original equation by at least one. That is, the number of changes of sign increases at least as rapidly as the increase in the number of positive roots when such a multiplication is made. Let f(x)=0 represent any particular equation of the nth degree. The first sign of f(x) is always +. The remaining signs occur in successive groups of or signs which may contain only one sign each. If any term is lacking, its sign is taken to be the same as an adjacent sign. Thus the most general way in which the signs of f(x) may occur is represented in the following table, in which the dots represent an indefinite number of signs. The multiplication of f(x) by xa is represented schematically, only the signs being given. sign may occur The sign indicates that either the + or the according to the value of the coefficients and of a. The vertical lines denote where changes of sign occur in f(x). Assuming that all the ambiguous signs are taken so as to afford the fewest possible number of changes of sign, even then in (x − a)ƒ(x) there is a change of sign at each or between each pair of the vertical lines, and in addition, one to the right of all the vertical lines. Thus as we increase the number of positive roots by one the number of changes of sign increases at least by one, perhaps by more. The only possible variation that could occur in the succession of groups of signs in f(x), namely, when the last group is a group of signs, does not alter the validity of the theorem. 2 4 changes 1 2 0+8 0+2 4 5 changes — 2ƒ(x), 177. Negative roots. Since f(x) has roots opposite in sign to those of f(x) (p. 185), we can state DESCARTES' RULE OF SIGNS FOR NEGATIVE ROOTS. f(x) has no more negative roots than there are changes in sign in ƒ(— x). If by Descartes' rule it appears that there cannot be more than a positive roots and b negative roots, and if a + b < n, the degree of the equation, then there must be imaginary roots, at least n − (a + b) in number. EXERCISES 1. Prove Descartes' rule of signs for x2 + bx + c = 0 directly from the expression for b and c in terms of the roots (see § 115). 2. Find the maximum number of positive and negative roots and any possible information about imaginary roots in the following equations. (a) x3 + 2x2 + 1 = 0. Solution: Writing signs of f(x), ++++, there is no change, hence no positive root. Writing signs of ƒ (− x), +++, there is one change, hence no more than one negative root. Since there can be only one real root there must 178. Integral roots. In finding the rational roots of an equation we make use of the following (where the a's are integers) has any rational root, such root must be an integer. p Suppose be a fraction reduced to its lowest terms which զ satisfies the equation. Thus some factor of q is a factor of p", that is, of p (p. 52), which contradicts the hypothesis that is reduced to its lowest terms. Thus all the rational roots of the equation are integers, which as we know (§ 170) are factors of a„. 179. Rational roots. If we seek the rational roots of аoxn + + a2 = 0, where a 1, we can multiply the roots by a properly chosen constant (§ 175) and obtain an equation of form (1) above whose integral roots may easily be found by synthetic division. Since by Descartes' rule of signs equation (1) has only one positive root and no negative root, we do not need to carry the table further than to test for a positive root. x 3 y 92 52 Form a table of values for equation (2) by synthetic division. We need only to try the factors of 126 (§ 170). Thus (2) has the root 3. Hence the original equation (1) has the root 3 ÷ 3 = 1. RULE. To find all the rational roots of an equation, transform the equation so that the first coefficient is +1. Find the maximum number of positive and negative roots by Descartes' rule of signs. Find the integral roots of this equation by trial, and the roots of the original equation by dividing the integral roots found by the constant by which the roots were multiplied. By the Theorem § 178 we are assured that all the rational roots can be found in this way. EXERCISES Find all the rational roots of the following equations. 7. 4x3 4x2 + x 6 = 0. 9. 4x3 16 x2 9x360. 11. 6x3 – 47 x2 + 71 x + 70 = 0. 13. 6x3- 29 x2 + 53 x - 45 = 0. 15. 27 x3 + 63 x2 + 30 x − 8 = 0. 17. 3 x3- 26 x2 + 52 x − 24 = 0. 19. 18x+81x2 + 121x + 60 = 0. 14. 6x4x3 — 8x2 14x+12= 0. 21. 10 x 17 x3 – 16 x2 + 2 x - 20 = 0. 180. Diminishing the roots of an equation. In the preceding sections we have solved completely the problem of finding the rational roots of an equation. We now pass to the problem of finding the approximate values of the irrational roots of an equation. In carrying out the process that we shall develop it is desirable to form an equation whose roots are equal respectively to the roots of the original equation each diminished by a constant. ƒ(x)= α ̧x2 + а1x2−1 + · + a2 = 0, Let whose roots are a1, a2,..., a,. an equation whose roots are a1 (1) Let a be any constant. We seek a, a. ', an If we let a stand for any one of the roots of (1), since ƒ(a) = 0 (p. 33), we see that Thus to form the desired equation replace x by z+a. We obtain f(x)= f(z+a) = a。 (z + a)" + a1 (≈ + a)"−1 + ··· + a1 = 0. Developing each term by the binomial theorem and collecting like powers of, we get an equation of the form (2) where the A's involve the a's and a. This is the equation desired. We now seek a convenient method of finding the values of the coefficients A, A1, A2, ..., An when do, a1, a2, ..., a, are given numerically. Now A, is the remainder from the division of F(2) by 2. But since F(2) = f(x) and z = x - a, the remainder from dividing F(x) by z is identical with the remainder from dividing f(x) by xa. Thus A, is the remainder from dividing ƒ(x) by xa. Furthermore, since An-1 is the remainder from dividing f(x) - A by, it is also the remainder from dividing. a. The process may be continued for finding the other A's. may then diminish the roots of an equation by a as follows: F(2) by x We 2 A x-a n RULE. The constant term of the new equation is the remainder from dividing f(x) by x-a. |