(1.) Resolve aux2-c x2 into its component factors. Here a x2+b x2—c x2=x2 (a+b—c).

(2.) Transform the expression n3+2 n2+n into factors Here n3+2n2+n = n (n2+2n+1)

= n (n+1) (n+1) by (6)

= n (n+1)2.

(3.) Decompose the expression x2-x-72 into two factors.

By inspecting formula (3) we have —1——9+8, and — 72 = —9×8; hence x2-x-72 = (x—9) (x+8).

(4.) Decompose 5a2bc+10ab2c+15a b c2 into two factors. (5.) Transform 3m1 no—6m3 n3 p+3m2 n1 p2 into factors.

(6.) Transform 363 c—36 c3 into factors.

(7.) Decompose x2+8x+15 into two factors.

(8.) Decompose a3-2x2-15x into three factors. (9.) Decompose x2-x-30 into factors.

(10.) Transform a2—b2+2b c—c2 into two factors. (11.) Transform a2x-x3 into factors.

23. By the usual process of division we might obtain the quotient of a"-b1 jivided by a—b, when any particular number is substituted for n; but we sal here prove generally that a"-b" is always exactly divisible by a-b. and exhibit the quotient.

It is required to divide a"-b" by a—b.

Now it appears from this result, that a"-b" wili be exactly divisible by a—b, if a11—b" be divisible by a-b; that is, if the difference of the same powers of two quantities is divisible by their difference; then the difference of the powers of the next higher degree is also divisible by that difference.

But a-b2 is exactly divisible by a-b, and we have

And since a2-b2 is divisible by a-b, it appears from what has been just proved, that a3—63 must be exactly divisible by a-b; and hence, by putting 8 for n in formula (1), we get

Again, a1—b1 must be exactly divisible by a-b, since a3-b3 is divisible by a-b; hence, by writing 4 for n in formula (1), we have

a3-b3
a1 -b

Hence, generally, a"-b" will always be exactly divisible by a-b, and give

In a similar manner we find, when n is an odd number,

By substituting particular numbers for n, in the formulas (5), (6), (7), we may deduce various algebraical formulas, several of which will be found in the following deductions from the rules of multiplication and division.

USEFUL ALGEBRAIC FORMULAS.

(1.) a2—b2 = (a +b) (a−b).

(2.) a' b1 = (a2+b2) (a2—b2) = (a2+b2) (a+b) (a—b). (3.) a3-b3 = (a2+ab+b2) (a-b).

(4.) a3+b3 = (a2—ab+b2) (a+b).

(5.) a b = (a3+b3) (a3—b3) = (a3+b3) (a2+ab+b2) (a−b). (6.) a b = (a3+b3) (a3—b3) = (a3—b3) (a2—ab+b2) (a+b). (7.) a®—b® = (a3+b3) (a3—b3) = (a2—b2) (a1+a2 b2+b1).

(8.) a®—b® = (a+b) (a−b ) (a2+ab+b2) (a2—a𷆂‚2 (9.) (a-b2) + (a−b) = a+b.

(10.) (a-b3) + (a−b) = a2+ab+b2.

(11.) (a3+b3)+(a+b) = a2—ab+b2.

(12.) (a-b') + (a+b) = a3—a2b+a b2-b3.

(13.) (a3—b3) ÷ (a—b) = aa+a3b+a2 b2+a b3+b1.

(14.) (a3+b3) ÷ (a+b) = a^—a3b+a2b2—a b3+bʻ.

(15.) (a—b) + (a2—b2)=a1+a2b2+b1.

DIVISION BY DETACHED Coefficients.

24. Arrange the terms of the divisor and dividend according to the successive powers of the letter or letters common to both; write down simply the coefficients with their respective signs, supplying the coefficients of the absent terms with zeros, and proceed as usual. Divide the highest power of the omitted letters in the dividend by that of the suppressed letters in the divisor, and the quotient will give the literal part of the first term in the quotient. The literal parts of the successive terms follow the same law of increase or decrease as those in the dividend. The coefficients prefixed to the literal parts will give the complete quotient, omitting those terms whose coefficients

are zero.

EXAMPLES.

(1.) Divide 6a1-96 by 3a-6.

3−6) 6+ 0+ 0+0−96 (2+4+8+16

6-12

12

12-24

24

24-48

48-96
48-96

But a1÷a=a3, and the literal parts of the successive terms are therefore a3, a2, a1, ao, or a3, a2, a, 1; hence, 2a3+4a2+Sa+i6=quotient.

(2.) Divide 8a-4a1x—2a3 x2+a2 x3 by 4a2—x2.
4+0—1) 8—4—2+1 (2−1

8+0-2

−4+0+1
-4-0+1

Now, a3÷a2=a3; hence a3 and a2r are the literal parts of the terms in the quotient, for there are only two coefficients in the quotient; therefore 2a3-a2 x=quotient required.

(8.) Divide a1-3a x3-8a2x2+18a3 x-8a1 by x2+2a x—2a3. (4.) Divide 3y+3x y2—4x2 y—4x3 by x+y.

