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Precalculus Notes

[Last Updated 22 January 2011]


Roman Numerals
Prime and Composite Numbers
Fractions
Real and Rational Numbers
Binary Numbers
Calculating Square Roots
Algebra
Law of Cosines
Synthetic Division
Quantitative Reasoning


Roman Numerals

Arabic Roman
1
I
2
II
3
III
4
IV
5
V
6
VI
7
VII
8
VIII
9
IX
10
X
11
XI
12
XII
13
XIII
14
XIV
15
XV
16
XVI
17
XVII
18
XVIII
19
XIX
20
XX
30
XXX
40
XL
50
L
60
LX
70
LXX
80
LXXX
90
XC
100
C
500
D
1000
M
1492
MCDXCII
1925
MCMXXV
1969
MCMLXIX
2009
MMIX

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EXERCISES

Convert each of the following to Roman or Arabic numerals:

1.1. 8
1.2. XVII
1.3. 3
1.4. X
1.5. 5
1.6. XIII
1.7. 20
1.8. XV
1.9. 9
1.10. XVI

ANSWERS

1.1. VIII
1.2. 17
1.3. III
1.4. 10
1.5. V
1.6. 13
1.7. XX
1.8. 15
1.9. IX
1.10. 16

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EXERCISES

Convert each of the following to Roman or Arabic numerals:

2.1. 46
2.2. LXXXIX
2.3. 93
2.4. XXVIII
2.5. 76
2.6. LXII
2.7. 74
2.8. LX
2.9. 71
2.10. LV

ANSWERS

2.1. XLVI
2.2. 89
2.3. XCIII
2.4. 28
2.5. LXXVI
2.6. 62
2.7. LXXIV
2.8. 60
2.9. LXXI
2.10. 55

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EXERCISES

Convert each of the following to Roman or Arabic numerals:

3.1. 1729
3.2. MDCVI
3.3. 1922
3.4. DXXXIX
3.5. 1155
3.6. MDXLVII
3.7. 666
3.8. DCXXV
3.9. 1539
3.10. MLIV

ANSWERS

3.1. MDCCXXIX
3.2. 1606
3.3. MCMXXII
3.4. 809
3.5. MCLV
3.6. 1547
3.7. DCLXVI
3.8. 625
3.9. MDXXXIX
3.10. 1054

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Prime and Composite Numbers

PROBLEM

Find the prime factors of the following numbers:
(a) 33755
(b) 54321
(c) 48700
(d) 675
(e) 123456
(f) 347826
(g) 748647
(h) 450031
(i) 235550
(j) 76432
(k) 52744
(l) 11367
(m) 1958
(n) 1929
(o) 1800
(p) 46553

SOLUTION

(a) 33755 = 5 · 6751 = 5 · 43 · 157

(b) 54321 = 3 · 18107 =3 · 19 · 953

(c) 48700 = 2 · 24350 = 22 · 12175 = 22 · 5 · 2435 = 22 · 52 · 487

(d) 675 = 5 · 135 = 52 · 27 = 52 · 33

(e) 123456 = 2 · 61728 = 22 · 30864 = 23 · 15432 = 24 · 7716 = 25 · 3858 = 26 · 1929 = 26 · 3 · 643

(f) 347826 = 2 · 173913 = 2 · 3 · 57971 = 2 · 3 · 29 · 1999

(g) 748647 = 3 · 249549 = 32 · 83183 = 32 · 193 · 431

(h) 450031 = 37 · 12163

(i) 235550 = 2 · 117775 = 2 · 5 · 23555 = 2 · 52 · 4711 = 2 · 52 · 7 · 673

(j) 76432 = 2 · 38216 = 22 · 19108 = 23 · 9554 = 24 · 4777 = 24 · 17 · 281

(k) 52744 = 2 · 26372 = 22 · 13186 = 23 · 6593 = 23 · 19 · 347

(l) 11367 = 3 · 3789 = 32 · 1263 = 33 · 421

(m) 1958 = 2 · 979 = 2 · 11 · 89

(n) 1929 = 3 · 643

(o) 1800 = 2 · 900 = 22 · 450 = 23 · 225 = 23 · 5 · 45 = 23 · 52 · 9 = 23 · 32 · 52

(p) 46553 = 13 · 3581

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Fractions

PROBLEM

Add 1/5, 1/10, 1/15, 1/20, and 1/25.

SOLUTION

Define

S = 1/5 + 1/10 + 1/15 + 1/20 + 1/25

Write the denominators as products of prime factors.

S = 1/5 + 1/(2 · 5) + 1/(3 · 5) + 1/(22 · 5) + 1/52

The lowest common denominator is 22 · 3 · 52 = 300.

S = 60/300 + 30/300 + 20/300 + 15/300 + 12/300 = 137/300

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PROBLEM

Write each of the following fractions in the form 1/a + 1/b, where a and b are integers.
(a) 1/2
(b) 1/12
(c) 1/25
(d) 1/37

SOLUTION

(a) Assume that in general we can write a fraction 1/n in the form 1/(n+1) + 1/x, where x is an integer. Then

1/n = 1/(n+1) + 1/x => 1/x = 1/n - 1/(n+1) = (n+1) / [n(n+1)] - n / [n(n+1)] = 1 / [n(n+1)]
=> 1/n = 1/(n+1) + 1 / [n(n+1)]

Thus,

1/2 = 1/(2+1) + 1/[2(2+1)] = 1/3 + 1/6

(b) 1/12 = 1/(12+1) + 1/[12(12+1)] = 1/13 + 1/156

(c) 1/25 = 1/(25+1) + 1/[25(25+1)] = 1/26 + 1/650

(d) 1/37 = 1/(37+1) + 1/[37(37+1)] = 1/38 + 1/1406

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Real and Rational Numbers

PROBLEM

Express each of the following real numbers (where the underlined digits are repeated indefinitely) as a rational number:
(a) 0.13
(b) 0.123
(c) 0.17
(d) 0.414
(e) 2.67479
(f) 0.14583
(g) 3.321
(h) 3.14
(i) 0.191
(j) 0.4
(k) 0.7144

SOLUTION

(a) Let x = 0.13. Then

10x = 1.3 (1)
100x = 13.3 (2)

Subtract (1) from (2).

