Ray's Algebra, Part Second : An Analytical Treatise Designed for High Schools And Colleges

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We next subtract the square of by'' from the given polynomial, and then divide the first term of the re- mainder, — 2Qxy^, by lOj^^, which gives — 2ocy, the second term of the root. We then place — 2ay in the root and also in the divisor, and multiply the divisor, thus increased, by — ~xy, and subtract the product from the first remainder. We then double by'' — 23nj, the terms of the root already found, for a partial divi- sor, and divide the first term, lOy' of the divisor, into — 30xi/', the ...first term of the remainder, which gives — 3a; for the third term of the root. Completing the division, multiplying by — 3a;, end subtracting, we find there is nothing left.
Note. — The first remainder consists of all the terms after the first subtraction, and the second, of all the terms after the second subtraction. It is useless to bring down more terms than have corresponding terms in the quantity to be subtracted.
If the preceding example be arranged according to the powers 136 RAY'S ALGEBRA, PART SECOND.


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