A. Long Division and Synthetic Division
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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
add remainder
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.
add
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.
add
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
↓
add
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
answer.
use Ϫ2 as a “divisor”
Ϫ2
1
↓
1
The result shows
Check
ᮣ
3
Ϫ2
1
Ϫ4
Ϫ2
Ϫ6
Ϫ12
12
0
drop lead coefficient into place;
multiply by divisor, place result
in next column and add
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
and check your answer.
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
subtract (algebraic addition)
bring down next term
subtract (algebraic addition)
bring down next term
subtract (algebraic addition)
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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Section 4.1 Synthetic Division; the Remainder and Factor Theorems
391
DEVELOPING YOUR SKILLS
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
answer the following questions.
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
help answer the following questions.
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?
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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
your answer as x ϩ xx 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?