90x = 12 => x = 12/90 = 6/45 = 2/15

(b) Let

x = 0.123 (1)

Then

1000x = 123.123 (2)

Subtract (1) from (2).

999x = 123 => x = 123/999 = 41/333

(c) Let

x = 0.17 (1)

Then

100x = 17.17 (2)

Subtract (1) from (2).

99x = 17 => x = 17/99

(d) Let

x = 0.414 (1)

Then

1000x = 414.414 (2)

Subtract (1) from (2).

999x = 414 => x = 414/999 = 138/333 = 46/111

(e) Let

x = 2.67479 (1)

Then

100000x = 267479.67479 (2)

Subtract (1) from (2).

99999x = 267477 => x = 267477/99999 = 89159/33333 = (7)(47)(271) / (3)(41)(271)
= (7)(47) / (3)(41) = 329/123

(f) Let x = 0.14583. Then

10000x = 1458.3 (1)
100000x = 14583.3 (2)

Subtract (1) from (2).

90000x = 13125 => x = 13125/90000 = 2625/18000 = 525/3600 = 105/720 = 21/144 = 7/48

(g) Let

x = 3.321 (1)

Then

1000x = 3321.321 (2)

Subtract (1) from (2).

999x = 3318 => x = 3318/999 = 1106/333

(h) Let

x = 3.14 (1)

Then

100x = 314.14 (2)

Subtract (1) from (2).

99x = 311 => x = 311/99

(i) Let

x = 0.191 (1)

Then

1000x = 191.191 (2)

Subtract (1) from (2).

999x = 191 => x = 191/999

(j) Let

x = 0.4 (1)

Then

10x = 4.4 (2)

Subtract (1) from (2).

9x = 4 => x = 4/9

(k) Let x = 0.7144. Then

10x = 7.144 (1)
10000x = 7144.144 (2)

Subtract (1) from (2).

9990x = 7137 => x = 7137/9990 = 793/1110

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Binary Numbers

PROBLEM

Express each of the following decimal numbers as a binary number, accurate to twelve places to the right of the decimal point.
(a) 2 1/4
(b) 4 3/32
(c) 0.1
(d) 3/4
(e) 3.61

SOLUTION

The general procedure for converting a decimal number into a binary number is to write each decimal number as a linear combination of powers of 2, and then write the result as a binary (base two) number.

(a) 2 1/4 = 2 + 1/4 = 10.012

(b) 4 3/32 = 4 + 1/16 + 1/32 = 100.000112

(c) 0.1 = 1/10

The largest power of two which is less than 1/10 is 1/16. So

0.1 = 1/16 + x

where

x = 1/10 - 1/16 = 8/80 - 5/80 = 3/80 => 0.1 = 1/16 + 3/80

Since 3/80 is equal to 1/26, the largest power of two which is less than 3/80 is 1/32. So

0.1 = 1/16 + 1/32 + x

where

x = 3/80 - 1/32 = 6/160 - 5/160 = 1/160 => 0.1 = 1/16 + 1/32 + 1/160

The largest power of two which is less than 1/160 is 1/256. So

0.1 = 1/16 + 1/32 + 1/256 + x

where

x = 1/160 - 1/256 = 8/1280 - 5/1280 = 3/1280
=> 0.1 = 1/16 + 1/32 + 1/256 + 3/1280

Since 3/1280 is equal to 1/426, the largest power of two which is less than 3/1280 is 1/512. So

0.1 = 1/16 + 1/32 + 1/256 + 1/512 + x

where

x = 3/1280 - 1/512 = 6/2560 - 5/2560 = 1/2560
=> 0.1 = 1/16 + 1/32 + 1/256 + 1/512 + 1/2560

The largest power of two which is less than 1/2560 is 1/4096. So
0.1 = 1/16 + 1/32 + 1/256 + 1/512 + 1/4096 + x

where

x = 1/2560 - 1/4096 = 8/20480 - 5/20480 = 3/20480
=> 0.1 = 0.0001100110012

(d) 3/4 = 1/2 + 1/4 = 0.112

(e) 3.61 = 2 + 1 + 61/100

Since 61/100 = 1/1.639..., the largest power of two which is less than 61/100 is 1/2. So

3.61 = 2 + 1 + 1/2 + x

where

x = 61/100 - 1/2 = 61/100 - 50/100 = 11/100
=> 3.61 = 2 + 1 + 1/2 + 11/100

Since 11/100 = 1/9.09, the largest power of two which is less than 11/100 is 1/16. So

3.61 = 2 + 1 + 1/2 + 1/16 + x

where

x = 11/100 - 1/16 = 176/1600 - 100/1600 = 76/1600 = 19/400
=> 3.61 = 2 + 1 + 1/2 + 1/16 + 19/400

Since 19/400 = 1/21.052..., the largest power of two which is less than 83/800 is 1/32. So

3.61 = 2 + 1 + 1/2 + 1/16 + 1/32 + x

where

x = 19/400 - 1/32 = 38/800 - 25/800 = 13/800
=> 3.61 = 2 + 1 + 1/2 + 1/16 + 1/32 + 13/800

Since 13/800 = 1/61.538..., the largest power of two which is less than 13/800 is 1/64. So

3.61 = 2 + 1 + 1/2 + 1/16 + 1/32 + 1/64 + x

where

x = 13/800 - 1/64 = 26/1600 - 25/1600 = 1/1600
=> 3.61 = 2 + 1 + 1/2 + 1/16 + 1/32 + 1/64 + 1/1600

The largest power of two which is less than 1/1600 is 1/2048. So

3.61 = 2 + 1 + 1/2 + 1/16 + 1/32 + 1/64 + 1/2048 + x

where

x = 1/1600 - 1/2048 = 32/51200 - 25/51200 = 7/51200
=> 3.61 = 2 + 1 + 1/2 + 1/16 + 1/32 + 1/64 + 1/2048 + 7/51200

