ52 Ϫ 6

ϭ 625 Ϫ 375 Ϫ 200 Ϫ 25 Ϫ 6

ϭ 625 Ϫ 606

ϭ 19

4

3

387

2

Now try Exercises 35 through 44 ᮣ

B. You’ve just seen how

we can use the remainder

theorem to evaluate

polynomials

Since p1Ϫ52 ϭ 19, we know 1Ϫ5, 192 must be a point of the graph of p(x). The ability

to quickly evaluate polynomial functions using the remainder theorem will be used

extensively in the sections that follow.

C. The Factor Theorem

As a consequence of the remainder theorem, when p(x) is divided by x Ϫ c and the

remainder is 0, p1c2 ϭ 0 and c is a zero of the polynomial. The relationship between

x Ϫ c, c, and p1c2 ϭ 0 are given in the factor theorem.

The Factor Theorem

For a polynomial p(x),

1. If p1c2 ϭ 0, then x Ϫ c is a factor of p(x).

2. If x Ϫ c is a factor of p(x), then p1c2 ϭ 0.

The remainder and factor theorems often work together to help us find factors of

higher degree polynomials.

EXAMPLE 6

Using the Factor Theorem to Find Factors of a Polynomial

Use the factor theorem to determine if

a. x Ϫ 2

b. x ϩ 1

c. 3x Ϫ 2

are factors of p1x2 ϭ 3x4 Ϫ 2x3 Ϫ 21x2 ϩ 32x Ϫ 12.

Solution

a. If x Ϫ 2 is a factor, then p(2) must be 0. Using the remainder theorem we have

2

3

3

Ϫ2

6

4

Ϫ21

8

Ϫ13

32

Ϫ26

6

Ϫ12

12

0

Since the remainder is zero, we know p122 ϭ 0 (remainder theorem) and

1x Ϫ 22 is a factor (factor theorem).

b. Similarly, if x ϩ 1 is a factor, then p1Ϫ12 must be 0.

Ϫ1

3

3

Ϫ2

Ϫ3

Ϫ5

Ϫ21

5

Ϫ16

32

16

48

Ϫ12

Ϫ48

Ϫ60

Since the remainder is not zero, 1x ϩ 12 is not a factor of p.

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2

c. The zero of the divisor 3x Ϫ 2 ϭ 0 is x ϭ , and this value is used in the

3

synthetic division.

2

Ϫ2

Ϫ21

Ϫ12

3

32

3

Ϫ14

2

0

12

3

0

Ϫ21

18

0

2

2

Since the remainder is zero, pa b ϭ 0 (remainder theorem) and ax Ϫ b is the

3

3

related factor (factor theorem). The original factor 3x Ϫ 2 is found by noting

that the quotient polynomial q1x2 ϭ 3x3 Ϫ 21x ϩ 18 has a common factor of

2

three, which will be factored out and applied to x Ϫ . Starting with the

3

partially factored form we have

2

p1x2 ϭ ax Ϫ b13x3 Ϫ 21x ϩ 182

3

2

ϭ ax Ϫ b1321x3 Ϫ 7x ϩ 62

3

ϭ 13x Ϫ 221x3 Ϫ 7x ϩ 62

partially factored form

factor out 3

2

multiply 3ax Ϫ b

3

This form of simplification will always take place when the zero found using

synthetic division is a fraction and the coefficients of the polynomial are

integers.

Now try Exercises 45 through 56 ᮣ

As a final note on Example 6, there should be no hesitation to use fractions in the synthetic division process if the given polynomial has integer coefficients. The fraction

will be a zero only if all values in the quotient line are integers (i.e., all of the products

and sums must be integers).

EXAMPLE 7

Building a Polynomial Using the Factor Theorem

A polynomial p(x) has three zeroes at x ϭ 3, 12, and Ϫ12. Use the factor

theorem to find the polynomial.

