Degree, Dimension, Vinculum, Bar, Parenthesis, Similar Terms. 15. Each literal factor of a term is called a dimension, and the degree of a term is the number of its dimensions. The degree of a term is, therefore, found by taking the sum of the exponents of its literal factors. Thus, 7x is of one dimension, or of the first degree; 5a2 b c is of four dimensions, or of the fourth degree, &c. 16. A polynomial is homogeneous, when all its terms are of the same degree. Thus, 3a-2b+c is homogeneous of the first degree, 8 a3 b — 16 a2 b2 +64 is homogeneous of the fourth degree. 17. A vinculum or bar -, placed over a quantity, or a parenthesis ( ) enclosing it, is used to express that all the terms of the quantity are to be considered together. Thus, (a+b+c) xd is the product of a+b+c by d, √x2+y2, or √(x2 + y2) is the square root of x2 + y2. The bar is sometimes placed vertically. (a−2b+3c)x+(5 a2+3—2 d) x2+(−3 c+4 d—1) x3. 18. Similar terms are those in which the literal factors are identical. Reduction of Polynomials. 19. The terms of a polynomial which are preceded by the sign+are called the positive terms, and those which are preceded by the sign are called the negative terms. When the first term is not preceded by any sign it is to be regarded as positive. 20. The following rule for reducing polynomials, which contain similar terms, is too obvious to require demonstration. Find the sum of the similar positive terms by adding their coefficients, and in the same way the sum of the similar negative terms. The difference of these sums preceded by the sign of the greater, may be substituted as a single term for the terms from which it is obtained. When these sums are equal they cancel each other, and the corresponding terms are to be omitted. Thus, a2b-9 a b2 +8 a2b+5c-3 a2 b+8 a b2 + 2 a2b+c+ab2. 8 c is the same as 8 a2 b—2 c. 21. EXAMPLES. 1. Reduce the polynomial 10 a 3 a1 + 6 a1 — a1 — 5 a to its simplest form. Ans. 13 a4. 2. Reduce the polynomial 5 a1b+3✓ a b2 c-7 ab + 17 ab+2ab2c-6 a1 b-8ab2c-10 ab+9 a1b to its simplest form. Ans. 8a1b-3 √ a b2 c. 3. Reduce the polynomial 3 a-2 a−7ƒ+3ƒ+2 a +4f-3 a to its simplest form. Ans. 0. Addition. SECTION II. Addition. 22. Addition consists in finding the quantity equivalent to the aggregate or sum of several different quantities. 23. Problem. To find the sum of any given quantities. Solution. The following solution requires no demonstration. The quantities to be added are to be written after each other with the proper sign between them, and the polynomial thus obtained can be reduced to its simplest form by art. 20. 7. Find the sum of -6f, 9f, 13f, and -8f. b. Ans. 8f. -3b. 8. Find the sum of 126, 4b and 13 b. Ans. 9. Find the sum of +ax-ab, ab - √ + xy, ax+xy-4ab, √x + √x − x and xy + xy+ax. Ans. 2+3 ax-4ab+4xy-x. Subtraction. 10. Find the sum of 7 x2-6 √x+5x3 z +3-g 25. Subtraction consists in finding the difference between two quantities. 26. Problem. To subtract one quantity from another. Solution. Let A denote the aggregate of all the positive terms of the quantity to be subtracted, and B the aggregate of all its negative terms; then A – B is the quantity to be subtracted, and let C denote the quantity from which it is to be taken. If A alone be taken from C, the remainder C-A is as much too small as the quantity subtracted is too large, that is, as much as A is larger than A-B. The required remainder is, consequently, obtained by increasing C-A by the excess of A above A—B, that is, by B, and it is thus found to be C-A+B. The same result would be obtained by adding to C the quantity AB, with its signs reversed, so as to make it -AB. Hence, To subtract one quantity from another, change the signs of the quantity to be subtracted from + to —, and from - Multiplication of Monomials. to +, and add it with its signs thus reversed to the quantity from which it is to be taken. 9. From 3a-176-10613a-2 a take 6b-8a-b-2a+3d+9 a-5 h. Ans. 15 a 32b-3d+5h. 10. Reduce 32 a 3 b― (5 a +17 b) to its simplest form. Ans, 27 a - 14 b. 11. Reduce ab―(2a-3b) (5 a +7b) — (-13a+2b) to its simplest form. Ans. 7a-5 b. 28. Problem. several monomials. SECTION IV. Multiplication. To find the continued product of Solution. The required product is indicated by writing the given monomials after each other with the sign of multiplication between them, and thus a monomial is formed, which is the continued product of all the factors of the given |