Since 7/51200 = 1/7314.285..., the largest power of two which is less than 7/51200 is 1/8192. So

3.61 = 2 + 1 + 1/2 + 1/16 + 1/32 + 1/64 + 1/2048 + 1/8192 + x

where

x = 7/51200 - 1/8192 = 28/204800 - 25/204800 = 3/204800
=> 3.61 = 2 + 1 + 1/2 + 1/16 + 1/32 + 1/64 + 1/2048 + 1/8192 + 3/204800 = 11.1001110000102 Note: The lowest common denominator of two fractions is obtained by factoring the denominators of each of the fractions and calculating the product of the highest powers of each prime factor present in either of the denominators. For example, to obtain the lowest common denominator of 1/1600 and 1/2048, we first factor 1600 and 2048. We have

1600 = 16 · 100 = 24 · 102 = 24 · 22 · 52 = 26 · 52
2048 = 211

The lowest common denominator is

211 · 52 = 51200

Calculating Square Roots

PROBLEM

Calculate the square root of 4321 rounded off to two decimal places.

SOLUTION

Break up 4321 into groups of two digits.

----------------
| 43 21.00 00 00


Examine the first group of two digits, 43, and determine the largest integer which is less than or equal to its square root. The largest such integer is 6. Write 6 on top of the 43.

   6

----------------
| 43 21.00 00 00


Square the 6, write the result under the 43, subtract it from the 43 and bring down the next two digits, the 21.

   6
----------------
| 43 21.00 00 00
|-36
----------------
   7 21


Double the 6 and write the result with a "-" to the left of the "7 21".

      6
   ----------------
   | 43 21.00 00 00
   |-36
   ----------------
12-|  7 21


Determine how many times "12-", where "-" represents some digit between zero and nine, will go into 721 without exceeding it. Write this number, 5, at the top, above the 21, and where the "-" was.

      6  5
   ----------------
   | 43 21.00 00 00
   |-36
   ----------------
125|  7 21


Multiply 125 by 5 and write the result, 625, on the next line, subtract it, and bring down the next two digits.

      6  5
   ----------------
   | 43 21.00 00 00
   |-36
   ----------------
125|  7 21
   |- 6 25
   ----------------
        96 00


Double the 65 and write the result, 130, with a "-" to the left of the "96 00".

       6  5
    ----------------
    | 43 21.00 00 00
    |-36
    ----------------
 125|  7 21
    |- 6 25
    ----------------
130-|    96 00


Determine how many times 130-, where "-" is some digit between zero and nine, will go into 9600 without exceeding it. Write this number, 7, at the top, with a decimal point to the left of it, and where the "-" was.

       6  5. 7
    ----------------
    | 43 21.00 00 00
    |-36
    ----------------
 125|  7 21
    |- 6 25
    ----------------
1307|    96 00


Multiply 1307 by 7 and write the result, 9149, on the next line, subtract it, and bring down the next two digits.

       6  5. 7
    ----------------
    | 43 21.00 00 00
    |-36
    ----------------
 125|  7 21
    |- 6 25
    ----------------
1307|    96 00
    |  - 91 49
    ----------------
          4 51 00


Double the 657 at the top and write the result, 1314, with a "-" to the left of the "4 51 00".

       6  5. 7
    ----------------
    | 43 21.00 00 00
    |-36
    ----------------
 125|  7 21
    |- 6 25
    ----------------
1307|    96 00
    |  - 91 49
    ----------------
  1314-|  4 51 00


Determine how many times 1314-, where "-" is some digit between zero and nine, will go into 45100 without exceeding it. Write this number, 3, at the top and where the "-" was.

       6  5. 7  3
    ----------------
    | 43 21.00 00 00
    |-36
    ----------------
 125|  7 21
    |- 6 25
    ----------------
1307|    96 00
    |  - 91 49
    ----------------
  13143|  4 51 00


Multiply 13143 by 3 and write the result, 39429, on the next line, subtract it, and bring down the next two digits.

       6  5. 7  3
    ----------------
    | 43 21.00 00 00
    |-36
    ----------------
 125|  7 21
    |- 6 25
    ----------------
1307|    96 00
    |  - 91 49
    ----------------
  13143|  4 51 00
       |- 3 94 29
       -------------
            56 71 00


Double the 6573 at the top and write the result, 13146, with a "-" to the left of the "56 71 00".

       6  5. 7  3
    ----------------
    | 43 21.00 00 00
    |-36
    ----------------
 125|  7 21
    |- 6 25
    ----------------
1307|    96 00
    |  - 91 49
    ----------------
  13143|  4 51 00
       |- 3 94 29
       -------------
   13146-|  56 71 00


Determine how many times 13146-, where "-" is some digit between zero and nine, will go into 567100 without exceeding it. Write this number, 4, at the top and where the "-" was.

       6  5. 7  3  4
    ----------------
    | 43 21.00 00 00
    |-36
    ----------------
 125|  7 21
    |- 6 25
    ----------------
1307|    96 00
    |  - 91 49
    ----------------
  13143|  4 51 00
       |- 3 94 29
       -------------
   131464|  56 71 00


Multiply 131464 by 4 and write the result, 525856, on the next line and subtract it.

       6  5. 7  3  4
    ----------------
    | 43 21.00 00 00
    |-36
    ----------------
 125|  7 21
    |- 6 25
    ----------------
1307|    96 00
    |  - 91 49
    ----------------
  13143|  4 51 00
       |- 3 94 29
       -------------
   131464|  56 71 00
         |- 52 58 56
           ---------
             4 12 44


This confirms that 4 is the largest whole number of times 131464 goes into 567100. The square root of 4321, rounded off to two decimal places, is 65.73.

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PROBLEM

Calculate the square root of p, rounded off to three decimal places.

SOLUTION

      1. 7  7  2  4
   ----------------
   |  3.14 15 92 65
   |- 1
   ----------------
 27|  2 14
   |- 1 89
   ----------------
  347|  25 15
     |- 24 29
     --------------
    3542|  86 92
        |- 70 84
        -----------
   35444|  16 08 65
        |- 14 17 76
        -----------
            1 90 89


The square root of p, rounded off to three decimal places, is 1.772.