Solution

Using the factor theorem, the factors of p(x) must be 1x Ϫ 32, 1x Ϫ 122, and

1x ϩ 222. Computing the product will yield the polynomial.

p1x2 ϭ 1x Ϫ 321x Ϫ 122 1x ϩ 122

ϭ 1x Ϫ 321x2 Ϫ 22

ϭ x3 Ϫ 3x2 Ϫ 2x ϩ 6

Now try Exercises 57 through 64 ᮣ

Actually, the result obtained in Example 7 is not unique, since any polynomial of

the form a1x3 Ϫ 3x2 Ϫ 2x ϩ 62 will also have the same three zeroes for a ʦ ‫ޒ‬.

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Figure 4.2 shows the graph of Y1 ϭ p1x2 , as

well as graph of Y2 ϭ 2p1x2. The only difference

is 2p1x2 has been vertically stretched. Likewise,

the graph of Ϫ1p1x2 would be a vertical reflection, but still with the same zeroes. As in previous

graph-to-equation exercises, finding a unique

value for the leading coefficient a requires that

we use a point (x, y) on the graph of p and substitute these values to solve for a. For Example 7,

assume you were also told the graph contains the

point (1, 2), or x ϭ 1, p112 ϭ 2. This would

yield:

p1x2 ϭ a1x3 Ϫ 3x2 Ϫ 2x ϩ 62

p112 ϭ a3 1 Ϫ 3112 Ϫ 2112 ϩ 6 4

3

2

389

Figure 4.2

15

Ϫ4

4

Ϫ10

original function

substitute 1 for x

2 ϭ 2a

substitute 2 for p(1), simplify

1ϭa

result

Consistent with the graph shown, if the point (1, 4) were specified instead, a like calculation would show a ϭ 2.

EXAMPLE 8

Finding Zeroes Using the Factor Theorem

Given that 2 is a zero of p1x2 ϭ x4 ϩ x3 Ϫ 10x2 Ϫ 4x ϩ 24, use the factor theorem

to help find all other zeroes.

Solution

Using synthetic division gives:

use 2 as a “divisor”

2

1

1

1

2

3

Ϫ10

6

Ϫ4

Ϫ4

Ϫ8

Ϫ12

Since the remainder is zero, 1x Ϫ 22 is a factor and p can be written:

24

Ϫ24

0

x4 ϩ x3 Ϫ 10x2 Ϫ 4x ϩ 24 ϭ 1x Ϫ 22 1x3 ϩ 3x2 Ϫ 4x Ϫ 122

WORTHY OF NOTE

In Section R.4 we noted a

third degree polynomial

ax3 ϩ bx2 ϩ cx ϩ d is factorable if

ad ϭ bc. In Example 8,

11Ϫ122 ϭ 31Ϫ42 and the polynomial

is factorable.

C. You’ve just seen how

we can use the factor theorem

to factor and build polynomials

Note the quotient polynomial can be factored by grouping to find the remaining

factors of p.

x4 ϩ x3 Ϫ 10x2 Ϫ 4x ϩ 24 ϭ

ϭ

ϭ

ϭ

ϭ

1x Ϫ 221x3 ϩ 3x2 Ϫ 4x Ϫ 122

1x Ϫ 22 3 x2 1x ϩ 32 Ϫ 41x ϩ 32 4

1x Ϫ 22 3 1x ϩ 32 1x2 Ϫ 42 4

1x Ϫ 221x ϩ 321x ϩ 221x Ϫ 22

1x ϩ 321x ϩ 221x Ϫ 22 2

group terms (in color)

remove common

factors from each group

factor common binomial

factor difference of squares

completely factored form

The final result shows 1x Ϫ 22 is actually a repeated factor, and the remaining

zeroes of p are Ϫ3 and Ϫ2.

Now try Exercises 65 through 78 ᮣ

D. Applications

While the factor and remainder theorems are valuable tools for factoring higher degree

polynomials, each has applications that extend beyond this use.

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EXAMPLE 9

Using the Remainder Theorem to Solve a Discharge Rate Application

The discharge rate of a river is a measure of

the river’s water flow as it empties into a

lake, sea, or ocean. The rate depends on

many factors, but is primarily influenced

by the precipitation in the surrounding area

and is often seasonal. Suppose the discharge

rate of the Shimote River was modeled by

D1m2 ϭ Ϫm4 ϩ 22m3 Ϫ 147m2 ϩ

317m ϩ 150, where D (m) represents the

discharge rate in thousands of cubic meters of

water per second in month m 1m ϭ 1 S Jan2.

a. What was the discharge rate in June

(summer heat)?

b. Is the discharge rate higher in February

(winter runoff) or October (fall rains)?

