CHAPTER VII. PARENTHESES. 1. The use of parentheses has been briefly discussed in Ch. II., § 2, Arts. 8-9. It is frequently necessary to employ more than two sets of parentheses in the same chain of operations, and to distinguish them the following forms are used: Parentheses, (); Brackets, []; Braces, {}. A Vinculum is a line drawn over an expression, and is equivalent to parentheses inclosing it. E.g., (a + b)(cd)= a+b⋅c-d. If more forms of parentheses than the above are needed, one or more of them is made larger and heavier. Removal of Parentheses. 2. The principles given in Ch. II., § 2, Art. 8, are to be applied successively when several sets of parentheses are to be removed from a given expression. In removing parentheses we may begin either with the inmost or with the outmost. 3. The following examples will illustrate the method of removing parentheses, beginning with the inmost: Ex. 1. - Ex. 2. [b2 — {(a2 + b) a − (a2 — b) b — a2 (a — b)}]3 =[b2 — {a3 + ab — a2b + b2 — a3 + a2b}]3 = [b2 — { ab + b2} ]3 = · 4. The method of removing parentheses, beginning with the outmost, may be illustrated by the following example. Notice that the part which is free from parentheses is simplified at the same time that the next parentheses are removed. x2-12x2-[3x2+(4 x2-5 x2-1)]} {2 = x2 - 2x2+[3x2 +(4 x2 - 5 x2 - 1)], removing braces, x2 = x2+3x2+(4x2-5 x2 - 1), removing brackets, =6x2-5 x2+ 1, removing vinculum, = x2+1. EXERCISES I. Simplify the following expressions by removing parentheses: 1. a + 2b - [6 a − {3 b − (6 a − 6 b)}]. 6. a {5b [a (3c3b)+2c-(a-2bc)]}. 7. 12 a 13{10 [7 (4 a 3) - 6] - 9 a}. 8. x-{x+y- [x + y + z − (x + y + z + v)]} 9. 102x-5[3-2x-6 (4x-7)] −3 (5-2x)}. 10 7 am - {2 am - [an − 3 am + (5 am – 2 an) — 4 am] — 2 an}. 11. {[(x + y2) x − (2y-1)]x - (x2 - 2 y) x - x2y232. 12. [(x − y)2 + 6 xy] − [ (x2 + 2 xy) − {x2 — [2 xy — (4 xy — y2)]} 13. a2x+ax2 — {} ax2 — [− ‡ a2x − (} ax2 − a)]}. Find the values of the expressions in Exx. 1-13, 14. When a = — − 3, b = 4, c = − 5, m = 2, n = 1, x = 8, y = — 9, 2 = 7, - 4. 21. (2x2+1-x) (2 x2 − 1 + x) = 1 + x2(2x + 1) (2 x − 1). 22. (x − 1)(x2 + x + 1) − 6 (x2 −— 1)= — (2 — x)3. 27. 4{4[4(4x-3) — 3] — 3} − 3 = 1. 28. 4-4{4 — 4 [4 — 4(4 — x)]}= 44. 29. -4x-(5x-[6x-7x-(8 x 9)}])= 10. Insertion of Parentheses. 5. The principles for inserting parentheses in a given expression were proved in Ch. II., § 2, Art. 9. Ex. 1. Express 4(xy)+y-x as a product, of which one factor is x - y. · We have 4(xy) + y − x = 4 (x − y) − (x − y) = 3 (x − y). The sign or before a pair of parentheses can evidently be reversed from + to -, or from to, if the signs of the terms within the parentheses be reversed. Ex. 2. 7(x-1) 3 (1 − x) = 7 (x − 1)+3(x − 1) = 10 (x − 1). EXERCISES II. Write each of the following expressions as a product, of which the expression within the parentheses is one of the factors: Write each of the following expressions as a single product, of which the expression within the first parentheses is a factor: CHAPTER VIII. FACTORS AND MULTIPLES OF INTEGRAL ALGEBRAIC EXPRESSIONS. § 1. INTEGRAL ALGEBRAIC FACTORS. 1. Factors have already been defined in multiplication (Ch. II., § 3, Art. 12). The factors were there given, and their product was required. The converse process, given a product to find its factors, is equally important. 2. A product of two or more factors is, by the definition of division, exactly divisible by any one of the factors. An Integral Algebraic Factor of an expression is an integral expression by which the given one is exactly divisible. E.g., integral factors of 6 a2x are 6, a2x, 3 x, 2 a2, etc.; integral factors of a2 b2 are a + b and a b. The word integral, here as in Ch. III., § 1, Art. 1, refers only to the literal parts of the expression. E.g., a and x are integral algebraic factors of 6 a2x. 3. A Prime Factor is one which is exactly divisible only by itself and unity. E.g., the prime factors of 6 a2x are 2, 3, a, a, x. A Composite Factor is one which is not prime, i.e., which is itself the product of two or more prime factors. E.g., composite factors of 6 a2x are 6, ax, 2 a, 3 ax, etc. 4. Any monomial can be resolved into its prime factors by inspection. E.g., the prime factors of 4 ab2 are 2, 2, a, a, a, b, b. The Fundamental Formula for Factoring. 5. A multinomial whose terms contain a common factor can be factored by applying the converse of the Distributive Law for Multiplication. From Ch. III., § 3, Art. 14, we have ab + ac ad = a (b + c - d). (1) That is, if the terms of a multinomial contain a common factor, the multinomial can be written as the product of the common factor and the algebraic sum of the remaining factors of the terms. The relation (1) may be called the Fundamental Formula for Factoring. Ex. 1. Factor 2 x2y - 2 xy2. The factor 2 xy is common to both terms; the remaining factor of the first term is x, that of the second term is — y, and their algebraic sum is x y. Consequently 2x2y-2xy2 = 2xy (x − y). Ex. 2. ab2 + abc + b2c = b (ab + ac + bc). 6. In the fundamental formula the letters a, b, c, d may stand for binomial or multinomial expressions. Ex. 1. Factor a (x-2y) + b (x - 2y). The factor x 2 y is common to both terms; the remaining factor of the first term is a, that of the second term is b, and their algebraic sum is a + b. Consequently a (x − 2 y) + b (x − 2 y) = (x − 2 y)(a + b). Ex. 2. Factor 1 a +x (1 - a). We have 1 a + x(1 − a)=(1a)(1 + x). Ex. 3. Factor (x − y) (a2 + b2) — (x + y)(a2 + b2). (x − y)(a2 + b2)—(x + y) (a2 + b2) = (a2 + b2) [(x − y) −(x + y)] It frequently happens that the parts of a given expression have a common factor except for sign. |