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Algebra

PROBLEM

Twice the larger of two numbers is three more than five times the smaller, and the sum of four times the larger and three times the smaller is 71. What are the numbers?

[from the movie Mean Girls (2004)]

SOLUTION

Let (x, y) = the (smaller, larger) number. Then

2y = 5x + 3 (1)

and

4y + 3x = 71 (2)

From (1), we get

y = 5x/2 + 3/2

Subsituting this into (2),

4(5x/2 + 3/2) + 3x = 71 => 10x + 6 + 3x = 71 => 13x + 6 = 71 => 13x = 65 => x = 5
=> y = (5/2)(5) + 3/2 = 14

The two numbers are 5 and 14.

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PROBLEM

Find an odd three-digit number whose digits add up to 12. The digits are all different and the difference between the first two digits equals the difference between the last two digits.

[from the movie Mean Girls (2004)]

SOLUTION

Let (x, y, z) be the (units, tens, hundreds) digit. Then the required number satisfies the following four conditions:

(1) x = 1, 3, 5, 7, or 9
(2) x + y + z = 12
(3) x ≠ y ≠ z ≠ x
(4) z - y = y - x

There are 65 numbers that satisfy condition (2):

129, 138, 147, 156, 165, 174, 183, 192
219, 228, 237, 246, 255, 264, 273, 282, 291
309, 318, 327, 336, 345, 354, 363, 372, 381
408, 417, 426, 435, 444, 453, 462, 471, 480
507, 516, 525, 534, 543, 552, 561, 570
606, 615, 624, 633, 642, 651, 660
705, 714, 723, 732, 741, 750
804, 813, 822, 831, 840
903, 912, 921, 930

31 of these also satisfy condition (1) and are odd:

129, 147, 165, 183
219, 237, 255, 273, 291
309, 327, 345, 363, 381
417, 435, 453, 471
507, 525, 543, 561
615, 633, 651
705, 723, 741
813, 831
903, 921

28 of these also satisfy condition (3) and have all-different digits:

129, 147, 165, 183
219, 237, 273, 291
309, 327, 345, 381
417, 435, 453, 471
507, 543, 561
615, 651
705, 723, 741
813, 831
903, 921

Four of these also satisfy condition (4):

147, 345, 543, 741

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PROBLEM

Three apples and two pears cost $3.48. Four pears and one apple cost $4.46. How much does one apple cost? How much does one pear cost?

SOLUTION

Let x = cost of one apple and y = cost of one pear. Then

3x + 2y = $3.48 (1)
4y + x = $4.46 (2)

Multiply (2) by three.

12y + 3x = $13.38 (3)

Subtract (1) from (3).

10y = $9.90 => y = $0.99

Use (2) to solve for x in terms of y.

x = $4.46 - 4y = $4.46 - (4)($0.99) = $4.46 - $3.96 = $0.50

One apple costs $0.50 or 50¢. One pear costs $0.99 or 99¢.

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PROBLEM

Tommy has forty coins which are either quarters or dimes. The total value of the coins is $7.75. How many quarters and how many dimes does Tommy have?

SOLUTION

Let x = number of quarters and y = number of dimes. Then

25x + 10y = 775 (1)
x + y = 40 => y = 40 - x (2)

Substituting (2) into (1),

25x + 10(40 - x) = 775 => 25x + 400 - 10x = 775 => 15x = 375 => x = 375/15 = 25
=> y = 40 - 25 = 15

Tommy has 25 quarters and 15 dimes.

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PROBLEM

Jack can shovel the snow on someone's property in 5 hr. Joe can shovel the snow in 3 hr.
(a) How long will it take Jack and Joe, working together, to shovel the snow?
(b) What fraction of the entire job does each one do?
(c) Assume that each is paid at the same hourly rate they would get if they did the job alone and that the total each would get paid if they worked alone is $75. Who gets paid more if they work together?

SOLUTION

(a) The fraction of the job that (Jack, Joe) does per hour if working along is (1/5, 1/3). Let t be the time (in hours) it takes for both of them to do the job together. Then

(1/5 + 1/3)t = 1

Multiply this equation by 15.

(3 + 5)t = 15 => 8t = 15 => t = 15/8 hr = 1.875 hr

(b) The fraction of the job that (Jack, Joe) does is [(1/5)(15/8), (1/3)(15/8)] = (3/8, 5/8).

(c) (Jack, Joe) is paid [$75/(5 hr), $75/(3 hr)] = ($15/hr, $25/hr). The total (Jack, Joe) gets paid is [($15/hr)(15/8 hr), ($25/hr)(15/8 hr)] = ($225/8, $375/8) = ($28.125, $46.875) => ($28.13, $46.88). So Joe gets paid more.

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Law of Cosines

PROBLEM

A ranger in an observation tower can sight the north end of a lake 15 km away and the south end of the same lake 19 km away. The angle between these two lines of sight is 104°. How long is the lake?

SOLUTION

Draw triangle ABC as shown:

south        north
 end    a     end
  C+----------+B
    \        /
     \      /
     b\    /c
       \  /
        \/
        A observation tower


A is at the observation tower, B is at the north end of the lake, and C is at the south end. The distance from the tower to the north end is

AB = c = 15 km

The distance from the tower to the south end is

AC = b = 19 km

Angle A = 104°. Use the law of cosines to solve for a, the distance from the north end to the south end, which is the length of the lake.

a = sqrt(b2 + c2 - 2bc cos A) = sqrt[(19 km)2 + (15 km)2 - (2)(19 km)(15 km) cos 104°]
= 26.9 km

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Synthetic Division

Consider the polynomial

cnxn + cn-1xn-1 + cn-2xn-2 + ... + ckxk + ... + c2x2 + c1x + c0

where c0, c1,..., and cn are constants, and suppose that we want to divide this polynomial by the binomial x - a, where a is also a constant. We may proceed as in ordinary long division by writing

       ________________________________________________________________
x - a ) cnxn + cn-1xn-1 + cn-2xn-2 + ... + ckxk + ... + c2x2 + c1x + c0


While in ordinary long division we work first with the most significant digits in the dividend and work our way towards the less significant digits, in dividing the above polynomial by the binomial, we start with the term which is of highest order in x and work our way towards the lower-order terms.