Solution

a. To find the discharge rate in June, we

evaluate D at m ϭ 6.

Using the remainder theorem gives

6

Ϫ1

Ϫ1

22

Ϫ6

16

Ϫ147

96

Ϫ51

317

Ϫ306

11

150

66

216

In June, the discharge rate is 216,000 m3/sec.

b. For the discharge rates in February 1m ϭ 22 and October 1m ϭ 102, we have

2 Ϫ1 22 Ϫ147

317 150

40 Ϫ214 206

↓ Ϫ2

103 356

Ϫ1 20 Ϫ107

D. You’ve just seen how

we can solve applications

using the remainder theorem

10 Ϫ1

22 Ϫ147

317 150

120 Ϫ270 470

T Ϫ10

12 Ϫ27

47 620

Ϫ1

The discharge rate during the fall rains in October is much higher: 620 7 356.

Now try Exercises 81 through 84 ᮣ

4.1 EXERCISES

CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section if needed.

1. For

division, we use the

the divisor to begin.

of

2. If the

is zero after division, then the

is a factor of the dividend.

3. If polynomial P(x) is divided by a linear divisor of

the form x Ϫ c, the remainder is identical to

. This is a statement of the

theorem.

4. If P1c2 ϭ 0, then

must be a factor of

P(x). Conversely, if

is a factor of P(x),

then P1c2 ϭ 0. These are statements from the

theorem.

5. Discuss/Explain how to write the quotient and

remainder using the last line from a synthetic

division.

6. Discuss/Explain why (a, b) is a point on the graph

of P, given b was the remainder after P was divided

by x Ϫ a using synthetic division.

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391

Divide using long division. Write the result as

dividend ‫( ؍‬divisor)(quotient) ؉ remainder.

Compute each indicated quotient. Write answers in the

remainder

form dividend

divisor ‫ ؍‬quotient ؉ divisor .

7.

x3 Ϫ 5x2 Ϫ 4x ϩ 23

xϪ2

31.

2x3 ϩ 7x2 Ϫ x ϩ 26

x4 ϩ 3x3 ϩ 2x2 Ϫ x Ϫ 5

32.

x2 ϩ 3

x2 Ϫ 2

8.

x3 ϩ 5x2 Ϫ 17x Ϫ 26

xϩ7

33.

x4 Ϫ 5x2 Ϫ 4x ϩ 7

x2 Ϫ 1

9. 12x3 ϩ 5x2 ϩ 4x ϩ 172 Ϭ 1x ϩ 32

35. P1x2 ϭ x3 Ϫ 6x2 ϩ 5x ϩ 12

a. P1Ϫ22

b. P152

11. 1x Ϫ 8x ϩ 11x ϩ 202 Ϭ 1x Ϫ 52

2

12. 1x3 Ϫ 5x2 Ϫ 22x Ϫ 162 Ϭ 1x ϩ 22

Divide using synthetic division. Write answers in

remainder

two ways: (a) dividend

divisor ‫ ؍‬quotient ؉ divisor , and

(b) dividend ϭ (divisor)(quotient) ϩ remainder. For

Exercises 13–18, check answers using multiplication.

2x2 Ϫ 5x Ϫ 3

13.

xϪ3

3x2 ϩ 13x Ϫ 10

14.

xϩ5

15. 1x3 Ϫ 3x2 Ϫ 14x Ϫ 82 Ϭ 1x ϩ 22

16. 1x3 Ϫ 6x2 Ϫ 24x Ϫ 172 Ϭ 1x ϩ 12

x Ϫ 5x Ϫ 4x ϩ 23

x ϩ 12x ϩ 34x Ϫ 9

18.

xϪ2

xϩ7

3

17.

2

3

2

19. 12x Ϫ 5x Ϫ 11x Ϫ 172 Ϭ 1x Ϫ 42

3

2

20. 13x3 Ϫ x2 Ϫ 7x ϩ 272 Ϭ 1x Ϫ 12

Divide using synthetic division. Note that some terms of

a polynomial may be “missing.” Write answers as

dividend ‫( ؍‬divisor)(quotient) ؉ remainder.

