Page:Elementary algebra (1896).djvu/127

109 Rh ^ 4a;2-9y2 g x(2 g^ - 3 aa;) _ ^^ (a;?/ -3?/) 2 ' 4 a;^ 4- 6 xy ' a(4 a'-^x — 9 aj^) ' xhj'^ — 27 2/^ g 20 (x'^ - ?/3) ^Q x3 - 2 X7f ^2 a;2-5x 6 x^ + 5 x?/ + 5 1/'^ X* — 4 x^y'^ + 4 2/* x^ — 4 x — 5 ^g 3 x^ + 6 X j^g a;3y + 2 a;^^; ^ 4 ^^y j4 5a36 + 10a^6^ Ig 3 g^ + 9 a^b + 6 ^252 3 a-b-^ + 6 a&3 " ' a^^a^b-2 dW ^Y x4-14x'-^- 51 jg 2x2 + 17X + 21 g^ 3 x^ + 23 x + 14 X* - 2 x2 - 15 ' ' 3 x2 + 26 X + 35* ' 3 x- + 41 x + 20* j^g x^ + xy - 2 y2 g^ «%2 _ 16 (^2 ^^ 27 a + g^ x3 - ^3 ■ • ax?' + 9 «x + 20 «' ■ 18 « - 6 g2 + 2 d^'
 * x2 + 4 X + 4* ' x3 - 8

138. When the factors of the numerator and denominator cannot be determined by inspection, we find the highest common factor, by the rules given in Chapter xi.

Ex. Reduce to lowest terms {3x^3 - 13x2 + 23x - 21 }{15x3 - 38x2 -2x + 21 }

First Method. The H. C. F. of numerator and denominator is 3x-7.

Dividing numerator and denominator by 3x — 7, we obtain as re- spective quotients x2 — 2 x + 3 and 5 x2 — x — 3.

Thus {3x3-13x2 + 23x-21} {15x3 - 38x2 -2x + 21}

={(3x-7)(x2-2x + 3) }{ (3x-7)(5x2-x-3) }

={ x2-2x + 3  }{5x2- x-3 }

This is the simplest solution for the beginner ; but in this and similar cases we may often effect the reduction without actually going through the process of finding the highest common factor.

Second method. By Art. 116, the H. C. F. of numerator and denominator must be a factor of their sum 18x3 — 51x2 + 21 x, that is, of 3x(3x — 7)(2x — 1). If there be a common divisor it must clearly be 3 x — 7 ; hence arranging numerator and denominator so as to show 3 x — 7 as a factor,

the fraction

{x2(3x - 7)- 2x(3x - 7)+ 3(3x - 7)} {5x^2(3x-7)-x(3x-7)-3(3x-7)}

= {(3 x- 7)(x2-2x + 3) }{(3x-7)(5x2-x-3) } ={x2-2x4- 3} {5x2 -x-3}