The highest-order term in the quotient will necessarily be cnxn-1 since, when this is multiplied by x - a, the result includes the term cnxn. Thus, we write cnxn-1 above the horizontal line.

        cnxn-1
       ________________________________________________________________
x - a ) cnxn + cn-1xn-1 + cn-2xn-2 + ... + ckxk + ... + c2x2 + c1x + c0


We multiply the cnxn-1 by x - a and write the result at the bottom.

        cnxn-1
       ________________________________________________________________
x - a ) cnxn + cn-1xn-1 + cn-2xn-2 + ... + ckxk + ... + c2x2 + c1x + c0
        cnxn - acnxn-1


Now subtract the terms on the bottom line from the dividend and bring down the next term in the dividend.

        cnxn-1
       ________________________________________________________________
x - a ) cnxn + cn-1xn-1 + cn-2xn-2 + ... + ckxk + ... + c2x2 + c1x + c0
      - cnxn + acnxn-1
      ____________________________
      (cn-1 + acn)xn-1 + cn-2xn-2


Define dn-1 = cn-1 + acn. Then we can write

        cnxn-1
       ________________________________________________________________
x - a ) cnxn + cn-1xn-1 + cn-2xn-2 + ... + ckxk + ... + c2x2 + c1x + c0
      - cnxn + acnxn-1
      ____________________________
              dn-1xn-1 + cn-2xn-2


The next term in the quotient will be dn-1xn-2.

        cnxn-1 + dn-1xn-2
       ________________________________________________________________
x - a ) cnxn + cn-1xn-1 + cn-2xn-2 + ... + ckxk + ... + c2x2 + c1x + c0
      - cnxn + acnxn-1
      ____________________________
              dn-1xn-1 + cn-2xn-2


Multiply the dn-1xn-2 by x - a and write the result at the bottom.

        cnxn-1 + dn-1xn-2
       ________________________________________________________________
x - a ) cnxn + cn-1xn-1 + cn-2xn-2 + ... + ckxk + ... + c2x2 + c1x + c0
      - cnxn + acnxn-1
      ____________________________
              dn-1xn-1 + cn-2xn-2
              dn-1xn-1 - adn-1xn-2


Subtract the terms on the bottom line from the preceding line and bring down the next term in the dividend.

        cnxn-1 + dn-1xn-2
       ________________________________________________________________
x - a ) cnxn + cn-1xn-1 + cn-2xn-2 + ... + ckxk + ... + c2x2 + c1x + c0
      - cnxn + acnxn-1
      ____________________________
              dn-1xn-1 + cn-2xn-2
            - dn-1xn-1 + adn-1xn-2
            _________________________________
                (cn-2 + adn-1)xn-2 + cn-3xn-3


Define dn-2 = cn-2 + adn-1. Then we can write

        cnxn-1 + dn-1xn-2
       ________________________________________________________________
x - a ) cnxn + cn-1xn-1 + cn-2xn-2 + ... + ckxk + ... + c2x2 + c1x + c0
      - cnxn + acnxn-1
      ____________________________
              dn-1xn-1 + cn-2xn-2
            - dn-1xn-1 + adn-1xn-2
            _________________________________
                          dn-2xn-2 + cn-3xn-3


If we continue this process, we eventually get

        cnxn-1 + dn-1xn-2 + dn-2xn-3 + ... + dkxk-1 + ... + d3x2
       ________________________________________________________________
x - a ) cnxn + cn-1xn-1 + cn-2xn-2 + ... + ckxk + ... + c2x2 + c1x + c0
      - cnxn + acnxn-1
      ____________________________
              dn-1xn-1 + cn-2xn-2
            - dn-1xn-1 + adn-1xn-2
            _________________________________
                          dn-2xn-2 + cn-3xn-3
                          ...
                          ________________________________________
                                                        d2x2 + c1x


where dk = ck + adk+1, d3 = c3 + ad4, and d2 = c2 + ad3.

The next term in the quotient will be d2x.

        cnxn-1 + dn-1xn-2 + dn-2xn-3 + ... + dkxk-1 + ... + d3x2 + d2x
       ________________________________________________________________
x - a ) cnxn + cn-1xn-1 + cn-2xn-2 + ... + ckxk + ... + c2x2 + c1x + c0
      - cnxn + acnxn-1
      ____________________________
              dn-1xn-1 + cn-2xn-2
            - dn-1xn-1 + adn-1xn-2
            _________________________________
                          dn-2xn-2 + cn-3xn-3
                          ...
                          ________________________________________
                                                        d2x2 + c1x


Multiply the d2x by x - a and write the result at the bottom.

        cnxn-1 + dn-1xn-2 + dn-2xn-3 + ... + dkxk-1 + ... + d3x2 + d2x
       ________________________________________________________________
x - a ) cnxn + cn-1xn-1 + cn-2xn-2 + ... + ckxk + ... + c2x2 + c1x + c0
      - cnxn + acnxn-1
      ____________________________
              dn-1xn-1 + cn-2xn-2
            - dn-1xn-1 + adn-1xn-2
            _________________________________
                          dn-2xn-2 + cn-3xn-3
                          ...
                          ________________________________________
                                                        d2x2 + c1x
                                                        d2x2 - ad2x


Subtract the terms on the bottom line from the preceding line and bring down the next and last term in the dividend.

        cnxn-1 + dn-1xn-2 + dn-2xn-3 + ... + dkxk-1 + ... + d3x2 + d2x
       ________________________________________________________________
x - a ) cnxn + cn-1xn-1 + cn-2xn-2 + ... + ckxk + ... + c2x2 + c1x + c0
      - cnxn + acnxn-1
      ____________________________
              dn-1xn-1 + cn-2xn-2
            - dn-1xn-1 + adn-1xn-2
            _________________________________
                          dn-2xn-2 + cn-3xn-3
                          ...
                          ________________________________________
                                                        d2x2 + c1x
                                                      - d2x2 + ad2x
                                                      __________________
                                                        (c1 + ad2)x + c0


