the greatest common divisor of the first member is, (2 — 1)? ; hence, the required equation is, 22 - 3x + 2 = 0; which may be solved by known rules. The roots of the given equation are, 1, 1, 1, and 2. III. SOLUTION OF HIGHER EQUATIONS. 206. If an equation of the form (1) has equal roots, it may be freed from them by the preceding principle, and the resulting equation will be of a lower degree than the primitive one; which is always to be desired. We shall, in what follows, suppose that the equation in question has been freed of its equal roots. It was shown, in Art. 197, that the absolute term of an equation of the form (1), is equal to the continued product of the roots with their signs changed. When this absolute term is a whole number, and it may always be made so (Art. 198), and any root is a whole number, it may often be found by trial; for, we may resolve the absolute term into its factors, and then, by the process of synthetic division, we whether the first member of the given equation is exactly divisible by the unknown quantity, increased by any one of these factors; if so, that factor, with its sign changed, is a root (Art. 193); if the first member is not divisible by the unknown quantity, increased by any one of the factors, then the equation has no entire roots. see EXAMPLES. 1. Find the entire root in the equation, 200 + 3x2 + 9x - 38 = 0. The divisors of 38 are, #1, #2, +19, = 38. We see, at a glance, that neither + 1, nor – 1, will satisfy the equation ; hence, neither – 1 or + 1, can be a root. By applying the rule for synthetic division we see that the first member of the given equation is divisible by x — 2; hence, + 2 is a root. Dividing both members of the given equation by a · 2, we have, x2 + 5x + 19 = 0; which can be solved by known rules. Both of its roots are imagin – 5+ V – 51 ary, one equal to -5 – V – 51 and 2 2 2. Find the entire root of the equation, 203 – 12x2 + 4x + 207 = 0. The divisors of 207 are, #1, #3, # 9, § 23, $ 69, and = 207; testing each factor, we find that the first member of the given equation is divisible by x 9; hence, 9 is a root. When freed of this root, the given equation becomes, In the same way, the following equations may be solved : 3. 203 + 3x2 6x . 8 = 0. Ans. X = 2, x = -1, and a = - 4. a 207, An equation of the form (1), whose coefficients are entire, has no roots that are irreducible fractions. For, if so, suppose one of these roots to be of the form õ' an irreducible fraction. Substituting this fraction for X, in equation (1), and transposing all of the terms, except the first, to the second member, we have, multiplying both members by 2n-1, we have, the first member of this equation is an irreducible fraction, and the second member is entire: which is manifestly impossible; hence, the supposition that a root can be an irreducible fraction, is absurd. Imaginary Roots. 208. If an equation of the form (1) has any imaginary roots, each of such roots must be of the form a +5V = 1, or a-bv-1, (Art. 138). It may be shown that imaginary roots always enter by pairs, that is, if a + 6V - 1, is a root, then will a-6V - 1, be a root, also; for, substituting a+b7-1 for x, in equation (1), we have, (a + b − 1)” + 2(a + b - 1)^-+ + (a + b -1) + 4 = 0; . . . (@) if we develop the different powers of a +b V-1, by the binomial formula, and perform the operations indicated, the first member of the resulting equation will be made up of two kinds of terms, real and . imaginary. The imaginary terms arise from the odd powers of bv – 1, and consequently (Art. 138), they contain no other imaginary factor than V-1; hence, denoting the sum of the real quantities by m, and the sum of the coefficients of V-1, by n, the equation becomes, m+nv -1 = 0; (6) . but, from principle 2°, Art. 138, equation (6) can only be satisfied, by making m = 0, and n = 0. If we substitute a -5V-1, for x, in equation (1), and perform the indicated operations, the result will differ from that expressed by equation (6), only in the signs of the odd powers of bv - 1. Hence, the resulting equation will be, m - nv -1 = 0; (C) but, we have shown that m= 0, and n = 0; hence, equation (c) is satisfied; and consequently, the substi tution of a-bv-1 for x, in equation (1), satisfies it, which shows that a-bv-1 is a root. If a + b V – 1, is a root of equation (1), the first member of that equation must be divisible by (a + b V – 1); and from what has just been proved, it must also be divisible by x — (a – 6V — 1); and consequently, it must be divisible by the product of these factors: but, the product of these two factors is positive for all real values of x (principle 3°, Art. 138); we therefore, conclude that the number of imaginary roots is always even, since they enter by pairs; and that the product of all the binomial factors that correspond to imaginary roots is positive for all real values of X. The number of real roots, and consequently the number of imaginary roots, of an equation of the form (1), can be determined by a principle called Sturm's theorem. Object of Sturm's Theorem, 209. If we denote the real roots of equation (1), by a, b, c, &c., and suppose them arranged according to their values, so that a shall be the least algebraically, that is, nearest to b the next least, and so on; and if we denote the product of all the binomial factors corresponding to imaginary roots, which is always positive, by Y, equation (1) may be written, (2 — a) (x — b) (x — c)...Y = 0 .. (13) If we suppose x = cs, each of the factors (x – a), (x — b), &c., will be negative; and the first member will |