Divide the first term of the remainder by the first term of the divisor; the result is the second term of the quotient. Multiply the divisor by the second term of the quotient, and subtract the product from the first remainder. Repeat the process of dividing the first terms of the successive remainders by the first term of the divisor, and of subtracting the products of the divisor by the partial quotients from these remainders, until there is obtained a remainder equal to zero, or a remainder which cannot be divided by the divisor. The algebraic sum of the partial quotients is the quotient required. Examples of the division of polynomial algebraic quantities: 3d. Ans. c3. .....Ans, a2+2ab+b2. (3a+16a4b-33a3b2+14a2b3)÷(a2+7ab)? Ans. 3a3-5 a2b+2ab2. 4th. (a7-6a6b3+14a5b6-12a1b9)÷(a3—2a2b3)? Ans. a1-4a3b3+6a2b6. 5th. (a8-1628)÷(a2-222) ?......... Ans. a6+2a1z2+4a2z1+826. 6th. (2a4-13a3b+31a2b2—38ab3+24b1)÷(2a2—3ab+4b2? Ans. a2-5ab+6b2. 7th. (4c4-9b2c2+6b3c—b1)÷(2c2—3bc+b2) ? 8th. (-1+a3n3)÷(-1+an)?.. Ans. 2c2+3bc-b2. .......Ans. 1+an+a2n2. ......Ans. 1-6z+9z2. 9th. (a+2a3z3+z6)÷(a2—az+z2)?......Ans. a1+a3z+az3+z2. Ans. a1+4a3x+12a2x2+16ax3+16x1. 12th. (14a5-27a1b+21a3b2—3a2b3—2ab1)÷(2a2—3ab+2b2)? Ans. 7a3-3a2b-ab2. 13th. (51a2b2+10aa—48a3b—15b1+4ab3)÷(4ab-5a2+3b2)? Ans.-2a2+8ab-5b2. 14th. (21x3y2+25x2y3+68xy1—40y5—56x5—18x1y) ÷ (5y2—8x2 .Ans. — 8y3+4xy2-3x2y+7x3. 15th. (95a-73a2+56a1—25—59a3)÷(−3a2+5—11a+7a3)? -6xy)?.... Ans. 8a-5. 16th. (am+nbn-4ɑm+n~1b2n—27am+n-2b3n+42am+n-3f4n) ÷ (a" b"-7an-162")?............. Ans. am+3am-1bn—6am-272n. 20. When the terms of the dividend and divisor are arranged according to the powers of the same letter, it may happen that several terms are affected with the same power of that letter. In such a case it becomes necessary to arrange these terms according to the powers of some other letter. Suppose, for example, that a polynomial contains the terms abx2+a2x2b2x2, all of which contain x2, the coefficients of 2 may be arranged according to the powers of a, and the terms in 2 written either thus (a2+ab-b2) x2, a2 r2 or thus, Assuming that the dividend and divisor are arranged according to the powers of the same letter, a for example, and that the literal coefficients of the terms in x are arranged according to the powers of some other letter, the division is made by the rule for the division of polynomials; for in each partial division the first term of the dividend ought to be the product of the first term of the divisor by that term of the quotient which is to be found by that partial division. All the terms of the quotient ought thus to be obtained, arranged respectively in the same manner as the dividend and divisor. To prove this, let a polynomial dividend and divisor be arranged according to the descending powers of x. Let x be the highest power of r in the dividend, and let the terms which contain + be represented by A, those which contain 3 by B, x2 by C, x by D, and the terms into which r does not enter by E; that is, let the dividend be denoted by Ax++Bx+Cx2+Dx+E. Also, let the highest power of x in the divisor be r2, and let the divisor be expressed by A'x2+B'x+C'. Then, since the highest exponent of x in the dividend and divisor, respectively, are 4 and 2, the highest in the quotient must be 2; and the quotient ought to have the form A+B′′x+С”. Again, the product A'xx A′′x2 cannot be reduced with the other products of the divisor by the quotient (Art. 12'); it must therefore be equal to Art. Whence, if Art is divided by A'r2, the result must be the part Á"x2 of the quotient; and since x++x2=x2, the division of Ax+ by A'r2 is reduced to the division of A by A'. Whence, if A, A' are themselves polynomials, composed of one or several letters, their quotient must be found by the rule for the