Factoring and Algebra - A Selection of Classic Mathematical Articles Containing Examples and Exercises on the Subject of Algebra (Mathematics Series)

Various

Description

This book contains classic material dating back to the 1900s and before. The content has been carefully selected for its interest and relevance to a modern audience. Carefully selecting the best articles from our collection we have compiled a series of historical and informative publications on the subject of mathematics. The titles in this range include "Ratio and Proportion" "Simple Equations" "Simultaneous Equations" and many more. Each publication has been professionally curated and includes all details on the original source material. This particular instalment, "Factoring and Algebra" contains a selection of classic educational articles containing examples and exercises on the subject of algebra. It is intended to illustrate aspects of factoring and serves as a guide for anyone wishing to obtain a general knowledge of the subject. We are republishing these classic works in affordable, high quality, modern editions, using the original text and artwork.

Sujets

• Sciences formelles
• Mathématiques appliquées
• Mathématiques appliquées
• Sciences et techniques
• Algebra
• Mathematics
• Algèbre
• Algèbre

Informations

• Publié par: Read Books Ltd.
• Date de parution: 25 octobre 2016
• Nombre de lectures: 0
• EAN13: 9781473358775
• Langue: English
• Poids de l'ouvrage: 1 Mo

Extrait

Mathematics Series
Factoring and Algebra
A Selection of Classic Mathematical Articles Containing Examples and Exercises on the Subject of Algebra
By
Various Authors
Copyright 2011 Read Books Ltd. This book is copyright and may not be reproduced or copied in any way without the express permission of the publisher in writing
British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library
Contents
Bradbury Elementary Algebra - Designed for the Use of Schools and Academies . William F. Bradbury
Grammar School Algebra . Wm. M. Griffin
A University Algebra . Edward Olney
First Principles of Algebra . Edward Olney
Mathematics for the Practical Man . George Howe
An Academic Algebra . James M. Taylor
FACTORING.
62. F ACTORING is the resolving a quantity into its factors.
63. The factors of a quantity are those integral quantities whose continued product is the quantity.
N OTE .-In using the word factor we shall exclude unity.
64. A P RIME Q UANTITY is one that is divisible without remainder by no integral quantity except itself and unity.
Two quantities are mutually prime when they have no common factor.
65. The P RIME F ACTORS of a quantity are those prime quantities whose continued product is the quantity.
66. The factors of a purely algebraic monomial quantity are apparent. Thus, the factors of a 2 b x y z are a a b x y z .
67. Polynomials are factored by inspection, in accordance with the principles of division and the theorems of the preceding section.
CASE I.
68. When all the terms have a common factor.
1. Find the factors of a x - a b + a c .
OPERATION .
( a x - a b + a c ) = a ( x - b + c )
As a is a factor of each term it must be a factor of the polynomial; and if we divide the polynomial by a , we obtain the other factor. Hence,
RULE.
Write the quotient of the polynomial divided by the common factor in a parenthesis, with the common factor prefixed as a coefficient .
2. Find the factors of 6 x y - 72 x y 2 + 18 a x 2 y 3 .
Ans. 6 x y (1 - 12 y + 3 a x y 2 ).
N OTE .- Any factor common to all the terms can be taken as well as 6 x y ; 2, 3, x, y , or the product of any two or more of these quantities, according to the result which is desired. In the examples given, let the greatest monomial factor be taken.
