To ascertain if 2 is a root, arrange the work as follows : 1 + 4 - 1 - 16 - 12 [2 -1-6-11- 6 - 2 - 12 - 22
Hence 2 is a root.
Explanation. The first line contains the coefficients of the original equation, and the divisor 2. Dividing the last term, — 12, by 2 gives a quotient — 6 ; adding — 16, the coefficient of x, gives — 22. Divid- ing — 22 by 2 gives — 11 ; adding — 1, the coefficient of x^, gives —12. Dividing — 12 by 2 gives — 6 ; adding + 4, the coefficient of x^, gives — 2, which divided by 2 gives a final quotient of — 1, hence 2 is a root.
Since the equation can have no more than one positive root, we will only make trial of the remaining negative divisors, thus :
1 + 4-1-16- 12|-6 + 2 - 14 Hence — 6 is not a root.
1 +4_1 _ 16-121 -3 -1-1+4+ 4 + 3 + 3-12 Hence — 3 is a root.
121-4 1+4-1-16 + 3 -13 Hence — 4 is not a root.
1+4-1- 16- 12 i^ - 1 - 2 + 5 + 6 + 2 + 4-10 Hence — 2 is a root.
Solve roots : 1. x^- 2. x^- 3. 2.x3 4. 4.r.3 5. x^- 6. x^- 7. X^- 8. x^- 9. x^- 10. J-^ -
EXAMPLES XLVIII. i.
the following equations, which have one or more integral
0. 11. 12. 13. 14. 15. 16. 9.x2 + 26x-24 a: - 2 a:2 + 2 = 0. + 5 x'^ - 11a: + 4 = 0. - 16 x-2 + 31 a: - 14 = 0. . 2 a:-2 - 29 x + 30 = 0. - 8 x- + 5 X + 14 = 0. - 2 x3 - 7 a:2 + 8 X + 12 = 0. 17. 4.x3_i4:e2_(_36a;+4.5=0. 18. - 3 x^ - 42 X - 40 = 0. 19. - 10 y? - 20 X - 16 = 0. 20. 21. 2.^4 -x3- 29x2 + 34a 22. x5 - 3 X* - .x3 + 15 .x2 x^ + 8x3 + 9x2 -8x- 10 = 0. .'(^ + 2 x3 - 7 x2 - 8 X + 12 = 0. .ri - 3 .x2 _ 6 X - 2 = 0. X*- 2 x3- 12 x2+ 10 X + 3 = 0. 6x4- x3- 17x2+ 16x -4 = 0. x* - 2 x3 - 1 3 :c2 + 14 X + 24 = 0. .^•4_|. 4 a^3._ 22 x2 - 4 X + 21 = 0. a-4 + 2 x3 - 7 x2 - 8 X + 12 = 0. .-'•-^ - 6 .r2 - 16 X + 21 = 0.
- t^ + 4 .r3_ j.i _ 10 .,• _ 12 = 0.
+ 24 = 0. + 4 X - 12 = 0.
