When the divisor, as in this example, contains more than one term, it is generally convenient to follow the method of arithmetical long division. a + 5b) 3 a2 Thus— + 3 a2 + It will be seen that in the last two examples care has been taken to keep the terms of the divisor, dividend, and successive remainders arranged according to the ascending or descending powers of some letter. In these cases we have arranged the terms according to the descending powers of a, and, as therefore follows, according to the ascending powers of b. Want of care in this respect will often render the operation of finding the true quotient tiresome, if not impossible. The next two examples will illustrate this point. They may be attempted first by the student, keeping the terms in the order 10x + 2 10x + 2 -1 + y−1. Ex. 8. Divide x3 + y3 by x 2 Here we have the powers of x in the dividend descending, while in the divisor they are ascending. Arranging them in the divisor as in the dividend, the operation is easy. Thusy-1 +x-1)x2 + y3 (x3y - x2y2 + xy3 We shall work the next example in two ways to illustrate, firstly, the above point again; and, secondly, to show how the operation may be sometimes abbreviated by the use of brackets: Ex. 9. Divide x3 + y3 + 23 3 xyz by x + y + z. x + y + z)x3 − 3 xyz + y3 + z3(x2 − xy − xz + y2 − yz + x2 x3 + x2y + x2z -x2y-x2z-3 xyz -xyz-xz2 xy2 − xyz +xz2 + y3 + y3 + y2z Or thus, inclosing the last two terms of the divisor in brackets x + (y + z) )x3 − 3 xyz + y3 + ≈3(x2 − (y + z)x + (y2 = y≈ + x2) x2+(y+2) x2 It will be seen in both the above operations that we have brought down the terms of the dividend only when the subtrahends indicated they were required. This often prevents much useless repetition. Ex. VIII. Find the quotient of 1. 28 ab 7 ab2 + 14 63 by 7b; 3 x2y 3 xy. 2. 6 ab 15 ab- 20 ab3 by 3 ab; 4 x3y2 + 6 x2y3 5 х 3 and by 6 x + 4 y. y; 5. 1+ 2x + 3 x2 + 2 x3 + x1 by 1 + x + x2. 9: x 3 y; and by x+y. y; and x + y5 by x + y. 9. acx+" + adxy + bcx"y" + bdy" by cx" + dy". 10. ao + a3b2 + a2b3 + bo by a3 = a2b ab2 + b3. 11. b2 + ab + bc + ac by a + b ; 15. Show that the remainder, after the division of x1 – px3 + qx2 rx + s by x a, is a1 – pa3 + qa3 16. Divide x - y1 by x ̃3 x + x-1 y ra + s. -2 by x-2y-2, and x2 + x2+ 2 by 17. Show that the quotient of 1 by 1 + x, is 1 − x + x2 2+ &c. ad infinitum. 18. Show that the quotient of 1 by 1 - 3 x + 3 x2 - x3 is 1 + 3x + 6 x2 + 10 x2 + 15 x2 + &c. ad infinitum. — 2 (x + y)2 29. The ordinary method of finding the quotient of two algebraical quantities having been explained in the last article, we shall now proceed to show how, in certain cases, this method may be avoided, and the quotient written down at sight. It may be remarked at the outset, that the resolution of algebraical expressions into their elementary factors is a subject of very great importance, and one which the student will do well to thoroughly master. (I.) The form x2 + 2 ax + a2. = We have seen (Art. 26) that x2 + 2 ax + a2 (x + a)3. Hence the sum of the squares of two quantities, together with twice the product of the quantities, is equal to the square of the sum of the quantities. And hence, any algebraical expression, which can be thrown into the form (x2 + 2 ax + a2), is of necessity a perfect square. Thus 16 a2x2 - 56 abxy + 49 b2y2 = (4 ax)2 + 2 (4 ax) ( − 7 by) +(-7 by)2 = (4 ax- 7 by)2. = (a + b) (a - b). We have seen (Art. 26, II.) that a2 – b2 Hence the difference of the squares of two quantities is equal to the product of the sum and difference of the quantities. Thus (III.) The form x2 + px + q. - (c + d)} d). xy} This form evidently includes both the preceding, for the first form-viz., x2 + 2 ax + a2 is included, since q may be the square of half p; and the second is included-viz., x2-a2, since we may have p = = 0, and q a negative square quantity. Now, the resolution into elementary rational factors of the quantity x2 + px + q is not always possible; but, since (Art. 26, III.), =3 x2 + (a + b) x + ab (x + a) (x + b), we have the following rule, when the quantity admits of resolution. RULE. If the third term q of the quantity x2 + px + q can be broken up into two factors, a and b, such that the sum of these will give the coefficient of x, then the elementary factors of x2 + px + q are x + a and x + b. Thus, x+7x+12= (x+3)(x+4); for the product of 3 and 4 is 12, and their sum is 7, the coefficient of x. |