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EXAMPLE. Multiply 3ac+2bd+5c2, by 4ab.

Multiplicand,

3ac+2bd+5c2.

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ac+3a2b-ac2, by 4ab2.

Ans. 2a2b2c+2a3b3-3a2b2c2.

3. Multiply
4. Multiply a2+‡ab+b2, by 3ab.

Ans. 2ab+4a2b2+ab3.

5. Multiply 7a2c2—5bc2—4ab, by abc.

Ans. 51a3bc3-33ab2c3—3a2b2c. 28. CASE III. When both multiplicand and multiplier are compound quantities.

RULE. Multiply all the terms of the multiplicand by each term of the multiplier, and collect the several products into one sum by addition.

EXAMPLE. Multiply 3a2-6ab+362, by a-b.

Product by b,

Total product,

Multiplicand,

3a2-6ab+362.

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Product by α,

3a3-6a2b+3ab2.

-3a2b+6ab2-363. 3a3_9a2b+9ab2—3b3.

Ans. a2+2ab+b2.
Ans. a2-2ab+b2.

Ans. a2-b2

Ans. a3-b3.

Ans. a3+x3.

1. Multiply a+b, by a+b.

2. Multiply a-b, by a-b.
3. Multiply a+b, by a—b.
4. Multiply a2+ab+b2, by a-b.
5. Multiply a-ax+x2, by a+x.
6. Multiply 7ab+3ac+4d, by 3a-2c.

Ans. 21a2b+9a2c+12ad-14abc-6ac2-8cd.

7. Multiply 5a+3x+y, by 3a+2x.

Ans. 15a2+19ax+3ay+6x2+2xy.

8. Multiply x+4y-2, by 4x+3y.

Ans. 4x2+3xy-x+14y2—бy.
Ans. 4y.

2

9. Multiply 23+x2y+xy2+y3, by x-y. 10. Multiply x4x3+x2-x+1, by x2+x-1. Ans. 64x3-x2+2x-1. 11. Multiply ax + bx2+cx3, by 1+x+x2+x3. Ans. ax+(a+b) x2+(a+b+c)x3 +(a+b+c)x++(b+c) x5+cx6.

29. THEOREM I. The square of the sum of two quantities is equal to the sum of the squares of each of the quan

tities, together with twice their product. (See Art. 28, Example 1.)

30. THEOREM II. The square of the difference of two quantities is less than the sum of their squares by twice their product. (See Art. 28, Example 2.)

31. THEOREM III. The product of the sum and difference of two quantities is equal to the difference of their squares. (See Art. 28, Example 3.)

NOTE. The above theorems are very important, and should be committed to memory.

12. Write by Theorem I. the squares of the following quantities, and verify the results by multiplication, (x+y)2, (2a+x)2, (a+2b)2, (a2+x2)2, (ac+x)2, (3x+4y)2.

Ans. (x2+2xy+y2), (4a2+4ax+x2), (a2+4ab+462), (a1+2a2x2+x1), (a2c2+2acx+x2), (9x2+24xy+16y2).

13. Write by Theorem II. the squares of the following quantities, and verify the results by multiplication, (a—x)2, (2a-b)2, (3x-2y)2, (a2—b2)2, (ac—y)2, (3a2—4c)2.

Ans. (a2-2ax+x2), (4a2-4ab+b2), (9x2-12xy+4y2), (a1—2a2b2+b1), (a2c2—2acy+y2), (9a1—24a2c+16c2).

14. Write by Theorem III. the product of the following quantities, and verify the results by multiplication, (a+x) (a-x), (2a+x) (2a-x), (3x+2y)(3x-2y), (a2— b2)(a2+b2), (4a2-9c2) × (4a2+9c2).

Ans. (a2x2), (4a2x2), (9x2-4y2), (a1—b1), (16a-81c1).

15. Find the continued product of (a—x) (a+x) (a2+x2) (a1+x1). Ans. a8-x8. 16. Find the continued product of (2a+x) (2a—x) (4a2 +x2) (16α+x1). Ans. 256a8-x8. 17. Find the continued product of (x2+xy+ y2) (x-y) (23+33).