21. 1x3 ϩ 5x2 ϩ 72 Ϭ 1x ϩ 12

22. 1x3 Ϫ 3x2 Ϫ 372 Ϭ 1x Ϫ 52

23. 1x3 Ϫ 13x Ϫ 122 Ϭ 1x Ϫ 42

24. 1x3 Ϫ 7x ϩ 62 Ϭ 1x ϩ 32

25.

3x3 Ϫ 8x ϩ 12

xϪ1

27. 1n3 ϩ 272 Ϭ 1n ϩ 32

26.

2x3 ϩ 7x Ϫ 81

xϪ3

28. 1m3 Ϫ 82 Ϭ 1m Ϫ 22

29. 1x4 ϩ 3x3 Ϫ 16x Ϫ 82 Ϭ 1x Ϫ 22

30. 1x4 ϩ 3x2 ϩ 29x Ϫ 212 Ϭ 1x ϩ 32

x4 ϩ 2x3 Ϫ 8x Ϫ 16

x2 ϩ 5

Use the remainder theorem to evaluate P(x) as given.

10. 13x3 ϩ 14x2 Ϫ 2x Ϫ 372 Ϭ 1x ϩ 42

3

34.

36. P1x2 ϭ x3 ϩ 4x2 Ϫ 8x Ϫ 15

a. P1Ϫ22

b. P132

37. P1x2 ϭ 2x3 Ϫ x2 Ϫ 19x ϩ 4

a. P1Ϫ32

b. P122

38. P1x2 ϭ 3x3 Ϫ 8x2 Ϫ 14x ϩ 9

a. P1Ϫ22

b. P142

39. P1x2 ϭ x4 Ϫ 4x2 ϩ x ϩ 1

a. P1Ϫ22

b. P122

40. P1x2 ϭ x4 ϩ 3x3 Ϫ 2x Ϫ 4

a. P1Ϫ22

b. P122

41. P1x2 ϭ 2x3 Ϫ 7x ϩ 33

a. P1Ϫ22

b. P1Ϫ32

42. P1x2 ϭ Ϫ2x3 ϩ 9x2 Ϫ 11

a. P1Ϫ22

b. P1Ϫ12

43. P1x2 ϭ 2x3 ϩ 3x2 Ϫ 9x Ϫ 10

a. P1 32 2

b. P1Ϫ52 2

44. P1x2 ϭ 3x3 ϩ 11x2 ϩ 2x Ϫ 16

a. P1 13 2

b. P1Ϫ83 2

Use the factor theorem to determine if the factors given

are factors of f(x).

45. f 1x2 ϭ x3 Ϫ 3x2 Ϫ 13x ϩ 15

a. 1x ϩ 32

b. 1x Ϫ 52

46. f 1x2 ϭ x3 ϩ 2x2 Ϫ 11x Ϫ 12

a. 1x ϩ 42

b. 1x Ϫ 32

47. f 1x2 ϭ x3 Ϫ 6x2 ϩ 3x ϩ 10

a. 1x ϩ 22

b. 1x Ϫ 52

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48. f 1x2 ϭ x3 ϩ 2x2 Ϫ 5x Ϫ 6

a. 1x Ϫ 22

b. 1x ϩ 42

In Exercises 65 through 70, a known zero of the

polynomial is given. Use the factor theorem to write the

polynomial in completely factored form.

49. f 1x2 ϭ Ϫ2x3 Ϫ x2 ϩ 12x Ϫ 9

a. 1x ϩ 32

b. 12x Ϫ 32

65. P1x2 ϭ x3 Ϫ 5x2 Ϫ 2x ϩ 24; x ϭ Ϫ2

66. Q1x2 ϭ x3 Ϫ 7x2 ϩ 7x ϩ 15; x ϭ 3

50. f 1x2 ϭ 3x3 Ϫ 19x2 ϩ 30x Ϫ 8

a. 13x Ϫ 12

b. 1x ϩ 42

67. p1x2 ϭ x4 ϩ 2x3 Ϫ 12x2 Ϫ 18x ϩ 27; x ϭ Ϫ3

68. q1x2 ϭ x4 ϩ 4x3 Ϫ 6x2 Ϫ 4x ϩ 5; x ϭ 1

Use the factor theorem to show the given value is a zero

of P(x).

