same letter, the divisor and dividend should always be arranged (Art. 31) with reference to a certain letter.

The situation of the divisor in regard to the dividend, is a matter of arbitrary arrangement; by placing it on the right it is more easily multiplied by the respective terms of the quotient.

ART. 76. From the preceding we derive the following

RULE FOR THE DIVISION OF ONE POLYNOMIAL BY ANOTHER.— Arrange the dividend and divisor with reference to a certain letter, and place the divisor on the right of the dividend.

Divide the first term of the dividend by the first term of the divisor ; the result will be the first term of the quotient. Multiply the divisor by this term, and subtract the product from the dividend. Divide the first term of the remainder by the first term of the divisor; the result will be the second term of the quotient. Multiply the divisor by this term, and subtract the product from the last remainder. 'Proceed in the same manner, and if you obtain 0 for a remainder, the division is said to be exact.

REMARKS. 1st. When there are more than two terms in the quotient, it is not necessary to bring down any more terms of the remainder, at each successive subtraction, than have corresponding terms in the quantity to be subtracted.

2d. It is evident that the exact division of one polynomial by another will be impossible, when the first term of the arranged dividend is not exactly divisible by the first term of the arranged divisor; or when any remainder is not divisible by the first term of the divisor.

4. Divide 7x2y+5xy2+2x3+y3 by 3xy+x2+y2.

OPERATION.

2x3+7x2y+5xy2+y3 \x2+3xy+y2
2x2+6x2y+2xy2 2x+y Quotient.
x2y+3xy2+y3
x2y+3xy2+y3

In this example, neither divisor nor dividend being arranged with reference to either x or y, we arrange them with reference to x, and then proceed to perform the division.

5. Divide x2+x3—7xa+5x5 by x—x2. Division performed, by arranging both quantities according to the ascending powers of x. x2+x3-7x1+5x3|x—x2

Division performed, by arranging both quantities according to the descending powers of x.

5x-7x+x+x3|—x2+x

The learner will perceive that the two quotients are the same,

differently arranged.

EXAMPLES

FOR

PRACTICE.

but

6. Divide 6x2+5xy-4y2 by 3x+4y.

7. Divide x3-40x-63 by x-7.

Ans. 2x-y.

Ans. x2+7x+9.

8. Divide 3h5+16h1k—33h3k2+14h2k3 by h2+7hk.

Ans. 3h3-5h2k+2hk2.

9. Divide a5-243 by a-3. Ans. a4+3a3+9a2+27a+81. 10. Divide x-2a3x3+a® by x2-2ax+a2.

Ans. x+2ax+3a2x2+2a3x+a1.

11. Divide 1-6x55x6 by 1-2x+x2.

Ans. 1+2x+3x2+4x3+5x1.

12. Divide p2+pq+2pr-2q2+7qr-3r2 by p-q+3r.

Ans. p+2q-r.

13. Divide 4x+4x-x3 by 3x+2x2+2. Ans. 2x3-3x2+2x. 14. Divide x3-a6 by x3+2ax2+2a2x+a3.

Ans. x3-2ax2+2a2x-a3.

15. Divide m2+2mp-n2-2ng+p2-q2 by m-n+p-q.

16. Divide a+b+c3-3abc by a+b+c.

Ans. m+n+p+q.

Ans. a2+b2+c2-ab-ac-bc.

17. Divide

++x2y"+x3ym+ym+" by x"+y. Ans. xTM+y".

18. Divide ax3—(a2+b)x2+b2 by ax-b.

Ans. x2-ax-b.

19. Divide mpx3+(mq—np)x2—(mr+nq)x+nr by mx-n.

-2

Ans. px2+qx-r.

Ans. am-2c".

Ans. x3-x-3.

20. Divide a2m-3amcn2c2 by a"-c". 21. Divide x1+x ̄a—x2—x2 by x—x ̄1. 22. Divide a+a®b2+a1b1+a2b¤+b2 by a1+a3b+a2b2+ab3+b1. Ans. a-a3ba2b2—ab3+ba.

23. Divide a2+(a−1)x2+(a−1)x3+(a−1)x1—x3 by a-x.

24. Divide x(x-1)a3+(x3+2x—2)a2+(3x2—x3)a—x1 by a2x Ans. (x-1)a+x2.

+2a-x2.

25. Divide x3-8y3+125z3+30xyz by x-2y+52.

Ans. x2+2xy-5xz+4y2+10yz+25x2.

