continued to work till the whole was finished; the first received twice as much as the last: find the time occupied, each man's wages being proportioned to the time that he worked.

(19.) The rent-charge of a farm consists of a certain number of quarters of wheat and of rye; now, when the wheat is at 568. the quarter and rye at 248., the values of the wheat and the rye are the same; but when the wheat is at 608. and the rye at 258., the rent is increased by £9: find the amount of the rent-charge.

(20.) A person has 80 gallons of wine in a cask, and wishes to draw off half of it, but not having a 40-gallon cask at hand, he does it by means of two other measures, A and B, which together hold 80 gallons, in the following way :

1st. He fills the larger measure (4) from the cask.

2nd. He fills the smaller measure (B) from the larger (A), and empties (B) back into the cask, and then pours what is left in (A) into (B).

3rd. He fills (4) from the cask, and then pours from (4) into (B) till (B) is full; half of the whole quantity of wine will then be found in (A): find the capacity of the two measures. (21.) Find a quantity such, that whether it be divided into three or four equal parts, the continued product of the parts shall be the same. (22.) A farmer sells to one person 6 horses and 3 cows for £180, and to another he sells 4 horses and 4 cows for £144: find the price of a horse and of a cow.

(23.) If from half the sum of the three sides of a triangle, each side be subtracted separately, the three remainders are 64, 53, and 4 yards respectively find the length of each side of the triangle.

(24.) A bill of 25 guineas was paid with crowns and half-guineas, and twice the number of half-guineas exceeded three times the number of crowns by 17; how many were there of each coin?

(25.) Two labourers, A and B, dig for two hours, after which A goes away, and B digs the remaining part of the work in 2h. 48m.; but if B had left at the end of the two hours, and A had continued digging, it would have taken him 4h. 40m. to finish the work: find in what time each would have done it separately.

(26.) A general finds that his cavalry with half his artillery and infantry together, or his artillery with one-third of his cavalry and infantry together, or his infantry with one-fourth of his cavalry and artillery together, make up the same number of men, viz. 5950: how many men were there in each arm of the service?

CHAPTER X.

ON THE RESOLUTION OF ALGEBRAICAL EXPRESSIONS INTO FACTORS.

75. In the articles of this chapter we shall discuss some of the most useful methods of resolving compound expressions into their factors.

BINOMIALS.

By Art. (50) it appears that any Binomial which expresses the difference of the squares of two quantities can always be resolved into two factors. Some examples will here be given in this formula of a more complicated kind than those in Chap. VIII. The method of dealing with such expressions will be seen from the examples worked out below.

=

(a2+x2) (a+x) (α − x)

a8-x8 = (a+x1) (a2+x2) (a+x) (a−x)

In this example the principle has been applied three times in succession; first to as-x8, then to a1-x4, then to a2-x2.

(x+ 1)2 - 1; (a+b)1-a2b2; (x+1)* −4; (œa−1)2− ( x − 1)2;

it follows that any Binomial which consists of the difference of the cubes of two quantities, or of the sum of the cubes of two quantities, can always be resolved into two factors; and these will be of the forms given above.

In this example, the two methods of Arts. 50 and 76 have been used.

Examples XXXV.

Resolve the following into all their factors:

26-64y6; 8ao+27b6; 125x3+1; 64a® — b®; 64ab3+27c3d®;

1

125

-64c12; 56ab1c-7abc4; 64a12+729612; a4b4cd+ abc4d4;

6

27

b213

216

(x+y)+(2x+3y)3; (a-1)+8; (1+x+x2)3-x5; 27(x+2)3-64; (x2+y2)3 — (3xy)3; (x2+4y2)6+(4x2y2)3 ;

77. Any Binomial consisting of the difference of any the same powers of two quantities can always be resolved into two factors at least. Also a Binomial consisting of the sum of any the same odd powers of two quantities can See Div. always be resolved into two factors at least. page (29).

Ex. 2.

(y-1)+&c.}

=

(1)

b

b4 b3 b2

+1 may be resolved.

(x+1)2+(y−1)7 = {(x+1)+(y−1) } { (x+y)® − (x+y)5

= (x+y){(x+y) - (x+y)3 (y-1)+&c.}

Examples XXXVI.

Similarly the following may be resolved :

*±1; a*b*_®; l±1; ±128; æ y10±10g; (a+b)+_ ab ;

- ცნ ;

bc5

(x2+2xy+y3)7± (x2 = 2xy+y3)7; (x2+1+1)*+ (~2-1+1)";

Note. The expression x3±1 is read "x to the fifth, plus or minus 1." 23+1 is first to be resolved into factors, and then 25 — 1.

78. By Multiplication we find that

(x+2xy+2y) (x2-2xy+2y2) = x2+4y1

Hence, any quantity of the form x+4y+ can be resolved into two factors, which will be of the same form as the two factors given above.

which reduces, by applying the same process as above, and simplifying

24

625a8+4a1b1; x1+64; 8la1+4b1;+4; (a+b)*+4;

(a+b)+4(a−b)4; (x2+xy)8+4 (y2+xy)8;

(a+b+c)*+4 (a+b−c)1; 256 (a2+b2)1+324a1b1 ;

(x2+3xy+2y2)*+ (x2 - 3xy+2y2)*.

F