in the first horizontal row of Form I, a5 + 3 ab + 3 a3b2 + a2b3, are found as the first terms of the different rows of Form II; that is, they occur in the first diagonal row. The same is true for the other rows of partial products. 17. The arrangement of the partial products for a given example depends simply on which of the two given polynomials is chosen as multiplicand, and which as multiplier.

18. In either Form I or Form II, the terms of any particular row are obtained by multiplying successively the terms of the multiplicand by one of the terms of the multiplier.

E. g. To obtain the first row of partial products use as a multiplier the first term of the polynomial chosen as multiplier; for the second row use the second term of the polynomial multiplier; for the third row the third term, etc.

19. The first term of each row is obtained by multiplying the first term of the multiplicand by the term of the multiplier corresponding to the number of the row.

Similar results may be seen to be true by examining Form II. (§ 15). 20. It follows that, if we know the first term of any specified row, we may, by dividing it by the first term of the multiplicand, obtain the term of the polynomial multiplier corresponding in number to the number of the row.

E. g. In Form I (§ 13) we have:

From the first row of partial products.

which is the first term of the polynomial multiplier.

From the second row of partial products.

. 2 a1ba3 = 2 ab, (2)

which is the second term of the polynomial multiplier.

From the third row of partial products

(3)

which is the third term of the polynomial multiplier.

21. From the arrangement of the partial products in columns, the first partial product, a, is the first term as of the reduced product in either Form I or Form II. (§§ 13, 15).

22. If now, returning to Form I, we divide a by a3, we obtain as a quotient the first term a2 of the polynomial multiplier.

That is,

a÷a3 = a2. (See (1) § 20.)

23. If we multiply the terms of the multiplicand successively by this multiplier, a2, we obtain the first horizontal row of partial products as in Form I (§ 13). By subtracting the terms of this first row from the reduced product, we have as below:

24. The result of the subtraction, 2 ab + 7 a3b2 + 9 a2b3 +5 ab1 + b, we shall call the first partial remainder. From the nature of the case the first term, 2 ab, is the same as the first term of the second row of partial products in Form I (§ 13). It is also the sum of all of the partial products in Form I, except those in the first row which were subtracted.

25. Dividing the first term 2 a1b of the first partial remainder by as, we obtain 2 aba3 2 ab. (See (2) § 20.)

The term 2 ab thus obtained is equal to the second term of the original multiplier. Referring to Form I (§ 13) it may be seen that this step in the process consists in dividing the term of highest degree with reference to the letter of arrangement (that is, the first term of the second row of partial products in Form I) by the first term of the multiplicand above.

26. By multiplying the multiplicand, as in Form I (§ 13), by 2 ab as a multiplier, we obtain the second row of partial products 2ab+6 a3b2 + 6 a2b3 + 2 ab*.

27. Subtracting these terms from the first partial remainder obtained above in § 23 (i.), we have

First partial remainder,

2a4b+7 a3b2+9 a2b3 + 5 ab1+b5,

a3b2+3a2b3+3 aba + b3.

Step (ii.) Second row of partial products, 2 ab +6 a3b2+6a2b3 +2 aba,

Second partial remainder,

28. The first term al of the second partial remainder is the same as the first term of the third row of partial products in Form I ($13).

29. As we proceed in our work any particular remainder will,

from the nature of the case, be the sum of the rows of partial products below the last one subtracted. (See Form I, § 13.)

30. Dividing, as before, the first term a2 by the first term of the multiplicand a3, we have as in (3) § 20,

a3b2 ÷ a3 = b2,

which is the third term of the original polynomial multiplier. 31. As before, we use this as a multiplier with the entire multiplicand, and obtain the third row of partial products in Form I (§ 13), a3b2 + 3 a2b3 + 3 ab1 + 65.

32. Subtracting these terms from the second partial remainder, we find as below, that the third partial remainder vanishes, that is, it is zero:

Second partial remainder,

Step (iii.) Third row of partial products,

a8b2+3a2b3 + 3 ab1 + b5
ab2 + 3 a2b3 +3 aba + b5

33. The process stops here, as we should naturally expect, since we have recovered all of the terms of the original polynomial multiplier a2, +2 ab and b2. (See Form I, § 13.)

34. Summary. By this process we have succeeded in finding the original polynomial multiplier, when the reduced product and the original multiplicand were given. The process would have been exactly the same if we had started with the reduced product and the original multiplier, a2 + 2 ab + b2, to find the original multiplicand a + 3ab + 3 ab2 + b3. In that case, however, there would have been this exception, that the successive rows of partial products would have corresponded to the rows of partial products in Form II (§ 15), where the polynomial to be found is used as a multiplier.

35. We have thus developed a method for finding the original multiplier when the multiplicand and the reduced product are given. The method holds good also for finding the original multiplicand when the multiplier and the reduced product are given.

36. It will be seen from the discussion above that the reduced product takes the place, in our problem of division, of the dividend, the original multiplicand of the divisor, and the original multiplier of the quotient.

37. The rows of partial products in the process of multiplication correspond to those appearing in the process of division.

38. The Division Transformation. Whenever division is possible, we may carry out the different steps of the process as follows:

First arrange the terms of both dividend and divisor according to ascending or descending powers of some letter.

Place the divisor, for convenience, at the right of the dividend. Since the different terms of the quotient are to be used as multipliers during the process, it will be convenient to write the quotient, term by term, immediately below the divisor.

Divide the first term of the dividend by the first term of the divisor, and write the result as the first term of the quotient.

Multiply the whole divisor by this first term of the quotient, and write the resulting partial products under the dividend.

Subtract from the dividend the polynomial composed of the partial products, and bring down the result as the first partial remainder.

Divide the first term of the first partial remainder by the first term of the divisor as before, and write the result as the second term of the quotient.

Multiply the whole divisor by the second term of the quotient, subtract the resulting product from the first partial remainder, and write the result as a second partial remainder.

Repeat the operations above either until the remainder is 0, or until as many terms of the quotient are found as are desired. In the latter case, add algebraically a fraction having for numerator the remainder at this stage, and for denominator the divisor.

ARRANGEMENT OF THE WORK

39. A convenient arrangement of the different steps of the process is shown below:

Multiplying divi- a5+5a1b+10a3b2+10a2b3+5aba+b5 | a3+3a2b+3ab2+b8 a5+3a4b+3a3b2+ a2b3

2ab+ 7a8b2+ 9a2b3+5aba+b5
2ab+ 6a3b2+ 6a2b3+2aba

sor by a2 1st partial remainder. Mult'g divisor by 2ab. 2nd partial remainder. Multiplying divisor by b2.

a2+2ab+b2

We have carried out the process above on the assumption that the degree of the dividend, with reference to the letter of arrangement, was at least as high as that of the divisor, that is, that division was possible. It is the exception rather than the rule that we find an integral quotient when dividing one integral polynomial by another. We therefore apply the Principle of No Exception, and assert that division may be performed, if indeed it can be begun at all, by the steps of the process above.

40. In the division transformation two cases may arise:

First, it may be possible, or second, it may be impossible, to find as a quotient an integral function of x.

In the first case the division, if carried out, is said to be exact and there is no remainder. It may be shown that when division is exact, the form of the quotient will be the same whether the division. be carried out with both dividend and divisor arranged according to descending or ascending powers of some specified letter.

Ex. 1. Divide x2+10x + 21 by x + 3.

The process is shown below, at the left with the dividend and divisor arranged according to descending powers, and at the right with both arranged according to ascending powers.