division of polynomials, which requires their arrangement according to the powers of some letter contained in them. The term A" being obtained, the whole divisor A'x2+B'x+C' is multiplied by it, and the product subtracted from the dividend Ax++Bx3+ Cx2+Dx+É. The remainder is a second partial dividend, upon which the same process is to be repeated, and so on. Example. Given the dividend (1262-29bc+15 c2)a3 +(23b2—31b2c— 9bc2+15c)a2+(10b+-6b2c2)a, and the divisor (3b-5c)a+2b2; required the quotient? First partial division of the polynomial coefficient of the highest power of a in the dividend by the polynomial coefficient of the highest power of a in the divisor: 3b-5c)12b2-29bc+15c2(4b-3c 12b2-20bc -9bc+15c2 Since also a3+a=a2, it follows that (4b-3c)a2 is the first part of the quotient; And since (3b-5c)a × (4b—3c)a2=(12b2—29bc+15c2)a3, the term of the dividend containing a3 is destroyed by the product of the first terms of the divisor and quotient. Next, 2b2x (4b-3c)a2=(8b3—6b2c)a2. Subtracting this from the coefficient of a2 in the dividend, (1563-25b2c -9bc2+15c3)a2 is the first term of the second partial dividend. Second partial division of the polynomial coefficient of a2 by the coefficient of the highest power of a in the divisor: 3b-5c)15b3-25b3c—9bc2+15c3 (5b2—3c2 -9bc2+15c3 -9bc2+15c3 Therefore, since a2÷a=a, (5b2—3c2)a is the second part of the quotient. S Now (3b-5c)a × (5b2—3c2)a=(15b2—25b2c-9bc2+15c3)a2; therefore the term in a2 of the second partial dividend is destroyed by the subtraction of the product of the first term of the divisor by the second term of the quotient. The product of the second term of the divisor by the second term of the quotient, 262 × (5b2—3c2)a=(10b3—6b2c2)a. This product being equal to the term in a of the dividend, the second remainder is zero, and the exact quotient is (4b-3c)a2+(5b2—3c2)a. Calculation described in the preceding detail: Quotient. -3c2 a Divisor. Dividend. 3ba+262) -5c ) a Quotient. +4-1 | 3ba+b2-2b\ 6ba+-7b2 a3-3b3 \a2+4b3 \a+b2-2b (2a3-3ba2+4ba+1 -5 -10 +236 +2262 -962 +b2 -ab2 +a2b2+a3b2 -ab5 -ab3 +a2b3 21. If a polynomial dividend contain different powers of a letter which is not contained in the divisor, the dividend and divisor may be arranged according to the powers of some other letter which is common to both, and the division made by the methods already explained. Otherwise, suppose that the dividend contains different powers of x, and that this letter does not enter into the divisor. If the dividend is arranged according to the powers of a (the highest being a) it may be expressed by the polynomial A+B+C+D, in which A, B, C, D, are quantities monomial or polynomial, which do not involve x. Let the polynomial divisor, which is also independent of x, be represented by P. Then the divisor multiplied by the quotient must reproduce the dividend; and since the divisor does not contain x, the quotient must be a polynomial affected with the same powers of x as are found in the dividend. The quotient is therefore of the form A'x+B'x2+C'x+D. Suppose this quotient determined, and let each term of it be multiplied by the divisor; the products are A'Pr3+B'Pr2+CPx+D'P. These products, involving different powers of x, cannot be reduced with each other; whence they are equal, respectively, to the terms Ar3+Bx2+Cx+D of the dividend; Therefore A'P=A and A'=A+P; B'P B and B'=B÷P; C'P C and C'=C÷P. Whence, in order that a polynomial arranged with respect to a letter may be exactly divisible by a polynomial independent of that letter, it is necessary that each of the coefficients of the different powers of the first polynomial be exactly divisible by the second polynomial. The coefficients of the different powers of the letter in the quotient are the successive quotients obtained by dividing the coefficients of the polynomial dividend by the polynomial divisor. Ex. Divide 3a2b3—3abc3—2b3c2+b5—3a2bc2+3ab3c—a°c3+bc*+a2b2c_by b2-c2? The dividend (arranged according to the powers of a, and the coefficients according to the powers of b) is +bc* ; (3b3+b2c—3bc2—c3)a2+(3b3c—3bc3)a+b5—2b3c2 Now 3b3+b2c-3bc2-c3 b2-c2 =3b+c; b5-2b3c+be+ b2-c2 -=b3—bc2; consequently the quotient is (3b+c)a2+3bca+b3—bc2. 