112 ALGEBRA. Converse Use of Factors. [CHAP. 143. The actual processes of multiplication and division can often be partially or wholly avoided by a skilful use of factors. It should be observed that the formula which the student has seen exemplified in this chapter are just as useful in their converse as in their direct application. Thus the formula for resolving into factors the difference of two squares is equally useful as enabling us to write down at once the product of the sum and the difference of two quantities. Example 1. Multiply 2a+3b− c by 2a-3b+c. These expressions may be arranged thus: 2a +(3b-c) and 2a-(3b-c). Hence the product = {2a +(3b −c)} {2a -(3b−c)} Example 2. Find the product of x+2, x2, x2 - 2x+4, x2+2x+4. Taking the first factor with the third, and the second with the fourth, the product = {(x+2)(x2 − 2x+4)} {(x − 2)(x2 + 2x+4)} = (23+8)(x3-8) =26-64. Example 3. Divide the product of 2x2+x-6 and 6x2-5x+1 by 3x2+5x-2. Denoting the division by means of a fraction, by cancelling factors which are common to numerator and denomin ator. 155. When the factors of the numerator and denominator cannot be determined by inspection, the fraction may be reduced to its lowest terms by dividing both numerator and denominator by the highest common factor, which may be found by the rules given in Chap. XVIII. Example. Reduce to lowest terms 3x3-13x2+23x - 21 The H.C.F. of numerator and denominator is 3x-7. Dividing numerator and denominator by 3x-7, we obtain as respective quotients x2 -2x+3 and 5x2 - x - 3. Thus 3x3-13x2+23x - 21 _ (3x − 7)(x2 − 2x+3) _ x2 - 2x+3 ̧ = -= 156. If either numerator or denominator can readily be resolved into factors we may use the following method. Example. Reduce to lowest terms x3+3x2-4x The numerator = x(x2+3x-4) = x(x+4)(x − 1). Of these factors the only one which can be a common divisor is x-1. Hence, arranging the denominator so as to shew x1 as a factor, (2) x+y-z=0, x-y+z = 4, 5x+y+z=20; 159 12. A purse of sovereigns is divided amongst three persons, the first receiving half of them and one more, the second half of the remainder and one more, and the third six. Find the number of sovereigns the purse contained. 13. If h =-1, k = 2, l = 0, m = 1, n = −3, find the value of 15(p3+q3), 5(p2-pq+q2), 4(p2+pq+q2), 6(p2 --g3). 15. Find the square root of 18. A sum of money is to be divided among a number of persons; if 8s. is given to each there will be 3s. short, and if 7s. 6d. is given to each there will be 2s. over: find the number of persons. it must be remembered that the expression (b2-4ac) is the square root of the compound quantity b2-4ac, taken as a whole. We cannot simplify the solution unless we know the numerical values of a, b, c. It may sometimes happen that these values do not make b2-4ac a perfect square. In such a case the exact numerical solution of the equation cannot be determined. Example. Solve 5x2-13x-11= 0. Here a = 5, b = −13, c = -11; therefore by the formula we have (-13)(-13)-4.5(-11) X= 2.5 13+ √169+220 10 13±√389 Since 389 has not an exact square root this result cannot be simplified; thus the two roots are 196. Solution by Factors. There is still ore method of obtaining the solution of a quadratic which will sometimes be found shorter than either of the methods already given. Consider the equation x2+x= 2. 7 3 Clearing of fractions, 3x2+7x-6=0........... by resolving the left-hand side into factors we have (3x-2)(x+3)=0. ...(1); Now if either of the factors 3x-2, x+3 be zero, their product is Hence the quadratic equation is satisfied by either of the suppositions zero. It appears from this that when a quadratic equation has been simplified and brought to the form of equation (1), its solution can always be readily obtained if the expression on the left-hand Elementary Algebra for Schools BY H. S. HALL, M.A. FORMERLY SCHOLAR OF CHRIST S COLLEGE, CAMBRIDGE AND S. R. KNIGHT, B.A., M.B., CH.B. FORMERLY SCHOLAR OF TRINITY COLLEGE, CAMBRIDGE London MACMILLAN AND CO., LIMITED NEW YORK: TIE MACMILLAN COMPANY Seventh Edition, Revised and Enlarged. Globe 8vo. (bound in maroon-coloured cloth), 3s. 6d. With Answers (bound in green-coloured cloth), 4s. 6d. The distinctive features of the Seventh Edition are: (1) Greater prominence has been given to the fundamental Laws of Algebra (see Arts. 22, 29-32, 46-48). With this object parts of the chapters on Multiplication and Division have been re-written. (2) A short section on the use of Detached Coefficients has been given on page 7. (3) A fuller treatment of the Remainder Theorem and its applications will be found on pages 236, 237. |