3. Find the factors of x + x 2 . Ans. x (1 + x ).
4. Find the factors of 8 a 2 x 2 + 12 a 3 x 4 - 4 a x y .
Ans. 4 a x (2 a x + 3 a 2 x 3 - y ).
5. Find the factors of 5 x 4 y 2 + 25 a x 5 - 15 x 3 y 3 .
Ans. 5 x 3 ( x y 2 + 5 a x 2 - 3 y 3 ).
6. Find the factors of 7 a x - 8 b y + 14 x 2 .
7. Find the factors of 4 x 2 y 2 - 28 x 3 y 4 - 44 x 4 y 2 .
8. Find the factors of 55 a 2 c - 11 a c + 33 a 2 c x .
9. Find the factors of 98 a 2 x 2 - 294 a 3 x 2 y 2 .
10. Find the factors of 15 a 2 b 2 c d - 9 a b 2 d 2 + 18 a 3 c 2 d 4 .
CASE II.
69. When two terms of a trinomial are perfect squares and positive, and the third term is equal to twice the product of their square roots.
1. Find the factors of a 2 + 2 a b + b 2 .
OPERATION .
a 2 + 2 a b + b 2 = ( a + b ) ( a + b )
We resolve this into its factors at once by the converse of the principle in Theorem II. Art. 58
2. Find the factors of a 2 - 2 a b + b 2 .
OPERATION .
a 2 - 2 a b + b 2 = ( a - b ) ( a - b )
We resolve this into its factors at once by the converse of the principle in Theorem III. Art. 59. Hence,
RULE.
Omitting the term that is equal to twice the product of the square roots of the other two, take for each factor the square root of each of the other two connected by the sign of the term omitted .
3. Find the factors of x 2 - 2 x y + y 2 .
Ans. ( x - y ) ( x - y ).
4. Find the factors of 4 a 2 c 2 + 12 a c d + 9 d 2 .
Ans. (2 a c + 3 d ) (2 a c + 3 d ).
5. Find the factors of 1 - 4 x z + 4 x 2 z 2 .
Ans. (1 - 2 x z ) (1 - 2 x z ).
6. Find the factors of 9 x 2 - 6 x + 1.
Ans. (3 x - 1) (3 x - 1).
7. Find the factors of 25 x 2 + 60 x + 36.
8. Find the factors of 49 a 2 - 14 a x + x 2 .
Ans. (7 a - x ) (7 a - x ).
9. Find the factors of 16 y 2 - 16 a 2 y + 4 a 4 .
10. Find the factors of 12 a x + 4 x 2 + 9 a 2 .
11. Find the factors of 6 x + 1 + 9 x 2 .
CASE III.
70. When a binomial is the difference between two squares.
1. Find the factors of a 2 - b 2 .
OPERATION .
a 2 - b 2 = ( a + b ) ( a - b )
We resolve this into its factors at once by the converse of the principle in Theorem IV. Art. 60. Hence,
RULE.
Take for one of the factors the sum, and for the other the difference, of the square roots of the terms of the binomial .
2. Find the factors of x 2 - y 2 .
Ans. ( x + y ) ( x - y ).
3. Find the factors of 4 a 2 - 9 b 4 .
Ans. (2 a + 3 b 2 ) (2 a - 3 b 2 ).
4. Find the factors of 16 x 2 - c 2 .
5. Find the factors of a 2 b 4 c 2 - x 2 y 6 .
6. Find the factors of 81 x 4 - 49 y 2 .
7. Find the factors of 25 a 2 - 4 c 4 .
8. Find the factors of m 8 - n 10 ).
N OTE .- When the exponents of each term of the residual factor obtained by this rule are even, this factor can be resolved again by the same rule. Thus, x 4 - y 4 = ( x 2 + y 2 ) ( x 2 - y 2 ); but x 2 - y 2 = ( x + y ) ( x - y ); and therefore the factors of x 4 - y 4 are x 3 + y 2 , x + y , and x - y .
9. Find the factors of a 4 - b 4 .
Ans. ( a 2 + b 2 ) ( a + b ) ( a - b ).
10. Find the factors of x 8 - y 8 .
Ans. ( x 4 + y 4 ) ( x 2 + y 2 ) ( x + y ) ( x - y ).
11. Find the factors of a 4 - 1.
12. Find the factors of 1 - x 8 .
Ans. (1 + x 4 ) (1 + x 2 ) (1 + x ) (1 - x ).
13. Find the factors of a 7 - a 5 .
Ans. a 5 ( a + 1) ( a - 1).
14. Find three factors of x 9 - x 3 .
71. Any binomial consisting of the difference of the same powers of two quantities, or the sum of the same odd powers, can be factored. For
I. The difference of the same powers of two quantities is divisible by the difference of the quantities .
Let a and b represent two quantities and a b , and by actual division we find

and so on.
II. The difference of the same even powers of two quantities is divisible by the sum of the quantities .

