Method of finding Integral Roots. As this root must verify the given equation, we have + &c. + m = 0; whence, multiplying by q”-1, and transposing, we obtain and, therefore, as the second member is integral, the first member must also be integral, or we must have whence q = 1, x = pi that is, every commensurable root of the given equation must be an integer. Again, the substitution of x = P, in the given equation, produces p2 + a p2-1+&c. . . . + k p2 +1p+m=0; whence, dividing by p, and transposing, we obtain m :-1-kp- &c. ... a pr—2— p1-1; and, therefore, as the second member is integral, the first member must be so likewise; that is, every integral root must be a divisor of m. · k — i p — h p2 — g p3—&c.—a p1 — 3 —p”—2. so that this integral root must likewise be a divisor of m'. Method of finding Integral Roots. In the same way, if we use m", m'', m1v, &c. as follows: this integral root must be a divisor of m", m", m1v, &c. ; and the last condition to be satisfied is Hence to find all the commensurate roots of the given equation, write in the same horizontal line all the integral divisors of m, which are contained between the extreme limits of the roots. Write below these divisors all the corresponding values of m', m", &c., which are integral, remembering that a divisor cannot be a root, when the value which it gives for either m', m", m", &c., is fractional. Proceed in this way till the values of min-11 are obtained, and those divisors only are roots which give -p for the value of this quantity. 312. EXAMPLES. 1. Find the commensurable roots of the equation 25 19x334x2 + 12 x — 40 = 0. Solution. The extreme limits of the real roots are +7·4, and -6.9. Hence we have m P = m' = m" 2, -13, and, therefore, 2, 14, 1, 5; 1, and -5 are roots of the given equation, and its first member, divided by the factor (x-2)(x+1)(x+5) = x3 + 4 x2 — 7 x — 10, and, therefore, the remaining roots are those of the equation which are equal to each other, and each is x = 2. 2. Find the commensurable roots of the equation x3 — 3 x7 —10 x¤ −2 xa+6 x3+21 x2-3x-10=0. Ans. 5, 1, -1, and —2. 3. Find all the roots of the equation x2 + x3-24 x2 + 43 x — 21 = 0 which has commensurable roots. 5. Find all the roots of the equation x410 x3+35 x2-50 x + 24 = 0 which has commensurable roots. Ans. 1, 2, 3, 4. Commensurable Roots of any Equation. 6. Find all the roots of the equation x53x48x3+24 x2-9x+27=0 which has commensurable and equal roots. Ans. 3, -3, and 1. 7. Find all the roots of the equation x6 — 23 x1- 48 x3 +95 x2+400x375=0 which has commensurable and also equal roots. Ans. 3, 5, and -2±1. 313. Problem. To find the commensurable roots of an equation. Solution. Reduce the equation to the form Ax+Bx-1+ &c.... + Lx + M = 0, in which A, B, &c., are all integers, either positive or negative. which, multiplied by An-1, is yn+Byn-1+A Cyn-2+&c....+An-2 Ly+An-1 M= 0. The commensurable roots of this equation may be found, as in the preceding article, and being divided by A, will give the commensurable roots of the required equation. 314. Scholium. The substitution of Commensurable Roots of any Equation. is not always the one which leads to the most simple result. But when A has two or more equal factors, it is often the case that the substitution leads to an equation of the desired form, A' being the product of the prime factors of A, and each factor need scarcely ever be repeated more than once. 1. Find the commensurable roots of the equation 64 x4. − 328 x3 + 574 x2 - 393 x - 90 = 0. Solution. We have, in this case, A 64 26; = = hence we may take A' equal to some power of 2; and it is easily seen that the third power will do, so that we may make Hence the given equation becomes y441 y3574 y2-3144 y + 57600. The commensurable roots of which are found, as in art. 311, to be y = 4, 6, 15, and 16; so that the roots of the given equation are x= 1, 2, 17, and 2. 2. Find the commensurable roots of the equation 8x334x2-79 x + 30 = 0. |