52. P1x2 ϭ x3 ϩ 3x2 Ϫ 16x ϩ 12; x ϭ Ϫ6

53. P1x2 ϭ x Ϫ 7x ϩ 6; x ϭ 2

3

54. P1x2 ϭ x3 Ϫ 13x ϩ 12; x ϭ Ϫ4

55. P1x2 ϭ 9x3 ϩ 18x2 Ϫ 4x Ϫ 8; x ϭ

2

3

56. P1x2 ϭ 5x3 ϩ 13x2 Ϫ 9x Ϫ 9; x ϭ Ϫ

69. f 1x2 ϭ 2x3 ϩ 11x2 Ϫ x Ϫ 30; x ϭ 32

70. g1x2 ϭ 3x3 ϩ 2x2 Ϫ 75x Ϫ 50; x ϭ Ϫ23

51. P1x2 ϭ x3 ϩ 2x2 Ϫ 5x Ϫ 6; x ϭ Ϫ3

If p(x) is a polynomial with rational coefficients and a

leading coefficient of a ‫ ؍‬1, the rational zeroes of p (if

they exist) must be factors of the constant term. Use this

property of polynomials with the factor and remainder

theorems to factor each polynomial completely.

71. p1x2 ϭ x3 Ϫ 3x2 Ϫ 9x ϩ 27

3

5

A polynomial P with integer coefficients has the zeroes

and degree indicated. Use the factor theorem to write

the function in factored form and standard form.

57. Ϫ2, 3, Ϫ5; degree 3

58. 1, Ϫ4, 2; degree 3

59. Ϫ2, 23, Ϫ 23;

degree 3

60. 25, Ϫ25, 4;

degree 3

61. Ϫ5, 2 23, Ϫ2 23;

degree 3

62. 4, 3 22, Ϫ322;

degree 3

63. 1, Ϫ2, 210, Ϫ 210;

degree 4

64. 27, Ϫ27, 3, Ϫ1;

degree 4

4–12

CHAPTER 4 Polynomial and Rational Functions

72. p1x2 ϭ x3 Ϫ 4x2 Ϫ 16x ϩ 64

73. p1x2 ϭ x3 Ϫ 6x2 ϩ 12x Ϫ 8

74. p1x2 ϭ x3 Ϫ 15x2 ϩ 75x Ϫ 125

75. p1x2 ϭ 1x2 Ϫ 6x ϩ 92 1x2 Ϫ 92

76. p1x2 ϭ 1x2 Ϫ 12 1x2 Ϫ 2x ϩ 12

77. p1x2 ϭ 1x3 ϩ 4x2 Ϫ 9x Ϫ 3621x2 ϩ x Ϫ 122

78. p1x2 ϭ 1x3 Ϫ 3x2 ϩ 3x Ϫ 121x2 Ϫ 3x ϩ 22

WORKING WITH FORMULAS

Volume of an open box: V(x) ‫ ؍‬4x3 ؊ 84x2 ؉ 432x

An open box is constructed by cutting square corners from a 24 in. by 18 in. sheet of cardboard

and folding up the sides. Its volume is given by the formula shown, where x represents the length

of the square cuts.

79. Given a volume of 640 in3, use synthetic division

and the remainder theorem to determine if the

squares were 2-, 3-, 4-, or 5-in. squares and state

the dimensions of the box. (Hint: Write as a

function V(x) and use synthetic division.)

80. Given the volume is 357.5 in3, use synthetic

division and the remainder theorem to determine if

the squares were 5.5-, 6.5-, or 7.5-in. squares and

state the dimensions of the box. (Hint: Write as a

function V(x) and use synthetic division.)

