5.

From (ab) xy- (p + q) √ x + y — hx2
Take (2p-3q) (x + y)* —axy — (3 + h) x2

MULTIPLICATION.

CASE 1.

(19.) When both multiplicand and multiplier are simple quantities.

To the product of the coefficients annex the product of the letters, and it will be the whole product.

Thus, if it be required to multiply 6ax by 4b, we have 24 for the product of the coefficients, and abx for the product of the letters; consequently, 24abx is the whole product; that is, 6ax × 4b=24abx.

NOTE.-It must be particularly observed that quantities with like signs multiplied together, furnish a positive product whether the like signs be both + or both—; and that quantities with unlike signs furnish a negative product. This may be expressed in short by the precept that like signs multiplied together produce plus, and unlike signs minus. The truth of this may be shown as follows:

1. Suppose any positive quantity b, is to be multiplied by any other positive quantity a; then b is to be taken as many times as there are units in a, and, as the sum of any number of positive quantities must be positive, the sign of the product ab must be +.

2. Suppose now that one factor b is negative, and the other a positive: then, as before, the product of-b by a will be as many times-b as there are units in a; and, since the sum of any number of negative quantities must be negative, the product in this case must be ab.

3. If this last case be admitted, it will immediately follow that the product of-b and a must be ab, for if this be denied, the pro

+

duct must be ab, so that—b multiplied by a produces the

NOTE.-If powers of the same quantity are to be multiplied together, the operation is performed by simply adding the indices: thus, a2 × a3 = a5, for a2 = aaaaaaaaaa, or a5: also, am × an

=aa, and a3

aaa, therefore a2 × a3 =

=

am+n, for am = a × a

xa.... to m factors, and a" = axax a to n factors, and therefore am xa" = (aaa. to m factors) x (aaa to n factors), or (leaving out the sign ×):

....

=

to mn factors am+". From this it follows, that in the division of powers the indices are to be subtracted.* *

*This mode of proof does not apply when the quantities to be multiplied have fractional indices, although the rule still holds. Thus, let the product of a1 and a3 be required, then the exponents }, }, in a common denominator, are,; hence the proposed factors are the same as alo

6

and

alo; that is, the 5th power of the 10th root of a, and the 6th power of the same root; we have therefore, as above,

Although the product of the letters will be the same in value in whatever order we arrange them, yet in these examples the student, conformably to the usual custom, is expected to arrange them according to their order in the alphabet.

CASE II.

(20.) When the multiplicand is a compound quantity, and the multiplier a simple quantity.

Find the product of the multiplier and each term of the multiplicand separately, beginning at the left hand; connect these products by their proper signs, and the complete product will be exhibited.

(21.) When both multiplicand and multiplier are compound quantities.

Multiply each term in the multiplier by all the terms in the multiplicand.

Connect the several products by their proper signs, as in the last case, and their sum will be the whole product.

4. Find the four first terms in the product of
a+am-1x+am-2x2+am-3x2+&c. and a+x.
am+am-1x+am¬2x2+am¬3x3+&c.

6. Multiply a+b” by a—b.

Ans. an+1+ab” — a′′b —b”+1.

7. Multiply a+x+x2+x3+x4 by a-x.

Ans. a2+(a-1) x2 + (a−1) x2+(a—1)x1 — x5.

8. Multiply -x3+x2-x+1 by x2+x-1.

Ans. 26-24+x3-x2+2x-1.

9. Multiply 3x2+(x+y)*—7 by 2x2+√x+y.

Ans. 6x+(5x2—7) √ x+y—14x2+x+y.

10. Multiply ax + bx2+cx3 by 1+x+x2+x3.

(22.) When both dividend and divisor are simple quantities.

To the quotient of the coefficients annex the quotient of the letters, and it will be the whole quotient.

*

NOTE. The rule for the signs must be observed here, as well as in multiplication.

* The learner will readily discover the quotient of the letters by asking himself, what letters must I join to those in the divisor to make those in the dividend? Thus in ex. 3, next page, the dividend contains two x's, two y's, and two z's, while the divisor contains but one x and one z; so that to make up the letters in the dividend, I must join to those in the divisor one x, two y's, and one z; that is, the quotient of the letters will be xy2z.