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This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1738 Excerpt: ...is too much; I try with 20, and find it by somethiug too small, but ail find too big; wherefore I know that x must be = ao, with some Decimal Fraction annex'd 5 and at last I discover 20.7 to be the very Root sought, or at least one of the Roots of the given quation., . i C2-f LEMMA to Chap. V. How to find all the Divisors of a given Number or Quantity. V RULE. 1 J' ''. ' ' Divide the given Quantity or Number by the least of its Divisors that is f i and the Quotient by the least of its Divisors that is C 1, and so on till you have 1 for a Quotient, and you'll have all the prime Divisors of the given Number or Quantity; then multiply each two, three, four, &V. of the prime Divisors into themselves continually, and the several Products are the compound Divisors. Examples. I. So if all the Divisors of 60 were required. First divide it by 2, and the Quotient 30 by 2, and the Quotient 15 by 3, and the Quotient 5 by 5: then the prime Divisors are 1, X, 2, 3, 5; and the Compound ones produe'd by each two are 4, 6, 'io, 15 j by each three are 12, 20, 30 5 by each four (or by all) 60. H II. Again, If all the Divisors of 21 a b b were required. Divide it by 3, and the Quotient jabbbyy, and, jhctDuotient a b b by a, and the Quotient b b by b, and the-Quotient b by b: then the prime Divisors are 1, 3, 7, a, i, b 5 and the Compound ones produe'd by each two are 21, 3 a, 3 bt 7 at' 7 b, ab, b b; by each three, 21 a, 21 b, 3 ab, 3 b b, 7 ab, 7 b1, abb; by each four, ai ab, XI b b, $ abb, 7 abb; and bj all five, 21 abb. III. In like manner all the Divisors of a abb 6 a ac a re 1, 2, & --3 j i., ati--J r, i --ja'r, % abb--6 a a c.. CHAP. CHAP. V. John Kerse/s Method of finding the Roots of some ADFHCTED, CUB ICK, ...

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