Define d1 = c1 + ad2. Then we can write

        cnxn-1 + dn-1xn-2 + dn-2xn-3 + ... + dkxk-1 + ... + d3x2 + d2x
       ________________________________________________________________
x - a ) cnxn + cn-1xn-1 + cn-2xn-2 + ... + ckxk + ... + c2x2 + c1x + c0
      - cnxn + acnxn-1
      ____________________________
              dn-1xn-1 + cn-2xn-2
            - dn-1xn-1 + adn-1xn-2
            _________________________________
                          dn-2xn-2 + cn-3xn-3
                          ...
                          ________________________________________
                                                        d2x2 + c1x
                                                      - d2x2 + ad2x
                                                      __________________
                                                                d1x + c0


The next term in the quotient will be d1.

        cnxn-1 + dn-1xn-2 + dn-2xn-3 + ... dkxk-1 + ... + d2x + d1
       ________________________________________________________________
x - a ) cnxn + cn-1xn-1 + cn-2xn-2 + ... + ckxk + ... + c2x2 + c1x + c0
      - cnxn + acnxn-1
      ____________________________
              dn-1xn-1 + cn-2xn-2
            - dn-1xn-1 + adn-1xn-2
            _________________________________
                          dn-2xn-2 + cn-3xn-3
                          ...
                          ________________________________________
                                                        d2x2 + c1x
                                                      - d2x2 + ad2x
                                                      __________________
                                                                d1x + c0


Multiply the d1 by x - a and write the result at the bottom.

        cnxn-1 + dn-1xn-2 + dn-2xn-3 + ... dkxk-1 + ... + d2x + d1
       ________________________________________________________________
x - a ) cnxn + cn-1xn-1 + cn-2xn-2 + ... + ckxk + ... + c2x2 + c1x + c0
      - cnxn + acnxn-1
      ____________________________
              dn-1xn-1 + cn-2xn-2
            - dn-1xn-1 + adn-1xn-2
            _________________________________
                          dn-2xn-2 + cn-3xn-3
                          ...
                          ________________________________________
                                                        d2x2 + c1x
                                                      - d2x2 + ad2x
                                                      __________________
                                                                d1x + c0
                                                                d1x - ad1


Subtract the terms on the bottom line from the preceding line.

        cnxn-1 + dn-1xn-2 + dn-2xn-3 + ... dkxk-1 + ... + d2x + d1
       ________________________________________________________________
x - a ) cnxn + cn-1xn-1 + cn-2xn-2 + ... + ckxk + ... + c2x2 + c1x + c0
      - cnxn + acnxn-1
      ____________________________
              dn-1xn-1 + cn-2xn-2
            - dn-1xn-1 + adn-1xn-2
            _________________________________
                          dn-2xn-2 + cn-3xn-3
                          ...
                          ________________________________________
                                                        d2x2 + c1x
                                                      - d2x2 + ad2x
                                                      __________________
                                                                d1x + c0
                                                              - d1x + ad1
                                                              ___________                                                                  c0 + ad1



The c0 + ad1 is a remainder. The quotient is completed by adding this remainder divided by x - a. Define d0 = c0 + ad1. Then the remainder term is

R = c0 + ad1 = d0

and we can write

        cnxn-1 + dn-1xn-2 + dn-2xn-3 + ... + d2x + d1 + d0 / (x - a)     (1)
       ________________________________________________________________
x - a ) cnxn + cn-1xn-1 + cn-2xn-2 + ... + ckxk + ... + c2x2 + c1x + c0


Consider again the polynomial

cnxn + cn-1xn-1 + cn-2xn-2 + ... + ckxk + ... + c2x2 + c1x + c0

where c0, c1,..., and cn are constants, and suppose that we want to divide this polynomial by the binomial x - a, where a is also a constant. Now consider the following procedure.

1. Write the coefficients cn, cn-1,..., c0 from left to right:

cn  cn-1  cn-2  ...  ck  ...  c2  c1  c0


2. Write the constant a at the right end of the line as shown:

cn  cn-1  cn-2  ...  ck  ...  c2  c1  c0  | a
                                          +---


3. Leave enough space for an additional row of numbers and then draw a horizontal line at the bottom.

cn  cn-1  cn-2  ...  ck  ...  c2  c1  c0  | a
                                          +---

________________________________________


4. Write the first coefficient below the horizontal line.

cn  cn-1  cn-2  ...  ck  ...  c2  c1  c0  | a
                                          +---

________________________________________
cn


5. Multiply the coefficient by a and write the result directly below the next coefficient.

cn  cn-1  cn-2  ...  ck  ...  c2  c1  c0  | a
                                          +---
    acn
________________________________________
cn


6. Add cn-1 and acn, and write the result below the line.

cn  cn-1  cn-2  ...  ck  ...  c2  c1  c0  | a
                                          +---
    acn
________________________________________
cn  cn-1 + acn


7. Define dn-1 = cn-1 + acn.

cn  cn-1  cn-2  ...  ck  ...  c2  c1  c0  | a
                                          +---
    acn
________________________________________
cn  dn-1


8. Now multiply dn-1 by a and write the result under the cn-2.

cn  cn-1  cn-2  ...  ck  ...  c2  c1  c0  | a
                                          +---
    acn  adn-1
________________________________________
cn  dn-1


9. Add cn-2 and adn-1, and write the result below the line.

cn  cn-1  cn-2  ...  ck  ...  c2  c1  c0  | a
                                          +---
    acn  adn-1
________________________________________
cn  dn-1  cn-2 + adn-1


10. Define dn-2 = cn-2 + adn-1.

cn  cn-1  cn-2  ...  ck  ...  c2  c1  c0  | a
                                          +---
    acn  adn-1
________________________________________
cn  dn-1  dn-2


11. Continuing this general procedure, you will eventually get

cn  cn-1  cn-2  ...  ck  ...  c2  c1  c0  | a
                                          +---
    acn  adn-1 ...  adk+1 ... ad3 ad2 ad1  
________________________________________
cn  dn-1  dn-2  ...  dk  ...  d2  d1  d0


where dk = ck + adk+1, d2 = c2 + ad3, d1 = c1 + ad2, and d0 = c0 + ad1.