Ans. x6-y6. 18. Find the continued product of (x3-x2y+xy2 —y3) (x+y) (x+y). Ans. 10-xy+x+yб—y1o ̧

DIVISION.

32. DIVISION being the converse of multiplication, naturally gives the following Rules.

1st, If the signs of the dividend and divisor are like, the sign of the quotient is +; and if unlike, the sign of the quotient will be

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2d, Divide the coefficient of the dividend by that of the divisor, for the coefficient of the quotient.

3d, Since any quantity divided by itself gives 1 for quo

tient, and a quantity multiplied by 1 gives that same quantity for product, it follows, that the letters which are common to the divisor and dividend, with the same exponent in each, will not appear in the quotient.

4th, The letters which are in the dividend, and not in the divisor, will appear as multipliers in the quotient; while letters which are in the divisor, and not in the dividend, will appear as denominators of a fraction in the quotient.

5th, When the same letter is in both dividend and divisor, with different exponents, it will appear in the quotient with an exponent equal to the difference of its exponents, and in the denominator of a fraction, when the exponent of the divisor is the greater; thus, a6÷a1a2, as3÷a2=a3,

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33. CASE I. When the dividend and divisor are both simple quantities.

RULE. Place the dividend as the numerator of a fraction, and the divisor as its denominator, then divide by the above general rules.

EXAMPLE. Divide 14a2b3c, by 2a2bc.

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Ans. 13x2y

6. Divide 36x2y3z, by 20x2y2 34. CASE II. When the dividend is a compound, and the

divisor a simple quantity.

RULE. Divide each term of the dividend by the divisor, as in the last case, and the sum of the separate quotients will be the answer.

EXAMPLE. Divide 3a2b+4a3bc, by 4ab.

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3. Divide 48a1b2x-36a2b2x2+14ab3x4, by 2abx.

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a

Ans. 24a3b-18abx+762x3.

4. Divide 40p2x2y2—30p*x3y2—60p2x3y4, by —10p2xy. Ans. 4xy+3p2x2y +6x2y3.

5. Divide 7a12a3c-14ab2c, by 4a2c2.

Ans. — a3¿§—31§ç§—3‡a‡íž.

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6. Divide 4a2c2-8a2cx+4a2x2, by 4a2.

Ans. c2-2cx+x2.

35. CASE III. When both dividend and divisor are compound quantities.

RULE. Place the quantities as in division in arithmetic, arranging both dividend and divisor according to the powers of the same letter. Divide the first term of the dividend by the first term of the divisor, and put the result with its proper sign for the first term of the quotient. Multiply the terms of the divisor by this quotient, and subtract the product from the dividend; to the remainder bring down as many terms of the dividend as may be necessary, then divide as before, and so on till the work is finished.

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a+x)a3+3a2x+3ax2+x3 (a2+2ax+x2

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ax2+x3

Dividend.

Quotient.

a2-2ax+x2)a1-4a3x+6a2x2-4ax3 +x1 (a2—2ax+x2

a4-2α3x+a2x2

-2a3x+5a2x2-4ax3
-2a5x+4a2x2-2ax3

a2x2-2ax3 + x 1
a2x2-2ax + x +

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36. Sometimes the quotient never terminates, but be carried out in an infinite series.

Divisor. Dividend.

Quotient.
(1+x+x2+x3+x1+ &c.

may

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Here it is evident the quotient would never terminate, but go on to infinity, the power of x increasing by unity at each step.

This quotient gives several interesting series, by substituting fractional values for x; for example, if we make x successively equal to 1, 3, 4, 3, 2, we will have the following results :

(1.)

(2.)

(3.)

(4.)

(5.)

2

3

=2=1+1+1++, &c. to infinity.

3

=1+3+}+27+81 + &c. to infin.

4— = = 1 + 1 + 16 +81 +236 + &c. to infin.

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=4=1+2+18+1++ &c. to infin.

3

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15

(6.) 4th-2nd gives =}+&+,4+}},+&c. to infin.

2

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(7.) 5th-3rd gives =}+&+++ &c. to inf.

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