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. 121.)

116. We begin by working out examples illustrative of the algebraic 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 contains 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.

Ex. Find the highest common factor of

4x3-3x2 - 24 x 9 and 8x8 - 2 x2 - 53 x - 39.

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 2 x2 - 3x9 is in turn made the divisor, the quotient 2 is placed to the right; and so on. As in Arith metic, the last divisor x 3 is the highest common factor required.

117. 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.

and

Ex. Find the highest common factor of

24x4-2x3- 60x2 32 x and 18x4 6 x8-39 x2-18x.

We have 24 x4 - 2 x3 – 60 x2 - 32 x = 2 x (12 x3 — x2 – 30 x ~ 16), 18x46x3-39 x2-18x= 3 x(6 x3- 2x2 - 13 x − 6).

Also 2x and 3x have the common factor x. Removing the simple factors 2x and 3 x, and reserving their common factor x, we continue as in Art. 116.

118. So far the process of Arithmetic has been found exactly applicable to the algebraic 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. 116, I & II.]

Ex. 1. Find the highest common factor of

3x8-13x2+ 23 x

21 and 6x + x2 − 44 x + 21.

3x8-13x2+ 23x-21 | 6x+ x2-44x+21 |2

Here on making 27 x2 - 90 x + 63 a divisor, we find that it is not contained in 3 x3- 13x2 + 23 x 21 with an integral quotient. But noticing that 27 x2 −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(3x2 - 10x+7). But the two original ex

pressions have no simple factors, therefore their H. C. F. can have none. We may therefore reject the factor 9 and go on with divisor 3x2 - 10 x + 7. Resuming the work, we have

Therefore the highest common factor is 3x-7.

The factor 2 has been removed on the same grounds as the factor

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 — 7 x2 + 5 x + 2 will not divide 2 x3 + 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 Ex. 1.

From these 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.

H

NOTE. If, in Ex. 2, the expressions had been arranged in ascend ng powers of x, it would have been found unnecessary to introduce a numerical factor in the course of the work.

119. The use of the Factor Theorem (Art. 105) often lessens, in a very marked degree, the work of finding the highest common factor. Thus in Ex. 2 of the preceding article it is easily seen that both expressions become equal to 0 when 1 is substituted for x, hence a 1 is a factor. Dividing the first of the given expressions by x--1, we obtain a quotient 22+3x+2. It is evident that this will not divide the second expression, hence x-1 is the H. C. F.

120. When the Method of Division by Detached Coefficients (Art. 63) is employed in finding the H. C. F., the following is a convenient arrangement.

Ex. Find the H. C. F. of

x + 3x3 + 12 x 16 and 23 13 x + 12.

We write the literal factors of the dividend until we reach a term of the same degree as the first term of the divisor.

The addition of the terms in the third column gives 13 x2, which is of lower degree than the first term of the divisor, hence we can proceed no further with the division and have for a remainder 13x2 + 39 x 52. Removing from this remainder the factor 13, as it is not a factor of the given expressions, we have for a second divisor x2+3x-4. The first divisor, as written before the signs wer changed, forms the second dividend:

since there is no remainder, the last divisor, as written before the signs were changed, is the H. C. F. Thus x2+3x-4 is the H. C. F.

121. Let the two expressions in Ex. 2, Art. 118, be written in the form

-

2x2+x2-x-2 = (x − 1) (2 x2+3x+2), 3x3-2x2+x-2= (x − 1)(3x2+x+2).

1, and therefore

Then their highest common factor is x 2x2+3x+2 and 3x2+x+2 have no algebraic common divisor. If, however, we put x=6, then

and the greatest common measure of 460 and 580 is 20; whereas 5 is the numerical value x-1, the algebraic highest common factor. Thus the numerical values of the algebraic 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 algebraic 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 algebraic quantities.

EXAMPLES XI. c.

Find the highest common factor of the following expressions:

1. x3 + 2x2 – 13x + 10, x3 + x2.

10x+8.

2. x3- 5 x2 - 99x + 40, x3 — 6 x2 - 86 x + 35.

3. x32x2- 8x-16, x3 + 3x2 - 8 x — 24.

7. 2x3- 5x2 + 11x+7, 4x3 – 11 x2 + 25 x + 7.

8. 2 x3 + 4x2 - 7 x — 14, 6 x3 - 10 x2 - 21 x + 35.