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Example 2. Find the highest common factor of

ax2 + 2a-x +a’, 2ax2 – 4a x – 6a3, 3 (ax +a%). Resolving the expressions into factors, we have

axa + 2a-x +a=a (x2 + 2ax: +ao)
=a (x + a)?

(1),
2ax2 – 4a x – 6a3=2a (x2 – 2ax – 3a2)

= 2a (x +a) (x – 3a)... (2), 3 (ax +a%)? = 3a? (x + a)?

(3). Therefore from (1), (2), (3), by inspection, the highest common factor is a (c+a).

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Find the highest common factor of 1. a? +ab, a? 62.

2. (x+y), 22-y. 3. 2.x2 – 2xy, 23 2y. 4. 6x2 – 9xy, 4x2 9y2. 5. x3 + x2y, 23 +43

6. a’b-ab3, a5b2 a285. 7. a3 a-x, a3 ax?, at - ax3. 8. a? — 4x”, a2 + 2ax. 9. apbx +ab-x, aểb 63. 10.2.cy - 6.xy?, 22-9y. 11. a? – x?, a? ax, a-x ax4. 12. 4x2 + 2xy, 12x2y 343. 13. 20.x — 4, 50x2 – 2.

14. 6bx+4by, 9cx + 6cy. 15. .*+ , (+1), +1. 16. 7-9, *y-cy. 17. 22 – 2xy + y2, (x y)3. 18. 203 +ax, x4 – a4. 19.23 + 8y3, x2 + xy 2y2. 20. 24–27a’x, (x – 3a)2. 21. x2 +3x+2, 22 – 4. 22. 22 – X – 20, x2 – 9x+20. 23. 22 – 18x+45, 32 – 9. 24. 2x2 – 7x+3, 3.x2 – 7x – 6. 25. 12x2 + x– 1, 15x2 +8x+1. 26. 2.02 – 2 – 1, 3.x2 – 0 -2. 27. c2x2 – d, aca2 bcx +adx bd. 28. 2.5-xy, 23 + xy + xy +yo. 29. a'x abx 6ab2x, a-bx2 4ab2x2+363x2. 30. 2x2 +9x +4, 2.x2 + 11x +5, 2x2 – 3x – 2. 31. 3.24 + 8x3+4x2, 3.06 +11x4 +6x3, 3x4 – 16x3 – 12x2. 32, 234 +5x3 + 3x2, 6x4 + 13x3 +6x2, 2x4 — 723 – 15x2. 33. 12:+6x2+6, 6x+3x2+3, 18x+3x2 +15. 34. 24+43c2 +3, ** +5x2 +6, 3x4 +11x2 +6. 35. 2a2+7ad +6d2, 2q2 +9ad+9d2, 6a2+11ad+3d2. 36. 2x2 +82y +6y2, 4x2 +14xy +6y2, 2.2 +10xy+12y2.

96. It will often happen that expressions cannot be readily resolved into factors. To find the highest common factor in such cases, we adopt a method analogous to that used in Arithmetic for finding the greatest common measure of two or more numbers.

Note. The term greatest common measure is sometimes used instead of highest common factor; but, strictly speaking, the term greatest common measure ought to be confined to arithmetical quantities; for the highest common factor is not necessarily the greatest common measure in all cases, as will appear later. [Art.101.]

97. We begin by working out examples illustrative of the algebraical process of finding the highest common factor, postponing for the present the complete proof of the rules we use. But we may conveniently enunciate two principles, which the student should bear in mind in reading the examples which follow.

I. If an expression contain a certain factor, any multiple of the expression is divisible by that factor.

II. If two expressions have a common factor, it will divide their sum and their difference; and also the sum and the difference of any multiples of them. Example. Find the highest common factor of

4.03 – 3.x2 – 24x – 9 and 8x3 – 2x2 – 53x – 39.
2 4.x3 – 3.x2 – 24x - 9 8x3 – 2x2 - 53x - 39 2
4.x3 – 5x2 - 21x

8x3 – 6x2 - 48.6 - 18
2.c 2.x2 – 3x - 9

4.x2 – 5x – 21 2
2.x2 - 6x

4.x2 6x – 18
3
3.x - 9

X-3
3x – 9
Therefore the H. C.F. is x – 3.

Explanation. First arrange the given expressions according to descending or ascending powers of x. The expressions so arranged having their first terms of the same order, we take for divisor that whose highest power has the smaller coefficient. Arrange the work in parallel columns as above. When the first remainder 4x2 – 5x – 21 is made the divisor we put the quotient x to the left of the dividend. Again, when the second remainder 2x2 – 3x – 9 is in turn made the divisor, the quotient 2 is placed to the right; and so on. As in Arithmetic, the last divisor 2 – 3 is the highest common factor required.

98. This method is only useful to determine the compound factor of the highest common factor. Simple factors of the given

expressions must be first removed from them, and the highest common factor of these, if any, must be observed and multiplied into the compound factor given by the rule. Example. Find the highest common factor of

24.x4 – 203 – 60x2 – 32x and 18x4 – 6.03 – 39x2 – 18.0. We have 24x4 – 2x3 – 60x2 – 32x=2x (12xco – x2 – 30x – 16), and 18.04 – 6.03 – 39x2 – 18x=3x (6x3 – 2x2 – 13x – 6).