The constants cn, dn-1, dn-2,..., dk,..., d2, and d1 that appear on the bottom line are the coefficients of xn, xn-1, and x in (1), while d0 is the remainder. Thus, the procedure just described is a faster way of obtaining the quotient in (1).

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PROBLEM

Use synthetic division to divide the given polynomial by the given binomial:

(a) 2x3 + 3x2 - x - 2, x - 1
(b) 2x3 + 3x2 - x - 2, x + 2
(c) 3x4 + x3 - 2x2 + x - 1, x + 1
(d) x3 - 3x2 - x + 2, x - 4
(e) 2x3 - 3x2 + 9x - 4, x - 1/2
(f) 2x4 - 5x3 - 5x2 + 3x + 2, x + 1/2
(g) x5 - 4x + 2, x - 1
(h) 3x4 + 2x3 + x - 1, x + 2
(i) 4x3 + 3x - 3, x - 1/2
(j) 3x3 + 7x2 - 3, x - 2/3

[from Cannon, Lawrence O. and Joseph Elich 1994, Precalculus, Second Edition (New York: HarperCollins College Publishers), Develop Mastery exercises 3.2.19-28]

SOLUTION

(a) 2x3 + 3x2 - x - 2, x - 1

2  3  -1  -2  | 1
   2   5   4  +---
____________
2  5   4   2
=> 2x2 + 5x + 4 + 2 / (x - 1)

(b) 2x3 + 3x2 - x - 2, x + 2

2  3  -1  -2 | -2
  -4   2  -2 +----
____________
2 -1   1  -4
=> 2x2 - x + 1 - 4 / (x + 2)

(c) 3x4 + x3 - 2x2 + x - 1, x + 1

3  1  -2  1  -1 | -1
  -3   2  0  -1 +----
_______________
3 -2   0  1  -2
=> 3x3 - 2x2 + 1 - 2 / (x + 1)

(d) x3 - 3x2 - x + 2, x - 4

1  -3  -1   2 | 4
    4   4  12 +---
_____________
1   1   3  14
=> x2 + x + 3 + 14 / (x - 4)

(e) 2x3 - 3x2 + 9x - 4, x - 1/2

2  -3   9 -4 | 1/2
    1  -1  4 +----
____________
2  -2   8  0
=> 2x2 - 2x + 8

(f) 2x4 - 5x3 - 5x2 + 3x + 2, x + 1/2

2  -5  -5  3  2 | -1/2
   -1   3  1 -2 +------
_______________
2  -6  -2  4  0
=> 2x3 - 6x2 - 2x + 4

(g) x5 - 4x + 2, x - 1

1  0  0  0  -4  2 | 1
   1  1  1   1 -3 +---
_________________
1  1  1  1  -3 -1
=> x4 + x3 + x2 + x - 3 - 1 / (x - 1)

(h) 3x4 + 2x3 + x - 1, x + 2

3  2  0   1  -1 | -2
  -6  8 -16  30 +----
_______________
3 -4  8 -15  29
=> 3x3 - 4x2 + 8x - 15 + 29 / (x + 2)

(i) 4x3 + 3x - 3, x - 1/2

4  0  3  -3 | 1/2
   2  1   2 +-----
___________
4  2  4  -1
=> 4x2 + 2x + 4 - 1 / (x - 1/2)

(j) 3x3 + 7x2 - 3, x - 2/3

3  7  0  -3 | 2/3
   2  6   4 +-----
___________
3  9  6   1
=> 3x2 + 9x + 6 + 1 / (x - 2/3)

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Quantitative Reasoning

PROBLEM

Consider three opaque boxes. One box contains all white balls, one all black balls, and one a mix of black and white balls. Each box is labeled, but the labels are all wrong. How many balls would you need to pull out to determine which box is which?

[from Kirsner, Scott 2010, "Seeking Work? Be Prepared for Tests Aimed At Separating Stars from Duds," Boston Sunday Globe, G1]

SOLUTION

Suppose the boxes are labeled as shown. Then, because we are told that the labels are all wrong, the possible types of balls in each box are as written below the drawing of the box (B = black, W = white, B/W = black and white).

+-----+  +-----+  +-----+
|  1  |  |  2  |  |  3  |
|     |  |     |  |black|
|black|  |white|  | and |
|     |  |     |  |white|
+-----+  +-----+  +-----+
   W        B        B
  B/W      B/W       W


Thus, box 1 actually contains either all white or mixed black and white balls, box 2 actually contains either all black or mixed black and white balls, and box 3 actually contains either all black or all white balls.

Suppose we draw one ball from box 3. If the ball is black, this means that box 3 contains all black balls, box 2 contains mixed black and white balls, and box 1 contains all white balls. If the ball drawn from box 3 is white, this means that box 3 contains all white balls, box 1 contains mixed black and white balls, and box 2 contains all black balls. Thus, by picking one ball from box 3, it is possible to determine what kind of balls is contained in all three of the boxes.

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PROBLEM

A car's odometer read 14632 miles on 26 October 2007, 16686 miles on 8 March 2008, and 20972 miles on 28 December 2008. Estimate the number of miles driven from 28 December 2007 to 28 December 2008 assuming that the miles driven each day is the same.

SOLUTION

From 26 October 2007 to 28 December 2007 is 63 days. 2008 is a leap year, so there are 29 days in February 2008. From 26 October 2007 to 8 March 2008 is therefore 134 days. On 28 December 2007, the odometer should have read about 14632 miles + (16686 miles - 14632 miles)(63 days) / (134 days) = 15597.69 miles. The number of miles driven from 28 December 2007 to 28 December 2008 was about 20972 miles - 15597.69 miles = 5374.313 miles => 5374 miles.

