Let us take the example of the preceding article. The arrangement of the work is as follows:
1 3-8- 5+26 -33 + 26 2 6+12-24 4 - 4- 8+16 -8 6+12- 24 3 -2 + 3 + 0 - 5 + 2
Explanation. The column of figures to the left of the vertical line consists of the coefficients of the divisor, the sign of each after the first being changed; the second horizontal line is obtained by multiplying 2,4, -8 by 3, the first term of the quotient. We then add the terms in the second column to the right of the vertical line; this gives - 2, which is the coefficient of the second term of the quotient. With the coefficient thus obtained we form the next horizontal line, and add the terms in the third column; this gives 3, which is the coefficient of the third term of the quotient.
By adding up the other columns we get the coefficients of the terms in the remainder.
566. In employing this method in the following articles our divisor will be of the form x \pm a, which enables us to still further simplify the work, as the following example shows:
Ex. Find the quotient and remainder when 3x^7 -x^6 + 21x + 5 is divided by x + 2.
3 -1 0 31 0 0 21 5 -2 6 14 -28 -6 12 -24 6 3 -7 14 3 -6 12 -3 11
Thus the quotient is 3x^6 - 7x^5 + 14x^4 + 3x^3 - 6x^2 + 12x -8, and the remainder is 11.
Explanation. The first horizontal line contains the coefficients of the dividend, zero coefficients being used to represent terms corresponding to powers of x which are absent. The divisor is written at the right of this line with its sign changed (Art. 564) and 1, the coefficient of x, omitted. The first term of the third horizontal line, which contains the quotient, is the result of dividing 3, the coefficient of x^7 in the dividend, by 1, the coefficient of x in the divisor. This is then multiplied by the divisor - 2, and the result is - 6, the first term of the second horizontal line; the sum of -1 and -6 gives -7, the