Also 2x and 3.c have the common factor x. Removing the simple factors 2.0 and 3x, and reserving their common factor x, we continue as in Art. 97.

2.C 6.03 - 2.x2 – 13.x - 6 12.03 – x? - 30x - 162
6.x3 - 8.ro - 8x

12.x3 - 4x2 – 26x - 12
2 6.ca 5.x - 6

3.x2 - 4x – 4 * 6x2 – 8x - 8

3.x2 + 220
3.0 +2

6.2 4
6x 4

2

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Therefore the H. C.F. is x (3x + 2).

99. So far the process of Arithmetic has been found exactly applicable to the algebraical expressions we have considered. But in many cases certain modifications of the arithmetical method will be found necessary.

These will be more clearly understood if it is remembered that, at every stage of the work, the remainder must contain as a factor of itself the highest common factor we are seeking. [See Art. 97, I & II.] Example 1. Find the highest common factor of

3x3 – 13x2 + 23x - 21 and 6.03 + x2 – 44x + 21.

3.x3 – 13.x2 +23.x – 21 | 6x3 + Xo - 44x + 21 2

6x3 – 26x” +46x – 42

27x2 – 90x + 03

Here on making 27x2 – 90x+63 a divisor, we find that it is not contained in 3x3 – 13x2 +23x – 21 with an integral quotient. But noticing that 27x2 – 90x+63 may be written in the form 9 (3x2 – 10x + 7), and also bearing in mind that every remainder in the course of the work contains the H. C.F., we conclude that the H. C.F. we are seeking is contained in 9 (3.02 – 10x + 7). But the two original expressions have no simple factors, therefore their H. C. F. can have none. We may therefore reject the factor 9 and go on with divisor 3.02 - 10x + 7. Resuming the work, we have

3 3x3 – 13x2 +23.x - 21 3x2 – 10x + 7 a
3.x3 – 17x2 + 7%

3x2 73
-1
30° +16x -21

3.3 +7-1
3x2 + 10.x – 7

3x +7 2 )6.3 – 14

31 – 7 Therefore the highest common factor is 3x – 7.

The factor 2 has been removed on the same grounds as the factor 9 above. Example 2. Find the highest common factor of 2x3 + xo - 2 - 2

(1), and 3x3 – 2.02 + x - 2

(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:

2

6.x – 4.c? + 2x 43 7

3.r? - 3.x 6 2x 14.x+ 7x - 7x – 14

7x* + 5x + 2
14x3 - 10.x2 - 4x

17
173
17.x" – 3x - 14

- 119.0+85x + 34 -7
17x2 – 17.3

- 119.co +21x + 98
14
14x – 14

64 )64.x - 64
14x - 14
Therefore the H. C.F. is x - 1.

After the first division the factor 7 is introduced because the first remainder – 7.x2 + 5x +2 will not divide 2.03 + –

- 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.

NOTE. Here the highest common factor might have been more easily obtained by arranging the expressions in ascending powers of x. In this case it will be found that there is no need to introduce a numerical factor in the course of the work.

100. 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.

2003 +

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6.x3 +

-6, then

101. Let the two expressions in Example 2, Art. 99, be written in the form

2.2.2+x2–3-2=(x-1) (2x +3.c +2),

3.x3 – 2.x2 + x -2=(x - 1) (3x2+x+2). Then their highest common factor is x – 1, and therefore 2.x2 + 3.c +2 and 3x2 +x+2 have no algebraical common divisor. If, however, we put

2.23 +22-2-2=460, and

3.03 – 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 XII. C.

Find the highest common factor of the following expressions: 1. 23+2x2 – 13x + 10, 203+x2 – 10x +8. 2.2.3 – 5x2 – 99%+40, 203 – 6x2 – 86x+35. 3. 23 + 2.x2 – 8x -- 16, 23 +382 – 8x – 24. 4. 23+4x2 – 5.0 – 20, 23+6x2 – 5x – 30. 5. 203 – 22 – 5x – 3, 03 — 4x2 – 11x – 6. 6. 23+3.x2 – 8x – 24, 23 +3.x2 – 3x – 9. 7. a? – 5a-x+ 7ax2 – 3x), a3 – 3ax2 + 2x?. 8. 2* — 2.2-3 – 4.0 – 7, 2:4 + 203 – 3.x2 – x+2. 9. 2.23 – 522 +11.0+7, 4.x3 – 11x2+25x+7. 10. 2.2-3 +4.x2 – 7x – 14, 6.2.3 – 10.x2 – 21x+35. 11. 3.74 – 3.2.3 – 2.2-2 – X – 1, 9.64 – 3.2-3 – X – 1. 12. 2.04 – 2.x3 + x2 + 3x – 6, 4.xt – 2.2:3 + 3x – 9.

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