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Section 4.1 Synthetic Division; the Remainder and Factor Theorems

APPLICATIONS

81. Tourist population:

During the 12 weeks

of summer, the

population of

tourists at a popular

beach resort is

modeled by the

polynomial

P1w2 ϭ Ϫ0.1w4 ϩ 2w3 Ϫ 14w2 ϩ 52w ϩ 5,

where P1w2 is the tourist population (in 1000s)

during week w. Use the remainder theorem to help

a. Were there more tourists at the resort in week 5

1w ϭ 52 or week 10? How many more tourists?

b. Were more tourists at the resort one week after

opening 1w ϭ 12 or one week before closing

1w ϭ 112. How many more tourists?

c. The tourist population peaked (reached its

highest) between weeks 7 and 10. Use the

remainder theorem to determine the peak

week.

82. Debt load: Due to a fluctuation in tax revenues, a

county government is projecting a deficit for the next

12 months, followed by a quick recovery and the

repayment of all debt near the end of this period. The

projected debt can be modeled by the polynomial

D1m2 ϭ 0.1m4 Ϫ 2m3 ϩ 15m2 Ϫ 64m Ϫ 3, where

D1m2 represents the amount of debt (in millions of

dollars) in month m. Use the remainder theorem to

a. Was the debt higher in month 5 1m ϭ 52 or

month 10 of this period? How much higher?

b. Was the debt higher in the first month of this

period (one month into the deficit) or after the

eleventh month (one month before the

expected recovery)? How much higher?

393

c. The total debt reached its maximum between

months 7 and 10. Use the remainder theorem

to determine which month.

83. Volume of water: The volume of water in a

rectangular, inground, swimming pool is given by

V1x2 ϭ x3 ϩ 11x2 ϩ 24x, where V(x) is the volume

in cubic feet when the water is x ft high. (a) Use

the remainder theorem to find the volume when

x ϭ 3 ft. (b) If the volume is 100 ft3 of water, what

is the height x? (c) If the maximum capacity of the

pool is 1000 ft3, what is the maximum depth (to the

nearest integer)?

84. Amusement park attendance: Attendance at an

amusement park depends on the weather. After

opening in spring, attendance rises quickly, slows

during the summer, soars in the fall, then quickly

falls with the approach of winter when the park

closes. The model for attendance is given by

A1m2 ϭ Ϫ14 m4 ϩ 6m3 Ϫ 52m2 ϩ 196m Ϫ 260,

where A(m) represents the number of people

attending in month m (in thousands). (a) Did more

people go to the park in April 1m ϭ 42 or June

1m ϭ 62? (b) In what month did maximum

attendance occur? (c) When did the park close?

In these applications, synthetic division is applied in the

usual way, treating k as an unknown constant.

85. Find a value of k that will make x ϩ 2 a factor of

f 1x2 ϭ x3 Ϫ 3x2 Ϫ 5x ϩ k.

86. Find a value of k that will make x Ϫ 3 a factor of

g1x2 ϭ x3 ϩ 2x2 Ϫ 7x ϩ k.

87. For what value(s) of k will x Ϫ 2 be a factor of

p1x2 ϭ x3 Ϫ 3x2 ϩ kx ϩ 10?

88. For what value(s) of k will x ϩ 5 be a factor of

q1x2 ϭ x3 ϩ 6x2 ϩ kx ϩ 50?

EXTENDING THE CONCEPT

89. To investigate whether the remainder and factor

theorems can be applied when the coefficients or

zeroes of a polynomial are complex, try using the

factor theorem to find a polynomial with degree 3,

whose zeroes are x ϭ 2i, x ϭ Ϫ2i, and x ϭ 3.

Then see if the result can be verified using the

remainder theorem and these zeroes. What does the

result suggest? Also see Exercise 92.

90. Since we use a base-10 number system, numbers

like 1196 can be written in polynomial form as

p1x2 ϭ 1x3 ϩ 1x2 ϩ 9x ϩ 6, where x ϭ 10. Divide

p(x) by x ϩ 3 using synthetic division and write

3

2

ϩ 9x ϩ 6

ϭ quotient ϩ

3

remainder

divisor . For x ϭ 10, what is the value of

quotient ϩ remainder

divisor ? What is the result of dividing

1196 by 10 ϩ 3 ϭ 13? What can you conclude?

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A. Long Division and Synthetic Division

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