22. If x2-a2 is divided by x-a, the quotient is x+a. x-a, x-a, x-a, x2+ax+a2. In these quotients it is to be observed, 1st. That all the terms are additive. 2d. That the first and last terms are formed by subtracting 1 from the exponents of x and a in the dividend. 3d. That, of the intermediate terms beginning with the second, the exponents of x are successively diminished by 1, and the exponents of a augmented by 1, in such a manner that the sum of the exponents of x and a in each term is constant. Denoting by m any additive whole number, it may be inferred by analogy -=xTM-1+xTM-2a+xm¬3 a2+.......+xaTM-2+am-l ̧ that x-a" x-a In effect, any number of the terms of this quotient can be found by division; but, without resorting to this kind of proof, the accuracy of the result can be verified by multiplication; for -xa-2-xaTM Now, it is evident that the second term of the first partial product, or first line, is equal and of a contrary sign to the first term of the second partial product or second line, and that the sum of these two terms is zero. In like manner the third term of the upper line is destroyed by the second term of the lower line, the fourth by the third, and so on to +xaTM-1, the last term of the upper line, which is destroyed by -xam-1, the last term but one of the lower line. Whence, if the quotient of x-a" is multiplied by the divisor x-a, the dividend x”—aTM is reproduced. Consequently 23. Let it be required to divide x +pxTM-1+qxTM-2+...+tx+u by x-a? Calculation: x-a) xTM+pxm-1+qxm-2 + ... + tx+u (xm-1+axm-2 + a2 xm 3+...+am-1 -2 +p +ap +9 +paTM-2 +9am-3 Dividing by x, the first term of the quotient is −1. The product of the divisor by this quotient is x-ax-1; subtracting this product from the dividend, that term of the remainder which is affected with the highest power of x is (a+p)xTM 1. Dividing (a+p)-1 by x, the quotient is (a+p).xm-2; and that term of the remainder which is affected with the highest power of x is (a2+ap+q)xTM Dividing this term by x, the third term of the quotient is (a2+ap+q)x 3. Proceeding in this manner, and regarding always as a single term all the quantities which involve the same power of x, it becomes evident that each term of the quotient is obtained from the preceding term by multiplying it by a, adding to the product the term of the dividend which involves the same power of x, and then dividing by x. Hence it follows that the part of the quotient which is independent of x is a”¬1+paTM¬2+qa”-3+......+t. m-2 If the divisor is multiplied by this quantity, its product by z will destroy the part in x of the last partial dividend, and the remainder will be aTM+paTM-1+qam 2+...+ta+u. Here the division ends, and the remainder is the original dividend, except that x is replaced by a. When this remainder is zero the division is exact. Whence, if a is a quantity such that, being substituted for x, it renders the polynomial x31+pxm-1+qxm-2+....tx+u=0, the polynomial is divisible by x-a. SECTION III. SIMPLIFICATION OF FRACTIONAL EXPRESSIONS. GREATEST COMMON MEASURE. REDUCTION OF FRACTIONAL EXPRESSIONS TO THE SAME DENOMINATOR. ADDITION, SUBTRACTION, MULTIPLICATION, AND DIVISION OF FRACTIONAL EXPRESSIONS. 24. When the division of one algebraic quantity by another cannot be executed, it is indicated by making the dividend the numerator and the divisor the denominator of an algebraic fraction. |