and this divided again by x-x', will give for the second re X'm-1 It is unnecessary to continue this process further, to see that these successive remainders are the coefficients of the transformed equation (1') beginning with the absolute term, or the coefficient of yo. The divisor to be employed is x-x' if the roots of the transformed equation are to be less, in value, than those of the given equation by the constant difference x'; if greater, the divisor must be x + x'. Hence, an equation may be transformed into another of which the roots are greater, or less, than those of the given equation by the following RULE.-I. Divide the first member of the given equation (the second member being zero) by x plus the constant difference between the roots of the two equations, continuing the operation until a remainder is obtained which is independent of x; then divide the quotient of this division by the same divisor, and so on, until m divisions have been performed. II. Write the transformed equation, making these successive remainders the coefficients of the different powers of the unknown quantity, beginning with the zero power. It must be borne in mind that the term plus in this rule is used in its algebraic sense. By a little reflection, it will be seen that the mth quotient will be the coefficient of am in the original equation, and that this will also be the coefficient of the highest power of the unknown quantity in the transformed equation. into another of which the roots shall be less by 2. This is example 5 of the last article. Make x = 2 + y, or y = x — 2 ; or, Hence, the transformed equation is y1 + 4y3± Oy3 — 24y + 0 = 0; y1 + 4y3 — 24y = 0, as before. 2. Transform the equation, x-12x3 + 17x2- 9x+7= 0, into one having roots less by 3. Here xy +3, or y=x -3. OPERATION. -3)x-12x3 + 17x2 — 9x + 7 (1⁄23 — 9x2 — 10x — 39 6х x-3)x3-9x210x-39 (x2-6x-28 - 37= — Hence, y0ys 37y2-123y-1100, is the transformed equation. We shall have 4 remainders, if we operate on an equation of the 4th degree; 5 remainders with an equation of the 5th degree; and, in general, n remainders with an equation of the nth degree. The transformation of equations by division, treated of in this article, if performed by the ordinary rule, would be too laborious for practical application; but by a modified method of division, called Synthetic Division, it becomes expeditious and easy. As preliminary to the explanation of this method of division, we must explain the process of MULTIPLICATION AND DIVISION BY DETACHED COEFFICIENTS. 440. It has been seen that when two polynomials are homogeneous their product is also homogeneous, and the number which denotes its degree is the sum of the numbers denoting the degrees of the factors. It is evident that if the polynomials contain but two letters, and both are arranged with reference to the same letter, the product will be arranged with reference to that letter. Since, in the operation of multiplying the terms of the multiplicand by the terms of the multiplier, the products of the coefficients are not affected by the literal parts to which they are prefixed, these coefficients may be detached and written down with their signs in their proper order, and the multiplication performed as with polynomials. The partial products, numerial or literal, being carefully arranged as if undetached, are then reduced and the literal parts annexed. 1 +3 +3 +1 Coefficients of product. Now by annexing the proper literal parts to the several terms thus obtained, we have x3 + 3a2x + 3αx2 + x3, Ans. This method of multiplication may be employed when the tw~ polynomials contain but one letter. When any of the powers of the letters, between the highest and lowest, do not appear in either factor, the terms corresponding to such powers must be supplied with the coefficient 0. 3. Multiply 3 + 2x2-1 by x2 + 2. The factors completed are x3 + 2x2 + 0x − 1 and x2 + 0x + 2. Hence the operation is and the product, or, 1+2+01 1+0+2 1+2+0-1 2+4+02 1+2+2+3+0−2 2, x5 + 2x2 + 2x3 + 3x2 + 0x x + 2x2 + 2x3 + 3x2 2. 4. Multiply 3x2-2x-1 by 4x+2. 5. Multiply 32-5x-10 by 2x-4. 6. Multiply x2+xy + y2 by x2—xy + y2. 7. Multiply 234x2+5x-2 by x2+4x-3. Ans. 12x3 — 2x2-8.x-2. 441. Now, if detached coefficients can be used in multiplication, so in like cases, they may be employed for division. When the dividend and divisor contain but two letters and are homogeneous, the degree of the quotient will be the excess of the degree of the dividend over that of the divisor. EXAMPLES. 1. Divide a1-3a3x-8a2x2+18ax3+16x4 by a2-2ax—2x2. |