and so on.
It follows from the two preceding statements that
The difference of the same even powers of two quantities is divisible by either the sum or the difference of the quantities .
III. The sum of the same odd powers of two quantities is divisible by the sum of the quantities .

and so on.
1. Find the factors of x 5 - y 5 .
OPERATION .
( x 5 - y 5 ) ( x - y ) = x 4 + x 3 y + x 2 y 2 + x y 3 + y 4
By I. of this article, the difference of the same powers of two quantities is divisible by the difference of the quantities; therefore x - y must be a factor of x 5 - y 5 ; and dividing x 5 - y 5 by x - y gives the other factor x 4 + x 3 y + x 2 y 2 + x y 3 + y 4 .
2. Find two factors of c 6 - d 6 .
OPERATION .
( c 6 - d 6 ) ( c + d ) = c 5 - c 4 d + c 3 d 2 - c 2 d 3 + c d 4 - d 5
By II. the difference of the same even powers of two quantities is divisible by the sum of the quantities; therefore c + d must be a factor of c 6 - d 6 ; and dividing c 6 - d 6 by c + d gives the other factor c 5 - c 4 d + c 3 d 2 - c 2 d 3 + c d 4 - d 5 .
3. Find the factors of m 5 + n 5 .
OPERATION .
( m 5 + n 5 ) ( m + n ) = m 4 - m 3 n + m 2 n 2 - m n 3 + n 4
By III. the sum of the same odd powers of two quantities is divisible by the sum of the quantities; therefore m + n must be a factor of m 5 + n 5 ; and dividing m 5 + n 5 by m + n gives the other factor m 4 - m 3 n + m 2 n 2 - m n 3 + n 4 .
4. Find the factors of a 3 - x 3 .
Ans. ( a - x ) ( a 2 + a x + x 2 ).
5. Find the factors of a 5 + x 5 .
N OTE . - In Example 2, the factors of c 6 - d 6 there obtained are not the only factors; for by I. c 6 - d 6 is divisible by c - d ; and dividing c 6 - d 6 by c - d gives another factor,
c 5 + c 4 d + c 3 d 2 + c 2 d 3 + c d 4 + d 5 ;
or by Art. 70,
c 6 - d 6 = ( c 3 + d 3 ) ( c 3 - d 3 ).
But
c 5 - c 4 d + c 3 d 2 - c 2 d 3 + c d 4 - d 5 ,
c 5 + c 4 d + c 3 d 2 + c 2 d 3 + c d 4 + d 5 ,
c 3 + d 3 ,
c 3 - d 3 ,
are not prime quantities; for the first can be divided by c - d , and the quotient thus arising can be divided by c 2 c d + d 2 ; the second can be divided by c + d , and the quotient thus arising will be the same as after the division of the first quantity by c - d , and can be divided by c 2 c d + d 2 ; the third can be divided by c + d , and the fourth by c - d . Performing these divisions, by each method we shall find the prime factors of c 6 - d 6 to be
c + d , c - d , c 2 + c d + d 2 , and c 2 - c d + d 2 .
In finding the prime factors, it is better to apply first the principle of Art. 70 as far as possible.
6. Find the prime factors of x 10 - y 10 .
x 10 - y 10
= ( x 5 + y 5 ) ( x 5 - y 5 ).
x 5 + y 5
= ( x + y ) ( x 4 - x 3 y + x 2 y 2 - x y 3 + y 4 ).
x 5 - y 5
= ( x - y ) ( x 4 + x 3 y - x 2 y 2 + x y 3 - y 4 ).
Ans. ( x + y ) ( x - y ) ( x 4 - x 3 y + x 2 y 2 - x y 3 + y 4 ) ( x 4 + x 3 y + x 2 y