Page:Elementary algebra (1896).djvu/115

97 Rh 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—10x+7. Resuming the work, we have

x| 3x3 — 13x2 + 23x —21 | 3x2-10x+7|a 3x3 —10x2+ 7x 3x2 — 7x -1 — 3x2+ 16x - 21 3x + 7 -1 — 3x2 + 10x— 7 — 3x + 7 2 6x—14 3x— 7

Therefore the highest common factor is 3x — 7.

The factor 2 has been removed on the same grounds as the factor 9 above.

Ex. 2. Find the highest common factor of 2x3 + x2 - x - 2 (1), and 3x3 - 2x2 + 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:

2x3 + x2 - x - 2 6x3 - 4x2 + 2x - 4 7 6x3 - 3x2 - 3x - 6 —2x|14x3 + 7x2 - 7a—14 7x2+ 5x+ 2 14x3 —10x- 4x 17 17x2 - 3x—14 — 119x2 + 85x+ 34 —7 17x2 — 17x — 119x2+ 21x +98 14 14x —14 64)64 x — 64 14x—14 x- 1

Therefore the H.C.F. is x — 1.

After the first division the factor 7 is introduced because the first remainder — 7x2+ 5x-42 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 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.