(5.) 10+2d2 + 2e3) ÷ (4a-4c−2b)—(6 + 4d-26) (6d-4c.) Ans. 411. (8ab2-c2-d2)÷ (3e2-d)+12cd-8de.

(6.)

28819.

Ans.

(1.) √b2 + √u2 ÷√d2 +2√d2÷ 4√a2 + 6√⁄a2 ÷ 362 Ans. 41

(2.) 5a85-3c4÷6+b÷c÷d+d2 ÷ 62. Ans.

92232,182

(3.) (a+b)+(d+5d)+d÷2b++4d+2cd+db

Ans. 841

4

=

When a 50, b=32, c=16, d=8, e=2, ƒ=0.

These are employed to include several terms which may sometimes be conveniently considered as one

quantity; if a positive sign precede the bracket it may be removed without affecting the value of the included terms, but if this be preceded by a negative sign the whole included quantity is negative, and therefore if the brackets be removed the sign of all the terms must be changed.

When a=4, b=6, c=2, d=8, =e0.

Find the value of

(1.) a (c+d-b)+a+2(c+d—b)+2+ad(b+c.)

Ans. 286.

(2.) 3 (a+bc-d)+ab+(c2d-a3de)—2a (b—a.)

Ans. 64.

(3.) ab÷2(c2+d+e2) — a2 (b2-d+e) ÷ 2 +abc

Ans. 64.

Áns. 12.

2(a3b-d—a2c)÷ad+3b(a+5cd-a3d)÷bab.

(a+c)÷2.

(4.)

(5.) {3a

3a + 2c
2c―d}+2+c− (2b+c―d}÷2

(6.) {1+(d ̄a)+2a} − {ɓa—;

26

Ans. 1.

6a-3b+1

}

(8.) {(a−b) — (c–d)} + {(e−d2) (b+c−2d)}

Ans. 528.

Such expressions as a+b, consisting of two terms are called Binomial, if the quantity consist of three terms it is called Trinomial as a+b+d, and if there are more than three we have a Multinomial quantity as a+b+d-f.

ADDITION.

RULE. Arrange the terms so as to have all like terms under each other, add the co-efficients or integers standing before the positive terms and do the same for the negative terms, putting down the proper sign.

(x+y) (xy)=-32 and conversely 2-y2=(x+y)

(x-y). (x+y) (x+2)= x2 + (y + z) x+yz and conversely x2+(y+z) x+yz=(x+y) (x+z)

Applying the last axioms to an illustration, x2 + 7x +16, we deduce that this=2 + (y + z) x+yz=(x+y) (x+z)=x2 (4+4)=x+4×4=(x+4) (x+4).

Resolve into factors

(1.) a2+2ab+b2. Ans. (a+b) (a+b).
(2.) a2-2ab+b2. Ans. (a-b) (a-b).
(3.) x2+3xy+2y. Ans. (2x+y) (x+2y.)
(4.) x3+x2y+xy2+y3. Ans. (x2+y2) (x+y).
(5.) -y Ans. (x+y) (xy) (x+y).
(6.) 3x3—8x2y—3xy2. Ans. (x2—3xy) (3x+y).

GREATEST COMMON MEASURE.-G.C.M.
DEFINITION.-See Arithmetic.

RULE. - Proceed as in ordinary arithmetic, or where possible resolve the quantities of which the G.C.M. is required into their factors; thus find the G.C.M. of -5æ3y2z and 10x1y3z2.

=

and of these quantities com 10x4y3z2=5x3y2z × 2xyz. 5x3y2z is common. or 5x3 y2z)10x4y3z2(2xyz The last divisor in this

10x4y3z2

case, being the first, forming the G.C.M.

a2+2a+1)a+2a2+2a+1(a
a3+2u2+a

a2+2a+1=(a+1)+(a+1)

a+1)a2+2a+1(a+1 a2+a

a+1

a +1

a3 2x2+2x+1=(a×1) (a2+2a+1)

so the G.C.M. is a +1.

(1.) 3x2y2+6x2y3 and 12xy2+9x2y2. Ans. 3xy3. (2.) x3+3x2+3x+1 and æ3 +1.

Ans. +1.

(3.) 2a2-a-1 and 6a2—a—2. Ans. 2a+1.

(14.) 2ab-bc-cd+3cb+4c-ba+cd.

Ans. ab+2bc+4cd.

(15.) 4a472-3a2y2—4b2x2 + 9u2y2 — 6a1y2 — 3b2x2. Ans. -2a4y2+6a2y2 —7b2x2

SUBTRACTION.

RULE.-With terms having unlike signs change the lower sign and add; with terms having like signs put down the difference with the proper sign.

(1.) From x+4x3y2 — 7x2y2 — 17xy4—12z5 take xa1+ 2x3y2-3x2y2-8xy+3z5.

Ans. 2a3y2-4x22y2 — 9xy4—15z5. (2.) From a2x-bx2+cy3 take 2a2x-3bx2+2cy3.

Ans.-a-x-2bx2-cy3.

(3.) From y3+2ay2—3by―r take y3 — ay2+by—r. Ans. Say2-4by.

(4.) From 6a1y2—3a2y2-8xy take 3a2y2—7xy+2a4y2 Ans. 4a+y2-6a2y2-2y.

(5.) From x+4x3y2—7x2y2—17xy1—12z3 take 2x3y2

Ans. x+2x3y2-3x2y2—8xy1+3z5.

(6.) From 3a2x-bx2+cy3 take a2x+2b.x2- cy3.

Ans. 2a2x-362x+2cy3.

(7.) From y3+2ay-3by-r take 3ay2-4by. Ans. 33—ay2+by—r.

(8.) From 6a+y2-3a2y2-Sxy take 4a1y2—6a2y2—xy.

Ans. 3ay2-7xy+2a4y2.