26. Divide 1-9x3-8x9 by 1+2x+x2.

Ans. 1-2x+3x2-4x3+5x1—6x5+7x6—8x7.

27. Divide 1+2x by 1-3x to 5 terms in the quotient.

Ans. 1+5x+15x2+45x+135x+&c. 28. Divide 1-3x-2x2 by 1-4x to 6 terms in the quotient. Ans. 1+x+2x2+8x3+32x2+128x3+&c.

DIVISION BY DETACHED COEFFICIENTS.

ART. 77. From Art. 62, it is evident that division sometimes may be conveniently performed, by operating on the coëfficients detached from the letters, and afterwards supplying the letters. Thus, if it be required to divide x2+2xy+y2 by x+y, we may perform the operation as follows:

1+2+1 1+1

Hence the coefficients of the quotient

1+1 1+1 are 1 and 1. Also, x2-x=x, and y2÷y=y; therefore the quotient is 1x+1y, or x+y.

+1+1

1+1

2. Divide 12a4-26a3b-8a2b2+10ab3-8b4 by 3a2-2ab+b2.

When any of the intermediate powers of the letters are wanting, the coëfficients of the corresponding terms must be supplied with zero, as in the following example.

4. Divide 6x+4x3y-9x2y2-3xy3+2y' by 2x2+2xy-y2.

Ans. 3x2-xy-2y2.

5. Divide m3 -5man +10m3n2 —10m2n3 +5mn1 —n3 by m3 Ans. m3-3m2n+3mn2—n3.

-2mn+n2.

6. Divide ao-3a1b2+3a2ba—b® by a3—3a2b+3ab2—b3.

Ans. a3+3a2b+3ab2+b3.

Most of the examples in the preceding article may be solved by this method.

CHAPTER II.

ALGEBRAIC THEOREMS,

DERIVED FROM MULTIPLICATION AND DIVISION.

REMARK.. One of the chief objects of Algebra is to establish certain general truths. The pupil has now obtained the necessary knowledge to prove the following theorems, which may be regarded as the simplest application of Algebra.

ART. 78. THEOREM I. The square of the sum of two quantities is equal to the square of the first, plus twice the product of the first by the second, plus the square of the second.

Let a represent one of the quantities, and b the other;

then a+b=their sum;

and (a+b)×(a+b), or (a+b)2=the square of their sum.
But (a+b)X(a+b)=a2+2ab+b2, which proves the theorem.

APPLICATION.

1. (2+5)=4+20+25=49.
2. (2m+3n)2=4m2+12mn+9n2.

3. (ax+by) a2x2+2abxy+b2y2.

4. (ax2+3xx3)2=a2x2+6ax31⁄23+9x21⁄23.

ART. 79. THEOREM II.-The square of the difference of two quantities is equal to the square of the first, minus twice the product of the first by the second, plus the square of the second.

Let a represent one of the quantities, and b the other;

then a-b-their difference;

and (a—b)X(a—b), or (a—b)2=the square of their difference. But (a-b)X(ab)=a2—2ab+b2, which proves the theorem.

APPLICATION.

1. (5-3)2-25—30+9=4.
2. (2x—y)2=4x2—4xy+y2.

3. (3x-5x)2=9x2-30xz+2522.

4. (az-3cx)2=a2z2-6 accz+9c2x2.

ART. 80. THEOREM III.-The product of the sum and difference

of two quantities, is equal to the difference of their squares.

Let a represent one of the quantities, and b the other; then a+b=their sum,

and a-b their difference.

And (a+b)(a-b)=a2-b2, which proves the theorem.

APPLICATION.

1. (7+4)(7—4)=49-16=33=11×3.
2. (2x+y)(2x-y)=4x2-y2.
3. (3a2+462)(3a2-4b2)=9a4-166.

4. (3ax+5by)(3ax—5by)—9a2x2—25b2y2.

ART. 81. THEOREM IV.—The reciprocal of a quantity, is equal to the same quantity with the sign of its exponent changed.

If we divide a3 by a3, the quotient is expressed by, or by

as a3-5-a-2, since the rule for the exponents in division (Art. 70) requires that the exponent of the same letter in the divisor should be subtracted from that of the dividend. But

a3
a5

is a fraction, and

if we divide both terms by a3, which does not alter its value 1

(Ray's Arith., part 3d, Art. 147), it becomes; hence a