(5.) Divide 10a1—27a3 x+34a2 x2-18a x3-8x1 by 2a2—3a x+4x2 (6.) Divide a®+4a3—8a*—25a3+35a2+21a−28 by a2+5a+4.

ANSWERS.

(3.) x2-5a x+4a*
(4.) -4x2+3y2.

(5.) 5a2-6a x—2x2.

(6.) a1—a3—7a2+14a−7,

SYNTHETIC DIVISION.

25. In the common method of division, the several terms in the divisor are multiplied by the first term in the quotient, and the product subtracted from the dividend; but subtraction is performed by changing all the signs of the quantities to be subtracted, and then adding the several terms in the lower line to the similar terms in the higher. If, therefore, the signs of the terms in the divisor were changed, we should have to add the product of the divisor and quotient instead of subtracting it. And since the process would be the same for every step in the operation, the successive products of the divisor and the several terms in the quotient would all become additive. By this process, then, the second dividend would be identically the same as by the usual method; but the second term in the quotient is found by dividing the first term of the second dividend by the first term of the divisor; and since the sign of the first term in the divisor has been changed, it is obvious that the sign of the second term in the quotient will also be changed. To avoid this change of sign in the quotient, the sign of the first term in the divisor might remain unchanged, and then omit altogether the products of the first term in the divisor by the successive terms in the quotient; because in the usual method the first term in each successive dividend is cancelled by these products. Omitting, therefore, these products, the coefficients of the first term in any dividend will be the coefficient of the succeeding term in the quotient, the coefficient in the first term of the divisor being unity; for in all cases it can be made unity, by dividing both divisor and dividend by the coefficient of the first term in the divisor. This being the case, the coefficients in the quotient are respectively the coefficients of the first terms in the successive dividends. The operation, thus simplified, may however be further abridged by omitting the successive additions, except so much only as is necessary to show the first term in each dividend, which, as before remarked, is also the coefficient of the succeeding term in the quotient, and writing the products of the modified divisor, and the several terms of the quotient as they arise, diagonally, instead of horizontally, beginning at the upper line. Hence the following

RULE.*

(1). Divide the divisor and dividend by the coefficient of the first term in the divisor, which will make the leading coefficient of the divisor unity, and the first term of the quotient will be identical with that of the dividend.

(2). Change all the signs of the terms in the divisor, except the first, and multiply all the terms so changed by the term in the quotient, and place the products successively under the corresponding terms of the dividend, in a diagonal column, beginning at the upper line.

(3). Add the results in the second column, which will give the second term of the quotient; and multiply the changed terms in the divisor by this result, placing the products in a diagonal series, as before.

• The rule here given for Synthetic Division is due to the late W. G. Horner, Esq., of Bath, whose researches in science have issued in several elegant and useful processes, especially in the higher branches of algebra, and in the evolution of the roots of equation of all dimensions.

(4). Add the results in the third column, which will give the next term in the quotient, and multiply the changed terms in the divisor by this term in the quotient, placing the products as before.

(5). This process continued till the results become 0, or till the quotient is determined as far as necessary, will give the same series of terms as the usual mode of division when carried to an equivalent extent.

EXAMPLES.

(1.) Divide a3-5a1x+10a3 x2—10a2x2+5a x1—x3 by a2—2a x+x2.

In this example the coefficients of the dividend are written horizontally, and those of the divisor vertically, with all the signs changed, except the first. Then + 2 and - 1, the changed terms in the divisor, are multiplied by 1, the first term of the dividend or quotient; and the products + 2 and 1 are placed diagonally, under — 5 and +10, the corresponding terms of the dividend. Then by adding the second column we have - 3 for the second term in the quotient, and the changed terms + 2 and 1 in the divisor, multiplied by—3, give — 6 and +3, which are placed diagonally under + 10 and sum of the third column is + 3, the next term in the quotient, which multiplied into the changed terms of the divisor, gives +6—3, for the next diagonal column. The sum of the fourth column is — 1, and by this we obtain the last diagonal column-2+1. The process here terminates, since the sums of the fifth and sixth columns are zero; and the quotient is completed by restoring the letters, as in detached coefficients.

-

- 10. The

Having made the coefficient of the first term in the divisor unity, that coefficient may be omitted entirely, since it is of no use whatever in continuing the operation here described.

(2.) Divide x-5x+15x1—24x3+27x2—13x+5 by x1—2x3+4x2-2x+1. 1-5+15-24+27—13+5

1

(3.) Divide a3+2a1 b+3a3 b2—a2 b3—2a b1—3b3 by a2+2ab+3b2.

-2
-3

1+2+3-1-2-3

-2+0+0+2

-3+0+0+8

1+0+0-1

Hence a3+0a2 b+0⋅a b2-ba-b' quotient.

(4.) Divide 1-x by 1+x.

(5.) Divide 1 by 1-x.

(6.) Divide x7-yỉ by x—y.

Ans. 1-2x+2x2—2x3+, &c.
Ans. 1+x+x2+x3+, &c.

Ans. x+y+x1y2+x3y2+x2y*+xy3+y

(7.) Divide ao—3aa x2+3a2 x1—xo by a3—3a2x+3a xo—x3.

Ans. a3+3a2x+3a x2+x13