_x_³ _x_² _x_ 1 _x_² _x_ 1 _x_ 1 1 _p_ 0 0 0 _q_ 0 0 _r_ 0 3 _p_ _pa_ _pa_² _pa_³ _q_ _qa_ _qa_² _r_ _ra_ _p_ 2_pa_ 3_pa_² _q_ 2_qa_ _r_ _p_ 3_pa_ _q_ _p_ _px_³ + 3_pax_² + 3_pa_²_x_ + _pa_³ + _qx_² + 2_qax_ + _qa_² + _rx_ + _ra_ + _s_ = _px_³ + (3_pa_ + _q_)_x_² + (3_pa_² + 2_qa_ + _r_)_x_ + _pa_³ + _qa_² + _ra_ + _s_ Now, observe that all this might be done in one process, by entering _q_, _r_, and _s_ under their proper powers of _x_ in the first process, as follows _x_³ _x_² _x_ 1 _p_ _q_ _r_ _s_ _p_ _pa_ + _q_ _pa_² + _qa_ + _r_ _pa_³ + _qa_² + _ra_ + _s_ _p_ 2_pa_ + _q_ 3_pa_² + 2_qa_ + _r_ _p_ 3_pa_ + _q_ _p_ This process[65] is the one used in Appendix XI., with the slight alteration of varying the sign of the last letter, and making subtractions instead of additions in the last column. As it stands, it is the most convenient mode of writing _x_ + _a_ instead of _x_ in a large class of algebraical expressions. For instance, what does 2_x_⁵ + _x_⁴ + 3_x_² + 7_x_ + 9 become when _x_ + 5 is written instead of _x_? The expression, made complete, is, 2_x_⁵ + 1_x_⁴ + 0_x_³ + 3_x_² + 7_x_ + 9 1 0 3 7 9 2 11 55 278 1397 6994 2 21 160 1078 6787 2 31 315 2653 2 41 520 2 51 _Answer_, 2_x_⁵ + 51_x_⁴ + 520_x_³ + 2653_x_² + 6787_x_ + 6994. [65] The principle of this mode of demonstration of Horner’s method was stated in Young’s Algebra (1823), being the earliest elementary work in which that method was given. APPENDIX XI. ON HORNER’S METHOD OF SOLVING EQUATIONS. The rule given in this chapter is inserted on account of its excellence as an exercise in computation. The examples chosen will require but little use of algebraical signs, that they may be understood by those who know no more of algebra than is contained in the present work. To solve an equation such as 2_x_⁴ + _x_² - 3_x_ = 416793, or, as it is usually written, 2_x_⁴ + _x_² - 3_x_ - 416793 = 0, we must first ascertain by trial not only the first figure of the root, but also the denomination of it: if it be a 2, for instance, we must know whether it be 2, or 20, or 200, &c., or ·2, or ·02, or ·002, &c. This must be found by trial; and the shortest way of making the trial is as follows: Write the expression in its complete form. In the preceding case the form is not complete, and the complete form is 2_x_⁴ + 0_x_³ + 1_x_² - 3_x_ - 416793.