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PROBLEM

John has two $44 vouchers which can be used to pay for monthly subway passes. The cost of a subway pass increases from $44 to $59 next month. John has the option of purchasing additional $21, $31, $35, $44, or $50 vouchers which are deducted, pretax, from his paycheck. John is in the 21% income tax bracket; 21% of his paycheck is deducted for income tax. What additional vouchers should John purchase in order to buy two subway passes next month? Assume that no change is given if the voucher value exceeds the cost of the subway passes.

SOLUTION

John has $88 of vouchers. The total cost of two subway passes next month is (2)($59) = $118, so John needs $118 - $88 = $30 more. He can either buy one $21 voucher and pay $9 aftertax cash or buy a $31 voucher and forfeit $1 since no change is given.

If he buys the $21 voucher and pays $9 aftertax cash, his total additional pretax cost is $21 + $9/(1 - 0.21) = $32.39. If he buys the $31 voucher his total additional pretax cost is $31. So John is better off buying the $31 voucher and forfeiting $1.

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PROBLEM

Gasoline is purchased for a car on the dates shown in the table below. Each time gas is purchased, the tank is filled up. The table shows the mileage on the car's odometer, the price paid, and the number of gallons of gasoline purchased.

Date Odometer Reading (miles) Gasoline Purchased (gallons) Total Cost Cost per Gallon
2007 Apr 1
11640
9.553
$25.40
$2.659
2007 Apr 13
11778
12.983
$35.69
$2.749
2007 May 12
12040
15.255
$44.84
$2.939

From the data in the table, determine the miles per gallon and the cost per mile achieved by the car.

SOLUTION

From the data, the total number of miles driven during the period shown is 12040 miles - 11640 miles = 400 miles. The total gas used is 12.983 gal + 15.255 gal = 28.238 gal. The total cost of the gas used is $35.69 + $44.84 = $80.53. The miles per gallon achieved by the car is (400 miles) / (28.238 gal) = 14.17 mpg. The cost per mile is $80.53 / (400 miles) = $0.2013 per mile. Note that we neglected the initial purchase of gas in our calculations because it was used to fill up the tank prior to the period during which the 400 miles were driven.

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PROBLEM

Jack, a taxpayer, is in the 35% income tax bracket.

(a) If his annual income in 2007 was $75,000, and his income tax was deducted from his paycheck throughout the year at the rate corresponding to his tax bracket, how much income tax did he pay in 2007?

(b) Jack paid $20,000 in property tax in 2007. If this amount is deductible for income tax purposes (i.e., it counts as negative income), how much of an income tax refund should he get when he files his income tax forms?

(c) If the $20,000 in property tax is not deductible for income tax purposes but is taxed at 28% instead of 35%, how much of an income tax refund should he get?

SOLUTION

(a) Jack paid (0.35)($75,000) = $26,250 income tax in 2007.

(b) The income tax that Jack should pay for 2007 is (0.35)($75,000 - $20,000) = $19,250. Since $26,250 was deducted from his paycheck, he should get a $26,250 - $19,250 = $7,000 refund.

(c) The income tax that Jack should pay for 2007 is (0.35)($55,000) + (0.28)($20,000) = $24,850. Since $26,250 was deducted from his paycheck, he should get a $26,250 - $24,850 = $1,400 refund.

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PROBLEM

Joan, a taxpayer, earned $85,000 in 2007.

(a) Joan's income tax was deducted from her paycheck throughout the year at the rate corresponding to her tax bracket. If Joan is in the 35% tax bracket, how much income tax was deducted from her paycheck?

(b) Joan paid $10,000 in property tax in 2007. If this amount is taxed at 28% instead of 35%, how much of a refund should Joan get when she files her income tax forms?

(c) How would your answer to part (b) change if Joan was in the 25% tax bracket?

SOLUTION

(a) (0.35)($85,000) = $29,750 was deducted from Joan's paycheck.

(b) Joan should pay (0.35)($75,000) + (0.28)($10,000) = $29,050 income tax for 2007, so she should get a refund of $29,750 - $29,050 = $700.

(c) If Joan was in the 25% tax bracket, (0.25)($85,000) = $21,250 would have been deducted from her paycheck. The amount of income tax she should actually pay is (0.25)($75,000) + (0.28)($10,000) = $21,550. So she would owe $21,550 - $21,250 = $300 in additional income tax when she files her income tax forms.

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PROBLEM

If 0 < st < 1, then which of the following can be true?

(A) s < -1 and t > 0
(B) s < -1 and t < -1
(C) s > -1 and t < -1
(D) s > 1 and t < -1
(E) s > 1 and t > 1

[from Educational Testing Service advertisement in U.S. News and World Report, May 2009, p. 69]

SOLUTION

The question is asking us to select the choice which doesn't necessarily contradict the condition 0 < st < 1. We examine each choice.

(A) If s < -1 and t > 0, st must be negative, which contradicts 0 < st < 1.

(B) if s < -1 and t < -1, st must be greater than 1, which contradicts 0 < st < 1.

(C) If s > -1 and t < -1, possible choices for s and t are s = - 1/4 and t = - 3/2. This yields st = (- 1/4)(- 3/2) = 3/8 = 0.375, which satisfies the condition 0 < st < 1. Thus, (C) is not necessarily a contradiction, and (C) can be true.

(D) If s > 1 and t < -1, st < -1, which contradicts 0 < st < 1.

(E) If s > 1 and t > 1, st > 1, which contradicts 0 < st < 1.

Thus, the only choice which doesn't necessarily contradict 0 < st < 1 is (C).

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PROBLEM

If u > t, r > q, s > t, and t > r, which of the following must be true?

I. u > s
II. s > q
III. u > r

(A) I only
(B) II only
(C) III only
(D) I and II
(E) II and III

[from Educational Testing Service advertisement in U.S. News and World Report, May 2009, p. 69]

SOLUTION

Combining the inequalities, we have s, u > t > r > q. Thus, s > q and u > r, but u might not be greater than s. It follows that II and III are true, and the answer is (E).

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