and the highest common factor of these, if any, must be
observed and multiplied into the compound factor given by
the rule.
Ex. Find the highest common factor of 24x4 — 2x3 — 60x2 — 32x and 18 a4 — 6x3 — 39x2 — 18x.
We have 24x4 — 2x3 — 60x2 — 32x = 2x(12 x3 — x2 — 30x — 16), and 18x4 — 6x3 — 39x2 — 18x =3x(6 x3 — 2x - 13x -6).
Also 2x and 3x have the common factor x. Removing the simple factors 2x and 3x, and reserving their common factor x, we continue as in Art. 116.
2x 6x3 — 2x2 -13x—6 12x3 — x2 - 30x — 16 2
6x3 — 8x2 — 8x 12 x3 —4x2 — 26x — 12 2 6x2— 5x—6 8x2— 4x - 4 x 6x2— 8x - 8 3x2 + 2x 3x + 2 — 6x - 4 -2 — 6x - 4
Therefore the H. C. F. is x(3x + 2).
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 3x3 — 13x2 + 23x — 21 and 6x3 + x2 — 44x + 21.
3x3 — 13x2 + 23x — 21 6x3 — x2— 44x + 21 6x3 — 26x2 + 46x — 42 27x2 — 90x + 63
Here on making 27 x2 — 90x + 63x divisor, we find that it is not contained in 3x3 — 13x2 + 23x — 21 with an integral quotient. But noticing that 27 x2— 90 x+ 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-
