Therefore the highest common factor is 3x – 7. The factor 2 has been removed on the same grounds as the factor 9 above. and Example 2. Find the highest common factor of 2x3 + x2-x-2 (1), (2). As the expressions stand we cannot begin to divide one by the other without using a fractional quotient. The difficulty may be obviated by introducing a suitable factor, just as in the last case we found it useful to remove a factor when we could no longer proceed with the division in the ordinary way. The given expressions have no common simple factor, hence their H. C. F. cannot be affected if we multiply either of them by any simple factor. Multiply (2) by 2, and use (1) as a divisor: After the first division the factor 7 is introduced because the first remainder 7x2+5x+2 will not divide 2x3 + x2 − x − 2. At the next stage the factor 17 is introduced for a similar reason, and finally the factor 64 is removed as explained in Example 1. 144. From the last two examples it appears that we may multiply or divide either of the given expressions, or any of the remainders which occur in the course of the work, by any factor which does not divide both of the given expressions. 145. Let the two expressions in Example 2, Art. 143, be written in the form 2x3+x2-x-2=(x-1) (2x2+3x+2), 3x3-2x2+x-2= (x − 1) (3x2+x+2). Then their highest common factor is x-1, and therefore 2x2 + 3x+2 and 3x2+x+2 have no algebraical common divisor. If, however, we put x=6, then and 2x3+x2-x-2=460, 3x3-2x2+x-2=580; and the greatest common measure of 460 and 580 is 20; whereas 5 is the numerical value of x-1, the algebraical highest common factor. Thus the numerical values of the algebraical highest common factor and of the arithmetical greatest common measure do not in this case agree. The reason may be explained as follows: when x=6, the expressions 2x2+3x+2 and 3x2+x+2 become equal to 92 and 116 respectively, and have a common arithmetical factor 4; whereas the expressions have no algebraical common factor. It will thus often happen that the highest common factor of two expressions, and their numerical greatest common measure, when the letters have particular values, are not the same; for this reason the term greatest common measure is inappropriate when applied to algebraical quantities. EXAMPLES XVIII. b. Find the highest common factor of the following expressions: 1. x3+2x2-13x+10, x3+x2 – 10x+8. 8. x3-x2+x+3, x2+x3-3x2 −x+2. 9. 2x3-5x2+11x+7, 4x3-11x2+25x+7. 10. 2x3+4x2-7x-14, 6x3 – 10x2 − 21x+35. 11. 3x4 - 3x3- 2x2 - x − 1, 6xa – 3x3 — x2 − x − 1. 12. 2x1- 2x3+x2+3x-6, 4x4 – 2x3+3x − 9. 13. 3x3-3x2+2a2x-2a3, 3x3+12ax2+2a2x+8a3. 14. 2x3-9ax2+9a2x-7a3, 4x3-20ax2+20α2x – 16α3. 15. 2x3+7ax2+4a2x-3a3, 4x3 +9ax2 - 2α2x — a3. - 16. 6a3+13a2x – 9ax2 – 10x3, 9a3+12a2x−11ax2 – 10x3. 17. 24x+y+72x3y2 - 6x2y3 - 90xy*, 6x4y2+13x3y3 — 4x2у4 — 15xу5. 18. 4x5a2+10x1a3 – 60x3a1+54x2a5, 24x3a3+30x3α5 – 126x2α. 19. 4x+14x2+20x3+70x2, 8x7+28x6 – 8x3 — 12x2+56x3. 20. 72x3-12ax2+72a2x-420a3, 18x3+42ax2 - 282a2x+270a3. *146. The statements of Art. 141 may be proved as follows. I. If F divides A it will also divide mA. For suppose A=aF, then mA=maF. II. If F divides A and B, then it will divide mA ±nB. then mA±nB=maF±nbF Thus F divides mA±nB. *147. We may now enunciate and prove the rule for finding the highest common factor of any two compound algebraical expressions. We suppose that any simple factors are first removed. [See Example, Art. 142.] Let A and B be the two expressions after the simple factors have been removed. Let them be arranged in descending or ascending powers of some common letter; also let the highest power of that letter in B be not less than the highest power in A. Divide B by A; let p be the quotient, and C the remainder. Suppose C to have a simple factor m. Remove this factor, and so obtain a new divisor D. Further, suppose that in order to make A divisible by D it is necessary to multiply A by a simple factor n. Let q be the next quotient and E the remainder. Finally, divide D by E; let r be the quotient, and suppose that there is no remainder. Then E will be the H.C.F. required. First, to shew that E is a common factor of A and B. By examining the steps of the work, it is clear that E divides D, therefore also qD; therefore qD+E, therefore nA; therefore A, since n is a simple factor. Again, E divides D, therefore mD, that is, C. And since E divides A and C, it also divides pA+C, that is, B. Hence E divides both A and B. Secondly, to shew that E is the highest common factor. If not, let there be a factor X of higher dimensions than E. Then X divides A and B, therefore B - pA, that is, C'; therefore D (since m is a simple factor); therefore nA-qD, that is, E. Thus X divides E; which is impossible since, by hypothesis, I is of higher dimensions than E. Therefore E is the highest common factor. *148. The highest common factor of three expressions A,B,C may be obtained as follows. First determine F the highest common factor of A and B; next find G the highest common factor of F and C; then G will be the required highest common factor of A, B, C. For F contains every factor which is common to A and B, and G is the highest common factor of F and C. Therefore & is the highest common factor of A, B, C. CHAPTER XIX. FRACTIONS. [On first reading the subject, the student may omit the general proofs of the rules given in this Chapter.] 149. IN Chapter XII. we discussed the simpler kinds of fractions, using the ordinary Arithmetical rules. We here propose to give proofs of those rules, and shew that they are applicable to Algebraical fractions. 150. DEFINITION. If a quantity x be divided into b equal parts, and a of these parts be taken, the result is called the fraction of x. a If x be the unit, the fraction α a b fraction so that the fraction represents a equal parts, b of b which make up the unit. By by α b ma mb bmb' where a, b, m are positive integers. we mean a equal parts, b of which make up the unit ... (1); But that is, Conversely, α |