Suppose we follow this mode of forming p(x + a)³ + q(x + a)² + r(x + a) + s.
| x³ | x² | x | 1 | x² | x | 1 | x | 1 | 1 | |||
| p | 0 | 0 | 0 | q | 0 | 0 | r | 0 | 1 | |||
| p | pa | pa² | pa³ | q | qa | qa² | r | ra | ||||
| p | 2pa | 3pa² | q | 2qa | r | |||||||
| p | 3pa | q | ||||||||||
| p |
px³ + 3pax² + 3pa²x + pa³ + qx² + 2qax + qa² + rx + ra + s
= px³ + (3pa + q)x² + (3pa² + 2qa + r)x + pa³ + qa² + 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
| x³ | x² | x | 1 | ||
| p | q | r | s | ||
| p | pa + q | pa² + qa + r | pa³ + qa² + ra + s | ||
| p | 2pa + q | 3pa² + 2qa + r | |||
| p | 3pa + 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 2x⁵ + x⁴ + 3x² + 7x + 9 become when x + 5 is written instead of x? The expression, made complete, is,
| 2x⁵ + | 1x⁴ + | 0x³ + | 3x² + | 7x + | 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, 2x⁵ + 51x⁴ + 520x³ + 2653x² + 6787x + 6994.